SOLUTION: Find the equation of a rational function which has an oblique asymptote at y=2x+1, a vertical asymptote at x= -1, and a hole at (3,0). I know the denominator would be (x+1)(x-3)

Algebra ->  Rational-functions -> SOLUTION: Find the equation of a rational function which has an oblique asymptote at y=2x+1, a vertical asymptote at x= -1, and a hole at (3,0). I know the denominator would be (x+1)(x-3)      Log On


   

Question 985904: Find the equation of a rational function which has an oblique asymptote at y=2x+1, a vertical asymptote at x= -1, and a hole at (3,0).
I know the denominator would be (x+1)(x-3)
and the numerator would be something like (x-3)(something) and its quotient with the denominator would be 2x+1. I don't know how to utilize the y coordinate of the hole to solve the problem, and exactly how to use the given oblique asymptote to figure out the numerator.
Thank you so much for helping me.
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Need degree of numerator to be one more than degree of denominator, because you want the oblique asymptote.

%28%28x-3%29%28ax%5E2%2Bbx%2Bc%29%29%2F%28%28x-3%29%28x%2B1%29%29
The simplification for the factor of 1 would still contain
%28ax%5E2%2Bbx%2Bc%29%2F%28x%2B1%29 which represents both a rational expression AND a polynomial division, either with or without a remainder.

You need quotient to be %28ax%5E2%2Bbx%2Bc%29%2F%28x%2B1%29=2x%2B1%2Br, the remainder being some nonzero r.

ax%5E2%2Bbx%2Bx=%28x%2B1%29%282x%2B1%2Br%29
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RHS, 2x%5E2%2Bx%2Brx%2B2x%2B1%2Br
2x%5E2%2B2x%2Brx%2Br%2B1
2x%5E2%2B%282%2Br%29x%2B%28r%2B1%29
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If r=0 then ax%5E2%2Bbc%2Bc=%28x%2B1%29%282x%2B1%29=2x%5E2%2B3x%2B1 but the "1" does not need to be exactly this.
It can be a constant other than 1.

Should be any k constant ax%5E2%2Bbx%2Bc=2x%5E2%2B3x%2Bk for the quadratic factor in the numerator.

highlight%28y=%28%28x-3%29%282x%5E2%2B3x%2Bk%29%29%2F%28%28x-3%29%28x%2B1%29%29%29