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.
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.