Exercise 14 Page 526

We want find the holes, as well as the vertical and horizontal asymptotes, in the graph of the function.

Holes and Vertical Asymptotes

Consider the given function. Recall that division by zero is not defined. Therefore, the rational function is undefined where $(x+2)(x-3)=0.$ We will use the Zero Product Property to solve this equation. The function is not defined when $x=\text{-}2$ and $x=3.$ This means we have either a hole or a vertical asymptote. Let's now simplify the function. We have canceled out the factor $x+2.$ However, $x-3$ is still in the denominator. This indicates that the graph has a hole when $x=\text{-}2,$ and that $x=3$ is a vertical asymptote.

Horizontal Asymptotes

Let's recall how to find horizontal asymptotes in rational functions.

$y=\frac{p(x)}{q(x)}$
The degree of $p(x)$ is greater than the degree of $q(x)$	There are no horizontal asymptotes
The degree of $q(x)$ is greater than the degree of $p(x)$	$y=0$ is a horizontal asymptote
The degrees of $p(x)$ and $q(x)$ are equal	$y=\frac{a}{b}$ is a horizontal asymptote, where $a$ and $b$ are the leading coefficients of $p(x)$ and $q(x),$ respectively

Let's expand the denominator of the given function. Since the degree of the denominator is greater than the degree of the numerator, the line $y=0$ is a horizontal asymptote.