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I would like to know how to find the quadratic equation for a parabola that opens down and intersects the origin along with the vertex being in the first quadrant given the maximum of the parabola and what the x value of the second intercept is? Thanks

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A parabola that opens up or down has equation $$y = a(x-c)^2 + d$$ with $a,c,d$ real numbers. If $a\gt 0$, then the parabola opens up. If $a\lt 0$, then the parabolar opens down. The vertex of the parabola is at $(c,d)$.

If you know the parabola opens down, then you know $a\lt 0$. If the vertex is on the first quadrant, then you know that $c\gt 0$. If you know the maximum of the parabola, then you know the value of $d$ (it's the maximum). If you know that it goes through the origin, then you know that $0= a(0-c)^2 + d = ac^2+d$, so you know that $ac^2 = -d$. You need to know the value of $c$ (the location of the vertex) or the value of $a$ in order to completely determine the parabola.

If you happen to know that the other intersection with the $x$-axis occurs at a point $(b,0)$, $b\gt 0$, then the vertex will lie halfway between $(0,0)$ and $(b,0)$ (since the axis of the parabola is vertical), so $c = \frac{b}{2}$ and this gives you the necessary information ot find $a$, namely we have $a = -\frac{4d}{b^2}$, so the equation becomes $$ y = -\frac{4d}{b^2}\left(x - \frac{b}{2}\right)^2 + d,$$ where the parabola goes through $(0,0)$ and $(b,0)$, opens down, and has highest value $d$. You can verify that $y=0$ when $x=0$ and when $x=b$, and that the maximum value is attained at $x=\frac{b}{2}$, where we have $y=d$.

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That was exactly what I needed. Thanks –  dagronlund May 14 '12 at 22:37
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