SOLUTION: I am trying to work with an oblique asymptote and I am at a delimma. The problem is -x^3+2x^2+2x-5 /x^2+x-12. I have gotten to this for the denominator but I am stumped. x^2+x

Question 984101: I am trying to work with an oblique asymptote and I am at a delimma. The problem is -x^3+2x^2+2x-5 /x^2+x-12. I have gotten to this for the denominator but I am stumped. x^2+x-12 as (x+4)(x-3). Can you help me?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
When you do long division of the denominator into the numerator, you get (-x+3) in the quotient with remainder of -12x +31
If you multiply the denominator, unfactored, by (-x+3) and then add -12x +31, you will get the original function back.
The slant asymptote is -x+3
There are vertical asymptotes at x=-4 and x+3.
Note, in order to show the minus 1x slope, you need larger numbers on the calculator for x, like at least 50 or preferably 100.

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