# For what values of x, if any, does f(x) = 1/((x-6)(x^2-9))  have vertical asymptotes?

Jan 11, 2016

$x = 6 , x = 3 , x = - 3$

#### Explanation:

Vertical asymptotes occur when the denominator equals zero.

$\left(x - 6\right) \left({x}^{2} - 9\right) = 0$

$\left(x - 6\right) \left(x - 3\right) \left(x + 3\right) = 0$

$x = 6 , x = 3 , x = - 3$