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 October 14th, 2009, 09:19 PM #1 Newbie   Joined: Sep 2009 Posts: 3 Thanks: 0 Rational Function vertical asymptotes: x = ?1 and x = 2 horizontal asymptote: y = 3 x-intercept: (3, 0) I need to find a rational function that supports this. so far I have the denominator set to (x+1)(x-2) but I CAAAAANT figure out the numerator for the life of me. someone help me please!
 October 15th, 2009, 04:52 AM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Hint: try to find a solution in the form $f(x)=\frac{x-\alpha}{(x+1)(x-2)}+3.$ Then since the denominator of the fraction goes like $x^2$ and the numerator goes like $x$, you will have $f(x)\to3$ as $x\to\pm\infty.$ You just need to find the value of $\alpha$ that will give you $f(3)=0.$
 October 15th, 2009, 03:02 PM #3 Global Moderator   Joined: Dec 2006 Posts: 18,034 Thanks: 1393 If (3, 0) is to be the only x-intercept, use $^{f(x)\,=\,\frac{3(x\,-\,3)^2}{(x\,+\,1)(x\,-\,2)}}$ instead, which clearly works.

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