# How do you find vertical, horizontal and oblique asymptotes for 1/(x^2-2x-8)?

Feb 26, 2017

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

#### Explanation:

Let's factorise the denominator

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

As you canot divide by $0$, $\implies$, $x \ne 4$ and $x \ne - 2$

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

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

${\lim}_{x \to \pm \infty} \frac{1}{{x}^{2} - 2 x - 8} = {\lim}_{x \to \pm \infty} \frac{1}{x} ^ 2 = {0}^{+}$

The horizontal asymptote is $y = 0$

graph{(y-1/(x^2-2x-8))(y-10000(x-4))(y-1000(x+2))(y)=0 [-3.85, 7.25, -2.893, 2.66]}