Question

describe the vertical asymptotes and holes for the graph of y=(x-2)/[(x+2)(x-2)]

Answer 1

there is a hole at x=2 and x=-2 there is a vertical asymp.

Answer 2

now what do i do to figure that out?

Answer 3

Is it
\[\frac{x-2}{(x+2)(x-2)} = \frac{1}{x+2} ?\]

Answer 4

notice the x-2 cancel x-2=0 implies x cannot 2 since x+2 did not cancel x+2=0 means x=-2 is a veritcal asymp.

Answer 5

oh ok and how is it a vertical asymp.

Answer 6

plug in numbers close to 2 on the left and numbers close 2 on the right to see why

Answer 7

oops -2

Answer 8

as those numbers get close to -2 the bottom is getting closer to 0 which means it is approaching either negative infinity or positive infinity