0
$\begingroup$

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.

  • 0
    Like Peter, I wonder what a "vertex" has that is *not* in the middle of the parabola. Can you explain this choice of terminology?2011-03-09

2 Answers 2

3

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.

1

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.

  • 0
    Ah, good point. There is one for every choice of (positively oriented) orthogonal axes.2011-03-09