f(x) = 1/(x^2-x-2)

Question

does f(x) have two or three asymptotes
does f(x) have one or two discontinuities

We have f(x) = 1/(x^2-x-2) or
f(x)=1/[(x-2)(x+1)]
f has vert. asympt. at the roots of the denominator.
As x becomes unbounded in either direction you should be able to see that it approaches the x-axis.
So 2 vertical and 1 horizntal asymptote = 3 asymptotes, and 2 roots = 2 discontinuities.