(x^2+x-6)/(x^2-9) = (x+3)(x-2) / (x+3)(x-3)
so there is a hole at x = -3, where y = 0/0 (undefined)
Everywhere else, y = (x-2)/(x-3) which has
zero at x=2
asymptote at x=3
so, D
Describe the vertical asymptote and hole for the graph of (x^2+x-6)/(x^2-9).
a. asymptote: x=2; hole: x=-3
b. asymptote: x=3; hole: x=2
c. asymptote: x=-3; hole: x=3
d. asymptote: x=3; hole: x=-3
I know that it has to either be b or d because the asymptote is x=3. Can someone please help with the rest?
1 answer
1 answer