# For what values of x, if any, does f(x) = 1/((x+12)(x^2-36))  have vertical asymptotes?

Feb 18, 2016

$x = - 12$ and $x = \pm 6$

#### Explanation:

Vertical asymptotes occur when the function is undefined. In the case of a rational function like $f \left(x\right) = \frac{1}{\left(x + 12\right) \left({x}^{2} - 36\right)}$, they occur specifically when the denominator is $0$, as in $\frac{1}{0}$. To find where the denominator equals zero, we simply set the denominator equal to zero and solve for $x$:
$\left(x + 12\right) \left({x}^{2} - 36\right) = 0$

$x + 12 = 0 \to x = - 12$
${x}^{2} - 36 = 0 \to {x}^{2} = 36 \to x = \pm \sqrt{36} \to x = \pm 6$

Therefore we have vertical asymptotes at $x = 12 , 6 ,$and $- 6$. We can see this in the graph below, where the function is not defined at these $x$-values. graph{1/((x^2-36)(x+12)) [-10, 10, -5, 5]}