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I am trying to build an equation where I could start at (x,y) which are known and create a parabola from that starting point. I have no idea where it intercepts the X or Y.

I know where I want the line on the other side to go down at (the other root) well I know I want it to be (??) units between the two lines. I know roughly how high it should go.

Any idea of how I could do that.

EX: http://crappygraphs.com/user_graphs/?id=7101 I know the (x,y) of 1 and I know the (x,y) of (x) and I know the (x,y) of (y). I just need to make something follow this path.

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    No other constraints? There are infinitely many parabolas that can be made to pass though a given point.2011-11-28
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    Let's see, I know where I want the line on the other side to go down at (the other root). I know roughly how high it should go.2011-11-28
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    What do you mean by "starts at"? Where on the parabola is this point?2011-11-28
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    Can you probably make a sketch so that we have a better idea of what you want?2011-11-28
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    Sorry it sucks so much but http://crappygraphs.com/user_graphs/?id=7101 for example (though it may go the other way).2011-11-28
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    So you know the coordinates of three points, and you want to fit a parabola through them? If you also know that the parabola must have a vertical axis, then this is a job for [Lagrange interpolation](http://en.wikipedia.org/wiki/Lagrange_polynomial).2011-11-28
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    If you know the vertex (the high point) is at $(h,k)$, you can use $y=a(x-h)^2+k$ as your equation. Then, solve for the value of $a$ to make the parabola pass through the other point.2011-11-28

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Here's your problem worked with the vertex of the parabola (with vertical line of symmetry) and another point on the parabola specified:

Suppose your vertex is at $(2,3)$ and that $(3,1)$ is another point on the parabola.

The equation of the parabola has the form $$ y=a(x-2)^2+3. $$

Since $(3,1)$ is on the parabola $$ 1=a(3-2)^2+3\quad\iff\quad 1=a+3\quad\iff a=-2. $$

So your equation is $y=-2(x-2)^2+3$.