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I have three points (250, 0), (500,500) and (750, 0). To find a curve passing through these points all I have to do is plug-in these values into the general quadratic equation:

f(x) = ax^2 + bx + c


f(x) = -x^2/125 + 8x -1500

But, this is specific to this curve.

How would I generalize this to a curve that passes through (j, k), (p,q) and (u,v) such that the equation

f(x) = ax^2 + bx + c

has a, b and c expressed in terms of j,k,p,q,u and v; thus allowing me to plug-in arbitrary point values into this new general equation to get a specific equation for a curve?

I tried solving and eliminating variables, but all I end up with is an unsimplifiable mess that becomes too large to handle and I can't get very far.

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up vote 1 down vote accepted

If you plug j,k,p,q,u and v into the quadratic equation, you can get a system of 3 linear equations with 3 variables a,b,c and 6 parameters.

$$\begin{cases} k=j^2a+jb+c\\ q=p^2a+pb+c\\ v=u^2a+ub+c \end{cases}$$

It can be solved by various methods. The most suitable in this case is Cramer's rule. Look for the section called Explicit formulas for small systems.

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Thanks for the Cramer's rule link. It helped. I figured out the solution using that and mathematica finally achieving f(x) = (-x^2 x2 y1-x x2^2 y1+x^2 x3 y1+x2^2 x3 y1+x x3^2 y1-x2 x3^2 y1+x^2 x1 y2+x x1^2 y2-x^2 x3 y2-x1^2 x3 y2-x x3^2 y2+x1 x3^2 y2-x^2 x1 y3-x x1^2 y3+x^2 x2 y3+x1^2 x2 y3+x x2^2 y3-x1 x2^2 y3)/((x1-x2) (x1-x3) (x2-x3)) where the curve passes through points (x1,y1),(x2,y2), and (x3,y3). I would vote up your answer if had enough rep. Thanks! – Æ - Nov 27 '12 at 21:48

Essentially, you have 3 linear equations in 3 variables $a$, $b$ and $c$ as follows:

$aj^2 + bj + c = k$

$ap^2 + bp + c = q$

$au^2 + bu + c = v$

This is a system of linear equations and you can solve them using any of the methods given.

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