Finding a rational function from given extremas A function is given by 
$$f\left(x\right)=\frac{\left(ax^2+2bx+c\right)}{Ax^2+2Bx+C}$$
It has points of extrema at x=1 and x=-1 such that f(1)=2, f(-1)=3 and f(0)=2.5. Then find the value of a/A, prove that A=C and find the function f(x).
My attempt:
I didn't want to  solve 4-6 linear equations, so I analyzed it graphically. As x tends to infinity, a/A must be in between 2 and 3. The graph was looking pretty symmetric and thus I assumed a/A=2.5 which is the solution given in my book. And then I got stuck. Any clever idea to tackle this question? Or I have to solve hell amount of equations to get to my answer?
 A: Here is a good way to discover one solution. Showing that it's the only solution is a different matter.
We know that $f(x)$ has a max at $(1,3)$ and a min at $(-1,2)$. This means that $g(x)=f(x)-\frac52$ has a max at $(1,\frac12)$ and a min at $(-1,-\frac12)$. Let us make the assumption that $g$ is an odd function. In fact, we assume that $g$ is of the form: $$g(x)=\frac{\beta x}{Ax^2 + C}.$$ Since $g$ has the horizontal asymptote $y=0$, we can deduce the end behavior of $f$, so $\dfrac{a}{A}=\dfrac52$. As noted in a comment, we have one degree of freedom, so we take $a=5,A=2$
Now, $g$ (hence also $f$) has derivative: $$f'(x)=g'(x)=\frac{-\beta(Ax^2-C)}{(Ax^2+C)^2}.$$ If $x=\pm1$ are to be critical points, then $A=C$, as required.
Next, we want $g(1)=\dfrac12$, which will guarantee the correct value at $-1$ as well. Plugging in $A=C=2$, we have $\dfrac{\beta}{4}=\dfrac12$, so $\beta=2$. Now that we have $g$, we add $\dfrac52$ to obtain $f$.
As far as uniqueness, since there are four equations and five (homogeneous) variables, we should expect more solutions, indexed perhaps by one real parameter. We can certainly produce similar shapes with less symmetry, by considering first curves of the form $\dfrac{x}{A(x-h)^2+k}$. I'm not sure whether one of those, with non-zero $h$, could satisfy the needed conditions.

Edit: Indeed, dropping symmetry, we can find other solutions. Try $(9,-3,9,4,-2,4)$. This shows that the original problem cannot be solved uniquely without some additional assumption (although we do still have $A=C$, in this example). 
