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). Thanks!

Algebra ->  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). Thanks!      Log On


   

Question 897477: 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). Thanks!
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Something like a%28%28x%5E2-something%29%28x-3%29%29%2F%28%28x%2B1%29%28x-3%29%29

and you want to be able to perform a division (after simplification) that makes a quotient of 2x+1 and have a remainder of denominator of x+1.

The simplification would be to cancel the (x-3) so you have something of a%28x%5E2%2B-+something%29%2F%28x%2B1%29; you could ignore the factor "a" in doing the division. Dividend would be some variable x%5E2%2Bmx%2Bn, divisor is x%2B1, and you know that remainder will be some numerator%2F%28x%2B1%29; The quotient as said, must be 2x+1.