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

##### 1 Answer

Feb 26, 2017

#### Answer:

The vertical asymptotes are

The horizontal asymptote is

No oblique asymptote

#### Explanation:

Let's factorise the denominator

As you canot divide by

The vertical asymptotes are

As the degree of the numerator is

The horizontal asymptote is

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]}