# What are the asymptote(s) and hole(s), if any, of  f(x) =(x^2-x-2)/(x+2)?

Mar 31, 2018

Vertical asymptote of-2

#### Explanation:

An vertical asymptote or a hole is created by a point in which the domain is equal to zero i.e. $x + 2 = 0$

So either $x = - 2$

An horizontal asymptote is created where the top and the bottom of the fraction don't cancel out. Whilst a hole is when you can cancel out.

So lets factorise the top

$\frac{\left(x - 2\right) \left(x + 1\right)}{x + 2}$

So as the denominator can't be canceled out by dividing a factor in the top and the bottom it is an asymptote rather than a hole.

Meaning that $x = - 2$ is an vertical asymptote

graph{((x-2)(x+1))/(x+2) [-51.38, 38.7, -26.08, 18.9]}