# Draw a parabola-like graph according to 3 points

I'm sure this question has a simple answer, but I'm a very beginner to calculus...
This is the problem I have: Given a vertex point and two x-axis cutting points, how do I find a formula for parabola-like diagram?
Important part: the vertex point is not in the middle of the parabola.

So I know, there are infinite solutions - how can find one specific?

Thank you.

-
Can you clarify what you mean by "vertex point"? –  Peter Taylor Mar 9 '11 at 12:18
Like Peter, I wonder what a "vertex" has that is not in the middle of the parabola. Can you explain this choice of terminology? –  The Chaz 2.0 Mar 9 '11 at 14:01

In general, one of the many ways to start, write

$Ax^2+Bx +C=y$

Write three of these for your three pairs, i.e translate the information given to: $(x_1,y_1)$,$(x_2,y_2)$,$(x_3,y_3)$

You've got three equations in three unknowns. Solve it.

Or, if $a,b$ are your x-axis cutting points, then parabola will be of the form

$y= K (x-a) (x-b)$

Determine K by using dy/dx =0 at vertex coordinates.

-

EDIT in light of myself's comment. By "parabola-like graph" I assume you mean quadratic in $x$. Three points determines a quadratic in $x$ uniquely. See http://en.wikipedia.org/wiki/Polynomial_interpolation.

-
Well actually $(0,0)$, $(1,1)$ and $(4,2)$ are on both the parabola $6y = 7x-x^2$ and the parabola $y^2 =x$. –  Myself Mar 9 '11 at 10:23
Ah, good point. There is one for every choice of (positively oriented) orthogonal axes. –  charles.y.zheng Mar 9 '11 at 10:36