For what values of x, if any, does #f(x) = 1/((5x+3)(x-6) # have vertical asymptotes?
1 Answer
Jul 12, 2018
Explanation:
The denominator of f(x) cannot be 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 "(5x+3)(x-6)=0#
#x=-3/5" and "x=6" are the asymptotes"#
graph{1/((5x+3)(x-6)) [-10, 10, -5, 5]}
