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

Let's start by factoring the denominator.

y = \frac{1}{x ^{2} + 3 x - 1 0}

Write as a difference

y = \frac{1}{x ^{2} + 5 x - 2 x - 1 0}

Factor out $x & - 2$

y = \frac{1}{x ( x + 5 ) - 2 ( x + 5 )}

Factor out $(x + 5)$

y = \frac{1}{( x - 2 ) ( x + 5 )}

Recall that division by zero is not defined. Therefore, the rational function is undefined where

(x - 2) (x + 5) = 0 .

We will use the Zero Product Property to solve this equation.

x - 2 = 0 ⇕ x = 2 x + 5 = 0 ⇕ x = - 5

The function is not defined when

x = 2,

and

x = - 5 .

This means we have either a hole or a vertical asymptote. We cannot simplify the function. This indicates the lines

x = 2

and

x = - 5

are vertical 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 now consider the given function.

y = \frac{1}{x ^{2} + 3 x - 1 0} ⇕ y = \frac{x ^{0}}{x ^{2} + 3 x - 1 0}

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

y = 0

is a horizontal asymptote.

Exercise