# How do you find the vertical, horizontal or slant asymptotes for (x^2-4x-32)/(x^2-16)?

Apr 9, 2016

vertical asymptote x = 4
horizontal asymptote y = 1

#### Explanation:

First step is to factorise the function.

rArr( (x - 8)(x + 4))/((x - 4)(x + 4)) = ((x-8)cancel((x+4)))/((x-4)cancel((x+4))

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

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let the denominator equal zero.

solve : x - 4 = 0 → x = 4 is the asymptote

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} f \left(x\right) \to 0$

divide terms on numerator/denominator by x

$\frac{\frac{x}{x} - \frac{8}{x}}{\frac{x}{x} - \frac{4}{x}} = \frac{1 - \frac{8}{x}}{1 - \frac{4}{x}}$

As x to+-oo , 8/x" and 4/x to 0

$\Rightarrow y = \frac{1}{1} = 1 \text{ is the asymptote }$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here , hence there are no slant asymptotes.

Here is the graph.
graph{(x-8)/(x-4) [-10, 10, -5, 5]}