For what values of x, if any, does #f(x) = 1/((x+12)(x^2-36)) # have vertical asymptotes?

1 Answer
Ken C.
Feb 18, 2016

#x=-12# and #x=+-6#

Explanation:

Vertical asymptotes occur when the function is undefined. In the case of a rational function like #f(x)=1/((x+12)(x^2-36))#, they occur specifically when the denominator is #0#, as in #1/0#. To find where the denominator equals zero, we simply set the denominator equal to zero and solve for #x#:
#(x+12)(x^2-36)=0#

#x+12=0->x=-12#
#x^2-36=0->x^2=36->x=+-sqrt(36)->x=+-6#

Therefore we have vertical asymptotes at #x=12, 6,#and #-6#. We can see this in the graph below, where the function is not defined at these #x#-values. graph{1/((x^2-36)(x+12)) [-10, 10, -5, 5]}

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