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

1 Answer
Sep 13, 2017

Answer:

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]}