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

##### 1 Answer
Jul 28, 2015

$F \left(x\right) = \frac{{x}^{2} + x - 12}{{x}^{2} - 4} = \frac{\left(x + 4\right) \left(x - 3\right)}{\left(x - 2\right) \left(x + 2\right)}$

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

#### Explanation:

The numerator ${x}^{2} + x - 12 = \left(x + 4\right) \left(x - 3\right)$ is zero when $x = - 4$ or $x = - 3$.

The denominator ${x}^{2} - 4 = \left(x - 2\right) \left(x + 2\right)$ is zero when $x = \pm 2$.

Since the denominator is zero at $x = \pm 2$ 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]}