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

1 Answer
Oct 6, 2015

Answer:

We first factorize:

Explanation:

#x^2+x-12# may be written as #(x-4)(x+3)# and
#x^2-4# may be written as #(x-2)(x+2)#

If either of the denominator-factors get closer to #0# we have a vertical asymptote or a "hole".

The whole function now becomes:

#F(x)=((x-4)(x+3))/((x-2)(x+2))#

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

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

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