For what values of x, if any, does #f(x) = 1/((2x-3)(7x-6) # have vertical asymptotes?

1 Answer
Dec 19, 2016

Answer:

#"vertical asymptotes at " x=3/2" and " x=6/7#

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 : #(2x-3)(7x-6)=0rArrx=3/2"or " x=6/7#

#rArrx=3/2" and " x=6/7" are the asymptotes"#
graph{1/((2x-3)(7x-6) [-10.52, 10.525, -5.26, 5.26]}