How do you find all the asymptotes for function F(x)=(x^2+x-12)/(x^2-4)F(x)=x2+x12x24?

1 Answer
Oct 6, 2015

We first factorize:

Explanation:

x^2+x-12x2+x12 may be written as (x-4)(x+3)(x4)(x+3) and
x^2-4x24 may be written as (x-2)(x+2)(x2)(x+2)

If either of the denominator-factors get closer to 00 we have a vertical asymptote or a "hole".

The whole function now becomes:

F(x)=((x-4)(x+3))/((x-2)(x+2))F(x)=(x4)(x+3)(x2)(x+2)

Which cannot be further reduced.
So the vertical asymtotes are at x=-2 and x=+2x=2andx=+2

Since the grade and factor of the highest power are the same, the function will tend to 11 for very high values of xx
So the horizontal asymptote is y=1y=1

graph{(x^2+x-12)/(x^2-4) [-28.9, 28.85, -14.43, 14.45]}