What is the vertical asymptote of #(x + 4)/(x^2+x-6)#?

1 Answer
Jim H
Sep 1, 2015

There are two vertical asymptotes: #x=2# and #x=-3#

Explanation:

The denominator #x^2+x-6# is #0# when #x# is #2# or #-3#.

Neither of these values make the numerator equal to #0#, so we cannot reduce (simplify) the fraction.

Therefore, we have vertical asymptotes at #2# and at #-3#.

The equations of the lines are #x=2# and #x=-3#.

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