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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1Parabolas don't have a "first" and "last" point. Presumably, you are talking about a portion of the parabola. – Arturo Magidin Mar 15 '11 at 04:13
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I am talking about where it crosses the x axis. I edited the question to reflect that. – myusuf3 Mar 15 '11 at 04:14
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2@garbagecollector: Neither $(0,-100)$ nor $(7500,-250)$ are on the $x$-axis, so how could they possibly be the intersections of the parabola with the $x$-axis? – Arturo Magidin Mar 15 '11 at 04:19
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@Arutro I edited the wrong question on accident. – myusuf3 Mar 15 '11 at 04:22
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3@garbagecollector: Parabolas still don't have a "first" or "last" point. They are infinite curves. – Arturo Magidin Mar 15 '11 at 04:24
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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. – myusuf3 Mar 15 '11 at 04:26
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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.
Arturo Magidin
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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 have updated the question. Could you update your answer to reflect it. – myusuf3 Mar 15 '11 at 04:17
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$(7500, -250)$ ain't an x-intercept, yo. ;) The quadratic formula is excellent for figuring where a parabola crosses the horizontal axis, tho. – Mar 15 '11 at 04:20
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I accidentally edited the wrong question. If you could answer it as written now. It would be appreciated. – myusuf3 Mar 15 '11 at 04:24