# How do you find the vertical, horizontal or slant asymptotes for (x+3)/(x^2-9)?

May 28, 2016

This function has a horizontal asymptote $y = 0$, a vertical asymptote $x = 3$ and a removable singularity at $x = - 3$.

#### Explanation:

$f \left(x\right) = \frac{x + 3}{{x}^{2} - 9} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{x + 3}}}}{\left(x - 3\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 3\right)}}}} = \frac{1}{x - 3}$

excluding $x = - 3$

As $x \to \pm \infty$, $\frac{1}{x - 3} \to 0$. So there is a horizontal asymptote $y = 0$

As $x \to {3}^{+}$, $\frac{1}{x - 3} \to + \infty$

As $x \to {3}^{-}$, $\frac{1}{x - 3} \to - \infty$

$f \left(3\right)$ is undefined since division by $0$ is undefined.

So $f \left(x\right)$ has a vertical asymptote at $x = 3$

$f \left(- 3\right)$ is undefined, since both numerator $\left(x + 3\right) = 0$ and denominator $\left({x}^{2} - 9\right) = 0$. Note however, that both left and right limits exist at $x = - 3$ and are both equal to $- \frac{1}{6}$. So there is a removable singularity at $x = - 3$ (removable by redefining $f \left(- 3\right) = - \frac{1}{6}$).