# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)=(x^2-4x-32)/(x^2-16)?

Sep 8, 2016

vertical asymptote at x = 4
horizontal asymptote at y = 1

#### Explanation:

The first step is to factorise and simplify f(x)

$\Rightarrow f \left(x\right) = \frac{\left(x - 8\right) \cancel{\left(x + 4\right)}}{\left(x - 4\right) \cancel{\left(x + 4\right)}} = \frac{x - 8}{x - 4}$

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: $x - 4 = 0 \Rightarrow x = 4 \text{ is the asymptote}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by x

$f \left(x\right) = \frac{\frac{x}{x} - \frac{8}{x}}{\frac{x}{x} - \frac{4}{x}} = \frac{1 - \frac{8}{x}}{1 - \frac{4}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{1 - 0}{1 - 0}$

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

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no oblique asymptotes.
graph{(x-8)/(x-4) [-20, 20, -10, 10]}