Question

For what values of x, if any, does `f(x) = 1/((x+12)(x+6))` have vertical asymptotes?

Answer 1

`"vertical asymptotes at " x=-12" and "x=-6`

Explanation:

The denominator of f(x) cannot equal zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

`"solve "(x+12)(x+6)=0`

`rArrx=-12" and "x=-6" are the asymptotes"`