# How do you find the vertical, horizontal and slant asymptotes of: f(x)=1/(x^2-2x-8 )?

Jan 21, 2017

The vertical asymptotes are $x = - 2$ and $x = 4$
The horizontal asymptote is $y = 0$
No slant asymptote.

#### Explanation:

Let's factorise the denominator

${x}^{2} - 2 x - 8 = \left(x + 2\right) \left(x - 4\right)$

So,

$f \left(x\right) = \frac{1}{{x}^{2} - 2 x - 8} = \frac{1}{\left(x + 2\right) \left(x - 4\right)}$

As you cannot divide by $0$, $x \ne - 2$ and $x \ne 4$

The vertical asymptotes are $x = - 2$ and $x = 4$

As the degree of the numerator is $<$ than the degree of the denominator, there are no slant asymptote.

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} \frac{1}{x} ^ 2 = 0$

The horizontal asymptote is $y = 0$

graph{(y-1/(x^2-2x-8))(y)(y-50x-100)(y-50x+200)=0 [-9.84, 10.18, -5.175, 4.825]}