# How do you find the asymptotes for (x-4)/(x^2-3x-4)?

Jan 16, 2016

You start by factoring the denominator, which can be written as $\left(x + 1\right) \left(x - 4\right)$

#### Explanation:

Now the vertical discontinuities are clear, because that's when the denominator becomes zero:
$x = - 1 \mathmr{and} x = + 4$

If we near $x = 4$ from both sides, it will be seen that:
${\lim}_{x \to 4 +} y = {\lim}_{x \to 4 -} y = \frac{1}{5}$
So this is a 'repairable' discontinuity, while $x = - 1$ is not.

We can now do the following:
$\frac{\cancel{x - 4}}{\left(x + 1\right) \left(\cancel{x - 4}\right)} = \frac{1}{x + 1}$

The horizontal asymptote is when $x$ becomes very large. In this case the whole thing goes to zero.

$x = - 1$
$y = 0$