Functions and Curve Sketching

**kleyzam** · November 27th 2008, 10:12 AM

A curve is given by the equation $y = \frac{ax^2 + bx + c}{dx^2 + ex + f}$ . The equations of the asymptotes to the curve are x = 1, x = 3 and y = 1. The curve passes through the points (2,0) and the origin..

I figured how to find d, e and f by taking the x asymptotes and make them = 0 ...getting $\frac{ax^2 + bx + c}{(x-1)(x-3)}$ I can't seem to figure out how to find a b and c though..any help would be greatly appreciated.

Thanks in advance

**earboth** · November 27th 2008, 10:44 AM

Originally Posted by kleyzam

A curve is given by the equation $y = \frac{ax^2 + bx + c}{dx^2 + ex + f}$ . The equations of the asymptotes to the curve are x = 1, x = 3 and y = 1. The curve passes through the points (2,0) and the origin..

I figured how to find d, e and f by taking the x asymptotes and make them = 0 ...getting $\frac{ax^2 + bx + c}{(x-1)(x-3)}$ I can't seem to figure out how to find a b and c though..any help would be greatly appreciated.

Thanks in advance

1. Since f(0) = 0 follows c = 0

2. If $\lim_{|x|\rightarrow \infty}\left(\dfrac{ax^2 + bx}{x^2-4x+3} \right)=1~\implies~ a = 1$

3. If $f(2) = 0$ then $2^2+2b=0~\implies~b = -2$

4. Therefore $f(x)=\dfrac{x^2 -2x}{x^2-4x+3}$

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