Question

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

Answer 1

`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]}