# How do you find any asymptotes of f(x)=(x+3)/(x^2-4)?

Oct 16, 2016

There are two vertical asymptotes at $x = - 2$ and $x = 2$
There is a horizontal asymptote at 0 as $x \to \infty$
There is a horizontal asymptote at 0 as $x \to - \infty$

#### Explanation:

$f \left(x\right) = \frac{x + 3}{{x}^{2} - 4}$
$\therefore f \left(x\right) = \frac{x + 3}{{x}^{2} - {2}^{2}}$
$\therefore f \left(x\right) = \frac{x + 3}{\left(x + 2\right) \left(x - 2\right)}$

Vertical Asymptotes
These occur when the denominator is zero
$\therefore \left(x + 2\right) \left(x - 2\right) = 0$
$x = \pm 2$#

Horizontal Asymptotes
We also need to examine the behaviour as $x \to \pm \infty$
Now for large $x$,
$x + 3 \approx x$ and $x + 2 \approx x$ and $x - 3 \approx x$
We have, $f \left(x\right) = \frac{x + 3}{\left(x + 2\right) \left(x - 2\right)}$
and so, for large $x$, $f \left(x\right) \approx \frac{x}{x} ^ 2 \approx \frac{1}{x}$
so, as $x \to - \infty \implies f \left(x\right) \to {0}^{-}$
and, as $x \to \infty \implies f \left(x\right) \to {0}^{+}$

Summary
There are two vertical asymptotes at $x = - 2$ and $x = 2$
There is a horizontal asymptote at 0 as $x \to \infty$
There is a horizontal asymptote at 0 as $x \to - \infty$

The graph validates this;
graph{(x+3)/((x+2)(x-2)) [-10, 10, -5, 5]}