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I have a pair of points at my disposal. One of these points represents the parabola's maximum y-value, which always occurs at x=0. I also have a point which represents the parabola's x-intercept(s). Given this information, is there a way to rapidly derive the formula for this parabolic curve? My issue is that I need to generate this equation directly in computer software, but all the standard-formula definitions for a parabolic curve use its Vertex, not its intercepts. Is there some standard form of equation into which these intercepts can be 'plugged in' in order to produce a working relation? If not, what is the most computationally direct way to solve this problem?

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your equation is of the form $c(x^2-a^2)$ where the x intercepts are $\pm a$ and $c = -y_{max}/a^2$ – deinst Jul 20 '11 at 15:16
@GWLlosa: What do you mean you have a point which represents the parabola's $y$-intecepts? You said you already know the $y$ value at $x=0$, that is the $y$ intercept. – Eric Jul 20 '11 at 15:30
@Eric Naslund: Typo. I meant 'x' intercepts. That is to say, i have a parabola which crosses x twice and y once, and I know all three of those points. – GWLlosa Jul 20 '11 at 17:30

3 Answers

up vote 3 down vote

I'll assume you meant you know the $x$-intercepts and maximum height.

If you have any parabola with $x$-intercepts $a,b$, $a\neq b$, and maximum height $c$, then you can write it as $$y=k(x-a)(x-b)$$ where $$k=-c\left(\frac{4}{(a-b)^2}\right).$$

(Notice that the value $c$ must be positive)

If $a=b$ we actually can't specify $k$ without more information.

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Actually OP says that the parabola attaints its maximal value at $x=0$, hence you can suppose $b \neq a = -b$, so that you have $k = -c/a^2$ and $y = k(x^2 - a^2)$, as deinst said in his comment. Letting $x = 0$ means $c = -y(0)/a^2$ so you also know $c$. Come on, man. – Patrick Da Silva Jul 20 '11 at 15:38
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@Patrick: Two things: One is that I wanted this answer to be more general so that deinst's comment follows immediately when you let $a=-b$, and two you can't assume $b\neq a$ even in the OP's case. The polynomial $-kx^2$, for $k>0$ has its maximum value at $x=0$ and you cannot determine $k$ based on the information in the question. (I thought you would know better...) – Eric Jul 20 '11 at 15:41
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Okay you win... XD – Patrick Da Silva Jul 20 '11 at 15:45
up vote 2 down vote

To answer question found in the title: "... an equation for a parabola from its $x$ and $y$ intercepts", the correct equation is:

$$y = \frac{c}{ab}(x-a)(x-b),$$

where $a, b$ are the $x$-intercepts and $c$ is the $y$-intercept. We can prove this is correct by noting that $y = 0$ when $x=a$ or $x=b$ is substituted, and when $x=0$, we have $y = \frac{c}{ab}(-a)(-b) = c$.

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Of course, if you know the max. value is the $y$-intercept, then $b = -a$ and you get $y = -\frac{c}{a^2}(x^2 - a^2)$, as @deinst commented above. – Shaun Ault Jul 20 '11 at 19:46
up vote 1 down vote

the equation would look like this

$$ y = k(x-a)(x-b)$$

now we have to figure out what k is. We know what the maximum value is, call it c, and that it's x value is 0. Therefor we can plug this into the equation so that we get the following

$$c = k(-a)(-b)$$ $$c = kab$$ therefor $k = c/(ab)$ therefor, your equation is

$$y = c(x-a)(x-b)/(a*b)$$

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