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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. –  J. M. Nov 28 '11 at 15:13
    
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. –  Steven Nov 28 '11 at 15:14
    
What do you mean by "starts at"? Where on the parabola is this point? –  David Mitra Nov 28 '11 at 15:14
    
Can you probably make a sketch so that we have a better idea of what you want? –  J. M. Nov 28 '11 at 15:15
    
Sorry it sucks so much but crappygraphs.com/user_graphs/?id=7101 for example (though it may go the other way). –  Steven Nov 28 '11 at 15:24
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1 Answer

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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 $y=-2(x-2)^2+3$.

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