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

Question

1722 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-11T20:20:01

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

Vertical asymptotes occur when the denominator equals zero.

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

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

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

For what values of x, if any, does f(x)=1(x−6)(x2−9)f(x) = 1/((x-6)(x^2-9)) have vertical asymptotes?