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Given $(0,0)$ origin, point $(3,2)$ and $y\text{-max} = 5$, find the parabola. I tried to shift point $(3,2)$ down to $(3,0)$ so that it can become symmetric to origin. Then the vertex would be $(5,\frac 3 2)$ then using the equation $$(x-h)^2=4p(y-k).$$

However after doing this the point $(0,0)$ does not lie within the parabola.

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  • $\begingroup$ You’re assuming that the axis of symmetry is parallel to the $y$-axis. Without some assumption about this axis or other assumptions, there’s not enough information to solve the problem. $\endgroup$ – amd Mar 20 '19 at 4:29
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Let $$y = -a(x-b)^2 + c$$

We know that $c=5$.

At point $(0,0)$, we have $0=-ab^2+5$, that is $$ab^2=5$$

At point $(3,2)$, we have $2=-a(3-b)^2+5$, that is $$a(3-b)^2=3$$

Solve for $a$ and $b$.

To solve for $b$, note that we have $$5(3-b)^2=3b^2$$

which is just a quadratic equation, try to take it from here.

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