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

Feb 27, 2016

$x = 4$ and $x = 8$

#### Explanation:

Vertical asymptotes occur whenever the function is undefined at a particular $x$-value. When we're dealing with rational functions like the one here, this simply means that we get asymptotes whenever a zero is in the denominator (because $\frac{1}{0}$ is undefined).

Our denominator is $\left(x - 4\right) \left(x - 8\right)$. Luckily for us, it's already factored, allowing us to solve it immediately. We are finding when this is equal to zero, so we write:
$\left(x - 4\right) \left(x - 8\right) = 0$

Using the zero-product property, we can say that $x - 4 = 0$ and $x - 8 = 0$. Solving these two equations tells us that $x = 4$ and $x = 8$. Therefore, our vertical asymptotes are at $x = 4$ and $x = 8$ because at these $x$-values, $f \left(x\right)$ is undefined.