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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    $\begingroup$ Parabolas don't have a "first" and "last" point. Presumably, you are talking about a portion of the parabola. $\endgroup$ – Arturo Magidin Mar 15 '11 at 4:13
  • $\begingroup$ I am talking about where it crosses the x axis. I edited the question to reflect that. $\endgroup$ – garbagecollector Mar 15 '11 at 4:14
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    $\begingroup$ @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? $\endgroup$ – Arturo Magidin Mar 15 '11 at 4:19
  • $\begingroup$ @Arutro I edited the wrong question on accident. $\endgroup$ – garbagecollector Mar 15 '11 at 4:22
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    $\begingroup$ @garbagecollector: Parabolas still don't have a "first" or "last" point. They are infinite curves. $\endgroup$ – Arturo Magidin Mar 15 '11 at 4:24

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 $(x_3,y_3)$:


whose verification is left as an exercise.

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  • $\begingroup$ I have updated the question. Could you update your answer to reflect it. $\endgroup$ – garbagecollector Mar 15 '11 at 4:17
  • $\begingroup$ $(7500, -250)$ ain't an x-intercept, yo. ;) The quadratic formula is excellent for figuring where a parabola crosses the horizontal axis, tho. $\endgroup$ – user8276 Mar 15 '11 at 4:20
  • $\begingroup$ I accidentally edited the wrong question. If you could answer it as written now. It would be appreciated. $\endgroup$ – garbagecollector Mar 15 '11 at 4:24

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