# For what values of x, if any, does f(x) = 1/((x+12)(x^2-6))  have vertical asymptotes?

Sep 13, 2017

There are $3$ vertcical asymptotes $x = - 12$, $x = - \sqrt{6}$ and $x = \sqrt{6}$

#### Explanation:

The denominator must be $\ne 0$

Therefore,

$f \left(x\right) = \frac{1}{\left(x + 12\right) \left({x}^{2} - 6\right)} = \frac{1}{\left(x + 12\right) \left(x - \sqrt{6}\right) \left(x + \sqrt{6}\right)}$

To determine the vertical asymptotes,

We determine the following limits

${\lim}_{x \to - {12}^{-}} f \left(x\right) = - \infty$

${\lim}_{x \to - {12}^{+}} f \left(x\right) = + \infty$

A vertical asymptote is $x = - 12$

${\lim}_{x \to - {\sqrt{6}}^{-}} f \left(x\right) = + \infty$

${\lim}_{x \to - {\sqrt{6}}^{+}} f \left(x\right) = - \infty$

Another vertical asymptote is $x = - \sqrt{6}$

${\lim}_{x \to {\sqrt{6}}^{-}} f \left(x\right) = - \infty$

${\lim}_{x \to {\sqrt{6}}^{+}} f \left(x\right) = + \infty$

The last vertical asymptote is $x = \sqrt{6}$

graph{1/((x+12)(x^2-6)) [-16.02, 16.01, -8.01, 8.01]}