Question

For what values of x, if any, does `f(x) = 1/((x+12)(x^2-6))` have vertical asymptotes?

Answer 1

There are `3` vertcical asymptotes `x=-12`, `x=-sqrt6` and `x=sqrt6`

Explanation:

The denominator must be `!=0`

Therefore,

`f(x)=1/((x+12)(x^2-6))=1/((x+12)(x-sqrt6)(x+sqrt6))`

To determine the vertical asymptotes,

We determine the following limits

`lim_(x->-12^-)f(x)=-oo`

`lim_(x->-12^+)f(x)=+oo`

A vertical asymptote is `x=-12`

`lim_(x->-sqrt6^-)f(x)=+oo`

`lim_(x->-sqrt6^+)f(x)=-oo`

Another vertical asymptote is `x=-sqrt6`

`lim_(x->sqrt6^-)f(x)=-oo`

`lim_(x->sqrt6^+)f(x)=+oo`

The last vertical asymptote is `x=sqrt6`

graph{1/((x+12)(x^2-6)) [-16.02, 16.01, -8.01, 8.01]}