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

Jul 17, 2017

$\text{vertical asymptotes at " x=-12" and } x = - 6$

#### Explanation:

The denominator of f(x) cannot equal zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve } \left(x + 12\right) \left(x + 6\right) = 0$

$\Rightarrow x = - 12 \text{ and "x=-6" are the asymptotes}$