# How do you graph y=x/(x^2-9) using asymptotes, intercepts, end behavior?

Nov 11, 2016

The vertical asymptotes are $x = 3$ and $x = - 3$
The horizontal asymptote is $y = 0$
No slant asymptotes
Only one intercept at $\left(0 , 0\right)$

#### Explanation:

Let's factorise the denominator ${x}^{2} - 9 = \left(x + 3\right) \left(x - 3\right)$
As we canot divide by $0$, the vertical asymptotes are $x = 3$ and $x = - 3$

As the degree of the numerator is $<$ the degree of the denominator, we do not have a slant asymptote.

${\lim}_{x \to - \infty} y = {\lim}_{x \to - \infty} \frac{x}{x} ^ 2 = {\lim}_{x \to - \infty} \frac{1}{x} = {0}^{-}$

${\lim}_{x \to + \infty} y = {\lim}_{x \to + \infty} \frac{x}{x} ^ 2 = {\lim}_{x \to + \infty} \frac{1}{x} = {0}^{+}$

So, $y = 0$ is a horizontal asymptote

The intercept is, when $x = 0$$\implies$$y = 0$
graph{x/(x^2-9) [-11.25, 11.24, -5.63, 5.62]}