dear math whizes, can you help me find the hole of the graph

[pinkcalculator]

I'm 4/5 done with this problem, but I don't know how to find the hole, and I'm not sure if it can be done algebraically or if I have to guess based on the graph.

Here is what I have done:
(x squared -9)/(x squared -2x-3)

I factored that out and got
(x+3) (x-3)/(x+1)(x-3)
Then I cancelled the (x-3) out because its in the numerator and denominator,
So I've got (x+3)/(x+1)

I know how to find the asymptotes, but I'm clueless about the hole.
Any suggestions??
Thanks so much.

 

[wjm11]

I'm 4/5 done with this problem, but I don't know how to find the hole, and I'm not sure if it can be done algebraically or if I have to guess based on the graph.

Here is what I have done:
(x squared -9)/(x squared -2x-3)

I factored that out and got
(x+3) (x-3)/(x+1)(x-3)
Then I cancelled the (x-3) out because its in the numerator and denominator,
So I've got (x+3)/(x+1)

I know how to find the asymptotes, but I'm clueless about the hole.

Congrats, pinkcalculator! You're done. If nothing had cancelled, you'd have had two vertical asymptotes (at x = -1 and 3). Since the (x-3) terms canceled, you have a hole (removable discontinuity) at the point where x = 3. Just draw the graph of y = (x+3)/(x+1), but put an open hole where it crosses x = 3.