How to find possible equations given 2 points and the $y$-value of a vertex

Question

If I have 2 points (let’s say $(0,8)$ and $(6,0)$) and a line on which a vertex can be ($f(x)=16$), how can I find the possible quadratic equations that would intersect both the points and have the vertex lay on that line?

Answer 1

Let the vertex be V(h, 16). Then the quadratic curve is $y = a(x - h)^2 + 16$ for some negative value of $a$ and also $h$.

All this deduction comes from the knowledge of the locations P(0, 8), Q(6, 0), and V.

Since it must pass through P, then $8 = a(0 - h)^2 + 16$.

Similarly for Q, we have $0 = a(6 - h)^2 + 16$

Solving the two, we get $h^2 + 12h -36 = 0$, from which we get h = -14.48528137 after rejecting the other impossible value.