For what values of x, if any, does #f(x) = 1/((x+12)(x-3)) # have vertical asymptotes?

1 Answer
Feb 10, 2016

Answer:

In x=-12 and x=3

Explanation:

If we replace x by -12 or 3, the denominator will be zero, which means:

#f(x)=oo#

But what kind of infinity you have when x has those values?

If you use #3^-#, which is a number a little smaller than 3, you will obtain
#f(3^-)=(1/0^-)=-oo#

By the other hand if you use #3^+#, which is a number a little higher than 3, you will obtain
#f(3^-)=(1/0^+)=+oo#.
If by on side you have #-oo# and by the other #+oo#, you've got a vertica lasymptote.