SOLUTION: for the function Q(x) sate the behavior of the function at each of the asymptotes. [Oblique Asymptote (x-5), Vertical Asymptote x=-2 & x=-1] Q(x) = (x^3-2x^2-5x+6)/(x^2+3x+2)

Algebra ->  Graphs -> SOLUTION: for the function Q(x) sate the behavior of the function at each of the asymptotes. [Oblique Asymptote (x-5), Vertical Asymptote x=-2 & x=-1] Q(x) = (x^3-2x^2-5x+6)/(x^2+3x+2)      Log On


   

Question 1041997: for the function Q(x) sate the behavior of the function at each of the asymptotes. [Oblique Asymptote (x-5), Vertical Asymptote x=-2 & x=-1]
Q(x) = (x^3-2x^2-5x+6)/(x^2+3x+2)
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Q%28x%29=%28x%5E3-2x%5E2-5x%2B6%29%2F%28x%5E2%2B3x%2B2%29

%28x%5E3-2x%5E2-5x%2B6%29%2F%28%28x%2B1%29%28x%2B2%29%29
Vertical asymptotes x=-1 and x=-2, and possible sign change in Q(x) at each asymptote. Why vertical asymptotes? Undefined at those x values. 


              x  -5
          ________________________
x^2+3x+2  |   x^3-2x^2-5x+6
          |
          |   x^3+3x^2+2x
          |_______________
              0 -5x^2-7x+6               Bring down the 6.
                -5x^2-10x-10
                --------------
                  0  3x+16

Q%28x%29=x-5%2B%283x%2B16%29%2F%28x%5E2%2B3x%2B2%29

The non-remainder part of the quotient is x-5, and the remainder part is %283x%2B16%29%2F%28x%5E2%2B3x%2B2%29; this remainder will become increasingly small and approach 0 as x goes to the left or to the right unbounded. The oblique asymptote for Q is y=x-5.

The exact behavior for the vertical asymptotes, I did not analyze for you.