How to find the vertical asymptote for #F(x)=(x^2+x-12)/ (x^2-4)#?

1 Answer
Jul 28, 2015

Answer:

#F(x) = (x^2+x-12)/(x^2-4) = ((x+4)(x-3))/((x-2)(x+2))#

So the vertical asymptotes are #x=2# and #x=-2#

Explanation:

The numerator #x^2+x-12 = (x+4)(x-3)# is zero when #x=-4# or #x=-3#.

The denominator #x^2-4 = (x-2)(x+2)# is zero when #x=+-2#.

Since the denominator is zero at #x=+-2# and neither of these is a zero of the numerator, both are vertical asymptotes.

graph{(x^2+x-12)/(x^2-4) [-10, 10, -5, 5]}