Question

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

Answer 1

`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`.

`(x-4)(x+8)=0`

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

`{(x-4=0),(x+8=0):}`

Solve for both.

`{(x=4),(x=-8):}`

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

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