How do you find the vertical, horizontal and slant asymptotes of #a(x)=(2x^2-1) / (3x^3-2x+1)#?

Answer 1

There is a vertical asymptote at #color(red)(x=-1)#, a horizontal asymptote at #color(red)(y=0)# and #color(red)("no")# slant asymptote.

#a(x)=(2x^2-1)/(3x^3-2x+1)#

Step 1. Find the vertical asymptotes.

Set the denominator equal to zero and solve for #x#.

#3x^3-2x+1=0#

According to the rational root theorem, the rational roots of #f(x) = 0# must all be of the form #p/q# with #p# a divisor of #1# and #q# a divisor of #3#.

So the only possible rational roots are #±1,±1/3#.

We have to test all four possibilities.

The only one that works is

#-1|" "3" "color(white)(1)0" "-2" "" "1#
#" "color(white)(1)|" "color(white)(1)-3" "" "3" "-1#
#" "color(white)(1)stackrel("———————————————)#
#" "" "" "3color(white)(1)-3" "" "1" "" "color(red)(0)#

So #3x^3-2x+1=(x+1)(3x^2-3x+1)#

#x+1=0# and #3x^2-3x+1=0#

#3x^2-3x+1=0# has no real zeroes.

#x=-1# is a zero.

There is a vertical asymptote at #x=-1#.

Step 2. Find the horizontal asymptotes.

The degree of the numerator is lower than the degree of the denominator, so the #x#-axis is the horizontal asymptote.

The horizontal asymptote is at #y=0#.

Step 3. Find the slant asymptotes.

A slant asymptote occurs when the degree of the numerator is higher than the degree of the denominator.

Here, the numerator is 2nd degree, and the denominator is 3rd degree, so there are no slant asymptotes.

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Answer 2

To find the vertical asymptotes of the function a(x), set the denominator equal to zero and solve for x. The vertical asymptotes occur at the values of x that make the denominator zero.

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

To find the slant asymptote, perform polynomial long division to divide the numerator by the denominator. The quotient obtained represents the equation of the slant asymptote. If the degree of the numerator is one greater than the degree of the denominator, there is a slant asymptote. If the degrees are equal or the numerator's degree is less than that of the denominator, there is no slant asymptote.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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