P(x) = (2x^3+x^2-18x-9)/(x^2-x-6)
1.Find any holes in the graph of P(x)
2.Find any horizontal or oblique asymptote
3.Determine where it crosses the horizontal/oblique asymptote (if it does)
P(x) =(2x+1)(x+3)(x-3)/((x-3)(x+2)
= (2x+1)(x+3)/(x+2)
When x = 3 , in new P(3) = 7(6)/5 = 42/5
so there is a hole at (3, 42/5)
using P(x) = (2x+1)(x+3)/(x+2)
= (2x^2 + 7x + 3)/(x+2)
= 2x+3 - 3/(x+2)
so we have an oblique asymptote
y = 2x +3
see graph
http://www.wolframalpha.com/input/?i=plot+y+%3D+%282x%5E3%2Bx%5E2-18x-9%29%2F%28x%5E2-x-6%29+%2C+y+%3D+2x%2B3
no intersection of the asymptote with the function