# For what values of x, if any, does f(x) = 1/((x-4)(x+8))  have vertical asymptotes?

Jan 2, 2016

$x = - 8 , 4$

#### Explanation:

Vertical asymptotes occur at breaks in the domain of a function. This happens when plugging in a value for $x$ would cause an issue.

The only such issue that could occur in this function is dividing a number by $0$.

Here, if the denominator equals $0$, the graph will have a vertical asymptote.

Thus, to find the spots where the graph has vertical asymptotes, set the denominator of the function equal to $0$.

$\left(x - 4\right) \left(x + 8\right) = 0$

Just like in working with quadratics, recall that either one of the two parts being multiplied can be equal to $0$.

$\left\{\begin{matrix}x - 4 = 0 \\ x + 8 = 0\end{matrix}\right.$

Solve for both.

$\left\{\begin{matrix}x = 4 \\ x = - 8\end{matrix}\right.$

Vertical asymptotes occur at $x = - 8 , 4$.

graph{1/((x-4)(x+8)) [-14.51, 10.81, -5.22, 7.44]}