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?
What is the most direct way to derive an equation for a parabola from its x and y intercepts?
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functions
analytic-geometry
conic-sections
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2your equation is of the form $c(x^2-a^2)$ where the x intercepts are $\pm a$ and $c = -y_{max}/a^2$ – 2011-07-20
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0@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. – 2011-07-20
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0@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. – 2011-07-20