Which function has a removable discontinuity?

Reasoning:

The denominator in choice A factors to (x-2)(x+1). We have a pair of "x-2" terms that cancel leaving us with 1/(x+1). We cannot plug x = -1 into the simplified result because it leads the denominator x+1 to turn into 0.

Also, we cannot plug x = 2 into p(x) for choice A either. It doesn't lead to issues for 1/(x+1) but the original function before we simplify will have x = 2 lead to a division by zero error.

A vertical asymptote is located at x = -1

The removable discontinuity happens when x = 2, so there will be a hole here.

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351°f

step-by-step explanation: