Answer:
There is a hole at x = 3 and an asymptote at x = –1.
Explanation:
I got it on edge
f(x) = (x-3)/(x²-2x-3)
f(x) = (x-3)/(x-3)(x+1) = 1 / x+1 (x=3 is a hole)
asymptote while x+1 --> 0 and f(x) -->∞
x = -1 is vertical asymptote
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