# How do you find the asymptotes for (x+3)/(x^2-9)?

Feb 15, 2016

To find the asymptotes, you need to find out where the denominator approaches zero for the vertical asymptotes and then establish what happens as $x \to \pm \infty$

#### Explanation:

The denominator in this case equals the difference of two squares, ${x}^{2}$ and ${3}^{2}$, so
$\frac{x + 3}{{x}^{2} - 9} = \frac{x + 3}{\left(x - 3\right) \left(x + 3\right)}$
The $\left(x + 3\right)$ terms cancel out so there is a removable asymptote at $x = - 3$
There is a vertical asymptote at $x = 3$

As $x \to \pm \infty$, $\left(x + 3\right) \to x$ and $\left({x}^{2} - 9\right) \to {x}^{2}$

$\frac{x + 3}{{x}^{2} - 9} \to \frac{x}{x} ^ 2 = \frac{1}{x}$ which gives the slant asymptotes.