Question

For what values of x, if any, does `f(x) = 1/((x+12)(x-3))` have vertical asymptotes?

Answer 1

In x=-12 and x=3

Explanation:

If we replace x by -12 or 3, the denominator will be zero, which means:

`f(x)=oo`

But what kind of infinity you have when x has those values?

If you use `3^-`, which is a number a little smaller than 3, you will obtain
`f(3^-)=(1/0^-)=-oo`

By the other hand if you use `3^+`, which is a number a little higher than 3, you will obtain
`f(3^-)=(1/0^+)=+oo`.
If by on side you have `-oo` and by the other `+oo`, you've got a vertica lasymptote.