Question

For what values of x, if any, does `f(x) = 1/((x+12)(x^2-36))` have vertical asymptotes?

Answer 1

`x=-12` and `x=+-6`

Explanation:

Vertical asymptotes occur when the function is undefined. In the case of a rational function like `f(x)=1/((x+12)(x^2-36))`, they occur specifically when the denominator is `0`, as in `1/0`. To find where the denominator equals zero, we simply set the denominator equal to zero and solve for `x`:
`(x+12)(x^2-36)=0`

`x+12=0->x=-12`
`x^2-36=0->x^2=36->x=+-sqrt(36)->x=+-6`

Therefore we have vertical asymptotes at `x=12, 6,`and `-6`. We can see this in the graph below, where the function is not defined at these `x`-values. graph{1/((x^2-36)(x+12)) [-10, 10, -5, 5]}