Determine the equations of any horizontal and vertical asymptotes of $f ( x ) = \frac { x ^ { 2 } } { x ^ { 2 } + 16 }$.
A) horizontal: $y = 4$; vertical: $x = - 4$
B) horizontal: $x = 1$; vertical: none
C) horizontal: $y = - 4$; vertical: $x = 1$
D) horizontal: $y = 1$; vertical: none
E) horizontal: none; vertical: none

Question

Quizplus · Accepted Answer

The horizontal asymptote is found by comparing the degrees of the numerator and denominator. Since they are the same, the horizontal asymptote is $y = \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}} = \frac{1}{1} = 1$. There are no vertical asymptotes because the denominator $x^2 + 16$ cannot be zero for any real value of $x$, as $x^2$ is always non-negative and adding 16 ensures the denominator is always positive.

Determine the Equations of Any Horizontal and Vertical Asymptotes Of f