Chapter 3.7, Problem 15E

To determine

To find: equations of vertical and horizontal asymptotes, if any, of the given function.

Answer to Problem 15E

  $x=-6\left[\text{Vertical asymptote}\right]$

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{x^2}{x+6}$

Calculation:

The function $f\left(-6\right)$ is undefined; there may be a vertical asymptote at $x=-6$ . Let’s make a table of values for the given function.

$x$	$f\left(x\right)$
$-$0.5999	0.066643
$-$0.599	0.066432
$-$0.59	0.06434
$-$0.5	0.04545

Since, the value in the table suggested that $f\left(x\right)\to \infty$ as $x\to -6$ from right. So, the vertical asymptote is $x=-6$ .

Look at the degree of numerator and denominator to find the horizontal asymptotes as follows:

  $f\left(x\right)=\frac{a{x}^n}{b{x}^m}$

Case	Horizontal Asymptote
$n<m$	$y=0$
$n=m$	$y=\frac{a}{b}$
$n>m$	No horizontal asymptote

Since the degree of numerator and denominator in given function is 2 and 1. Thus, there is no horizontal asymptote.