Question

find the vertical asymptotes and holes.

y= (x+3)(x-2)/ (x-2)(x+1) (a fraction)

Answer 1

\[y=\frac{(x+3)(x-2)}{(x-2)(x+1)}\]

Answer 2

Since the factor x-2 cancels and since it would be 0 is x is 2, there is a hole at x = 2
Since x+1 would be 0 if x = -1, there is a vertical asymptote at x = -1