The problem asks to find the vertical asymptote(s) of the function $f(x) = \frac{x-4}{x^2-16}$.

1. Problem Description

The problem asks to find the vertical asymptote(s) of the function $f(x) = \frac{x-4}{x^2-16}$.

2. Solution Steps

To find the vertical asymptotes of a rational function, we need to find the values of $x$ for which the denominator is equal to zero, and the numerator is not zero. First, we factor the denominator:

$x^2 - 16 = (x-4)(x+4)$.

Now, we set the denominator equal to zero:

$(x-4)(x+4) = 0$.

This gives us two possible vertical asymptotes: $x=4$ and $x=-4$.

We need to check if these values make the numerator zero as well.

If $x=4$, the numerator is $x-4 = 4-4 = 0$. Thus, $x=4$ is a removable singularity (a hole), not a vertical asymptote.

If $x=-4$, the numerator is $x-4 = -4-4 = -8 \neq 0$. Therefore, $x=-4$ is a vertical asymptote.

3. Final Answer

$x = -4$