# How do you find the asymptotes for (x^2-4x-32)/(x^2-16)?

Nov 20, 2015

Always simplify first before solving the problem ...

#### Explanation:

First, factor ...

$y = \frac{\left(x - 8\right) \left(x + 4\right)}{\left(x + 4\right) \left(x - 4\right)}$

Cancel out $\left(x + 4\right)$ ...

$y = \frac{x - 8}{x - 4}$

[Note: The function is undefined at $x = - 4$, but no asymptote. It is called a removable discontinuity .]

Now, the denominator is zero when $x = 4$ resulting in a vertical asymptote at $x = 4$.

There is a horizontal asymptote at $y = 1$. This is found by dividing the coefficients on x in the numerator and denominator.

hope that helped