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I have the first point is (0, -100) and the second point is (7500, -250), and the maximum point is at (x, 210).

Is it possible to find X or the equation of the parabola using this information alone?

If so how?

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    @Arturo I am learning alot more than i bargained for. I have updated the question once more. Again, those are the only two points I am aware of.2011-03-15

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Assume you have $y=ax^2 + bx + c$. The maximum is achieved at $x = -\frac{b}{2a}$.

The values you have give you the value of $c$ ($-100$), and a relation between $a$ and $b$ obtained by plugging in $(7500,-250)$. You also know that the maximum is achieved at $-\frac{b}{2a}$, so plugging that will give you another relation between $a$ and $b$. Putting them together will give you a quadratic equation that $b$ must satisfied.

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Three points determine a parabola, since you have three constants to account for in $y=ax^2+bx+c$ . Construct the appropriate system of linear equations and you're golden.

Otherwise, if you're too cool for solving linear equations, there is a determinant expression for the parabola passing through three points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$:

$\begin{vmatrix}x^2&x&y&1\\x_1^2&x_1&y_1&1\\x_2^2&x_2&y_2&1\\x_3^2&x_3&y_3&1\end{vmatrix}=0$

whose verification is left as an exercise.

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    I accidentally edited the wrong question. If you could answer it as written now. It would be appreciated.2011-03-15