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

1 Answer
Jul 17, 2017

Answer:

#"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"#