Question

what are the points of discontinuity for y= x+1 over x^2-4?

Answer 1

you have vertical asymptotes when the denominator is zero

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

so you need to solve

x - 2 = 0

and

x + 2 = 0

there is a horizontal asymptote at y = 0

hope this helps

Answer 2

thank you! (: