For what values of x, if any, does #f(x) = 1/((x+12)(x+6)) # have vertical asymptotes?
1 Answer
Jul 17, 2017
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"#