since the denominator is (x-3)(x+2), there are asymptotes (and hence, holes) at x = -2,3
as x->∞, g(x)->1, so the horizontal asymptote is at g=1
g(-7/3) = 1
g(x)=(x^2+2x+1)/(x^2-x-6)
1. Find any holes in the graph of g(x)
2. Find any horizontal or oblique asymptote for the graph.
3. Determine where (if any at all) the graph of g(x) crosses the horizontal or oblique asymptote.
1 answer
1 answer