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

Oct 6, 2015

We first factorize:

#### Explanation:

${x}^{2} + x - 12$ may be written as $\left(x - 4\right) \left(x + 3\right)$ and
${x}^{2} - 4$ may be written as $\left(x - 2\right) \left(x + 2\right)$

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

The whole function now becomes:

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

Which cannot be further reduced.
So the vertical asymtotes are at $x = - 2 \mathmr{and} x = + 2$

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]}