Question

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

Answer 1

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

`((x-2)(x+1))/(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]}