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

1 Answer
Mar 31, 2018

Answer:

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]}