How do you find vertical, horizontal and oblique asymptotes for #(x^3+1)/(x^2+3x)#?

Question

James Allen · Accepted Answer

Vertical asymptotes are #x=0# and #x=-3#.
Oblique asymptote is #y=x# To find all the asymptotes for function #y=(x^3+1)/(x^2+3x)#, let us first start with vertical asymptotes, which are given by putting denominator equal to zero or #x^2+3x=0# or #x(x+3)=0# i.e. #x=0# and #x+3=0# or #x=-3#. As the highest degree of numerator is #3# and that of denominator is #2# (higher by just one), we have one oblique asymptote given by #y=x^3/x^2=x# or #y=x#. Hence, Vertical asymptotes are #x=0# and #x=-3#. Oblique asymptote is given by #y=x# graph{(x^3+1)/(x^2+3x) [-10, 10, -5, 5]}

Greyson Brown · Answer

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To find the horizontalTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

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To find the horizontal asymptoteTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote,To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotesTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compareTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes,To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compareTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degreesTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degreesTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator andTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numeratorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator andTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominatorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominatorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. IfTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degreeTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numeratorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numeratorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater thanTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater,To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, thereTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degreeTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there isTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontalTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator,To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, thereTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptoteTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there isTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is noTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontalTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degreeTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptoteTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

ToTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator isTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To findTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greaterTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater,To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the obliqueTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptoteTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote isTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomialTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ yTo find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial longTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y =To find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division orTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0To find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or useTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 \To find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use syntheticTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic divisionTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. IfTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division toTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degreesTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numeratorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees areTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator byTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal,To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divideTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominatorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leadingTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. TheTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficientsTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The obTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptoteTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numeratorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote isTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator andTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator toTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotientTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to findTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtainedTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained fromTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontalTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from thisTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this divisionTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptoteTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division whenTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $To find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

ToTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ xTo find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To findTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x \To find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find obTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find obliqueTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends toTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotesTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positiveTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes,To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negativeTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinityTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numeratorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator byTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. AfterTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominatorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the divisionTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator usingTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division,To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using longTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, ifTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long divisionTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division orTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degreeTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or syntheticTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic divisionTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less thanTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. TheTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotientTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degreeTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the obliqueTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptoteTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominatorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

InTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this caseTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the obTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the obliqueTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote existsTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. IfTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotesTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes:To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degreeTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: SetTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equalTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainderTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal toTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equalsTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zeroTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degreeTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero andTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve forTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $To find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominatorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ xTo find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator,To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x \To find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there isTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is aTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ xTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slantTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 +To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. IfTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3xTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degreeTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x =To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 \To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder isTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greaterTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
\[ xTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater thanTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
\[ x(xTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degreeTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
\[ x(x +To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
\[ x(x + To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
\[ x(x + 3)To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominatorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
\[ x(x + 3) =To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator,To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
\[ x(x + 3) = To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, thereTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
\[ x(x + 3) = 0To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there isTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
\[ x(x + 3) = 0 \To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is noTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no obTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no obliqueTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ xTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ x =To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptoteTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ x = To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ x = 0To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ x = 0 \To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomialTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ x = 0 \textTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial longTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ x = 0 \text{To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long divisionTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ x = 0 \text{ orTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ x = 0 \text{ or }To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ x = 0 \text{ or } xTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ x = 0 \text{ or } x =To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \fracTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ x = 0 \text{ or } x = -To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{xTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ x = 0 \text{ or } x = -3To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
\[ x = 0 \text{ or } x = -3 \To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 +To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. HorizontalTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{xTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptoteTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote:To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: CompareTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 +To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degreesTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3xTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x}To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numeratorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} =To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator andTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = xTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominatorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x -To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. SinceTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 +To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degreeTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \fracTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numeratorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{xTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3)To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x +To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) isTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greaterTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater thanTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3}To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} \To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
\[ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degreeTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree ofTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

ThereforeTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore,To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominatorTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, theTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the obTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the obliqueTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2),To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), thereTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptoteTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there isTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptote isTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is noTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptote is $To find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontalTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptote is $ yTo find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptote is $ y =To find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptoteTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptote is $ y = xTo find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptote.

To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptote is $ y = x -To find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptote.

3To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptote is $ y = x - To find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptote.

3.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptote is $ y = x - 3To find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptote.

3. ObTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptote is $ y = x - 3 \To find the vertical asymptotes of the function \( \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptote.

3. ObliqueTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptote is $ y = x - 3 $.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptote.

3. Oblique asymptTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptote is $ y = x - 3 $.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptote.

3. Oblique asymptoteTo find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptote is $ y = x - 3 $.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptote.

3. Oblique asymptote:To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, set the denominator equal to zero and solve for $ x $. There are no vertical asymptotes for this function.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division when $ x $ tends to positive or negative infinity. After performing the division, if the degree of the remainder is less than the degree of the denominator, the oblique asymptote exists. If the degree of the remainder equals the degree of the denominator, there is a slant asymptote. If the degree of the remainder is greater than the degree of the denominator, there is no oblique asymptote.

In this case, performing the polynomial long division:

$$ \frac{x^3 + 1}{x^2 + 3x} = x - 3 + \frac{9}{x + 3} $$

Therefore, the oblique asymptote is $ y = x - 3 $.To find the vertical asymptotes of the function $ \frac{x^3 + 1}{x^2 + 3x} $, identify the values of $ x $ that make the denominator zero but the numerator nonzero. These values will give vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient represents the oblique asymptote.

In this case:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $.
$$ x^2 + 3x = 0 $$
$$ x(x + 3) = 0 $$
$$ x = 0 \text{ or } x = -3 $$

2. Horizontal asymptote: Compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptote.

3. Oblique asymptote: Perform polynomial long division or synthetic division to divide $ x^3 + 1 $ by $ x^2 + 3x $. The quotient will represent the oblique asymptote.