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

Consider the given function.

y = \frac{x + 2}{( x + 2 ) ( x - 3 )}

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.

x + 2 = 0 ⇕ x = - 2 x - 3 = 0 ⇕ x = 3

The function is not defined when

x = - 2

and

x = 3 .

This means we have either a hole or a vertical asymptote. Let's now simplify the function.

y = \frac{x + 2}{( x + 2 ) ( x - 3 )}

Cancel out common factors

y = \frac{x + 2}{( x + 2 ) ( x - 3 )}

Simplify quotient

y = \frac{1}{x - 3}

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 = - 2,

and that

x = 3

is a vertical asymptote.

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.

\frac{x + 2}{( x + 2 ) ( x - 3 )}

Distribute $(x - 3)$

\frac{x + 2}{x ( x - 3 ) + 2 ( x - 3 )}

Distribute $x & 2$

\frac{x + 2}{x ^{2} - 3 x + 2 x - 6}

Add terms

\frac{x + 2}{x ^{2} - x - 6}

$a = a^{1}$

\frac{x ^{1} + 2}{x ^{2} - x - 6}

Since the degree of the denominator is greater than the degree of the numerator, the line

y = 0

is a horizontal asymptote.

Exercise