Where is the hole for the following function located? f(x) = x+3/(x-4)(x+3)
A. x= -3
B. y=-3
C. x= 3
D. y = 3

It would seem to be a good idea to cancel the (x + 3) factors to get f(x) = 1/(x - 4)
This is a hyperbolic graph with a vertical asymptote at x = 4.
However when x = -3 the factor (x + 3) = 0 so cannot be cancelled at this point. 0/0 is undefined.
The graph is drawn leaving a hole in the line at x = -3
The limits as we approach x = -3 from above or below that point get closer and closer to -1/7

The correct answer is A