For what values of x, if any, does $f(x) = 1/((x-6)(x^2-9))$ have vertical asymptotes?

Question

1715 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-11T20:20:01

$x=6, x=3, x=-3$

Vertical asymptotes occur when the denominator equals zero.

$(x-6)(x^2-9)=0$

$(x-6)(x-3)(x+3)=0$

$x=6, x=3, x=-3$

For what values of x, if any, does f(x) = 1/((x-6)(x^2-9)) f(x) = 1/((x-6)(x^2-9)) have vertical asymptotes?