# What are the asymptote(s) and hole(s), if any, of  f(x) =(-2x^2-6x)/((x-3)(x+3)) ?

Feb 15, 2018

Asymptotes at $x = 3$ and $y = - 2$. A hole at $x = - 3$

#### Explanation:

We have $\frac{2 {x}^{2} - 6 x}{\left(x - 3\right) \left(x + 3\right)}$.

Which we can write as:

$\frac{- 2 \left(x + 3\right)}{\left(x + 3\right) \left(x - 3\right)}$

Which reduces to:

$- \frac{2}{x - 3}$

You find the vertical asymptote of $\frac{m}{n}$ when $n = 0$.

So here,

$x - 3 = 0$

$x = 3$ is the vertical asymptote.

For the horizontal asymptote, there exists three rules:

To find the horizontal asymptotes, we must look at the degree of the numerator ($n$) and the denominator ($m$).

If $n > m ,$ there is no horizontal asymptote

If $n = m$, we divide the leading coefficients,

If $n <$$m$, the asymptote is at $y = 0$.

Here, since the degree of the numerator is $2$ and that of the denominator is $2$ we divide the leading coefficients. As the coefficient of the numerator is $- 2$, and that of the denominator is $1 ,$ the horizontal asymptote is at $y = - \frac{2}{1} = - 2$.

The hole is at $x = - 3$.

This is because our denominator had $\left(x + 3\right) \left(x - 3\right)$. We have an asymptote at $3$, but even at $x = - 3$ there is no value of $y$.

A graph confirms this:

graph{(-2x^2-6x)/((x+3)(x-3)) [-12.29, 13.02, -7.44, 5.22]}