Question

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

Answer 1

`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 `1/0` is undefined).

Our denominator is `(x-4)(x-8)`. 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:
`(x-4)(x-8) = 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(x)` is undefined.