How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x) = (4x) / (x^2+1)#?

Question

James Allen · Accepted Answer

The horizontal asymptote is #y=0# and the function has neither a vertical asymptote nor an oblique asymptote.

#f(x)=frac{4x}{x^2+1}# To find the vertical asymptote (VA), find the value of #x# that will make the denominator equal zero. #x^2+1=0# #x^2=-1# There is no real solution, so there is no VA. To find the horizontal asymptote(HA), compare the degree of the numerator to the degree of the denominator. #f(x)=frac{4x^color(red)1}{x^color(blue)2+1}# The degree of the numerator is #color(red)1# and the degree of the denominator is #color(blue)2#. If the degree of the numerator is less than the degree of the denominator, the HA is #y=0#. If the degree of the numerator is equal to the degree of the denominator, the HA is #y=# the leading coefficient of the numerator divided by the leading coefficient of the denominator. If the degree of the numerator is greater than the degree of the denominator, there is an oblique asymptote. In this example, the degree of the numerator is less than the degree of the denominator, so the HA is #y=0# and there is no oblique asymptote.

Elias Ackerman · Answer

undefined To find the vertical asymptotes of a function, we look for values of $ x $ that make the denominator of the function equal to zero, but do not make the numerator zero. So, for $ f(x) = \frac{4x}{x^2+1} $, the vertical asymptotes occur where $ x^2 + 1 = 0 $. However, since $ x^2 + 1 $ is always positive, there are no values of $ x $ that make the denominator zero. Hence, there are no vertical asymptotes for this function.

To find the horizontal asymptote, we examine the behavior of the function as $ x $ approaches positive or negative infinity. For the given function, as $ x $ approaches positive or negative infinity, the term $ x^2 + 1 $ becomes insignificant compared to $ 4x $. Therefore, the horizontal asymptote is $ y = 0 $.

Finally, to determine the oblique asymptote (also known as slant asymptote), we perform long division or polynomial division. By dividing $ 4x $ by $ x^2 + 1 $, we find the quotient. Since $ x $ is in the numerator and $ x^2 $ is in the denominator, the degree of the numerator is less than the degree of the denominator. Therefore, the oblique asymptote does not exist for this function.

In summary:
- There are no vertical asymptotes.
- The horizontal asymptote is $ y = 0 $.
- There is no oblique asymptote.

Sadie Alexander · Answer

undefined To find the vertical asymptotes of $ f(x) = \frac{4x}{x^2 + 1} $, set the denominator equal to zero and solve for $ x $. There are no real solutions, so there are no vertical asymptotes.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $.

There are no oblique asymptotes for this function.