Formula of parabola from two points and the $y$ coordinate of the vertex The parabola has a vertical axis of symmetry. Given two points and the $y$ coordinate of the vertex, how to determine its formula?
For example:


Formula of parabola is $y=ax^2+bx+c$. The task here is first to find the values of $a,b,c$. We know that the vertex happens at $x^*=-\frac{b}{2a}$ (right?). We plug the points we have in the formula for parabola:
\begin{align}
6&=5^2a+5b+c\\
8&=8^2a+8b+c
\end{align}
and for the vertex we have 
\begin{align}
10&=(-\frac{b}{2a})^2a+b(-\frac{b}{2a})+c\\
\end{align}
From the first equation we find that $c=6-25a-5b$. This we plug in the other two equations and simplify to obtain
\begin{align}
39a+3b-2&=0\\
5+25a+5b+\frac{b^2}{4a}&=0
\end{align}
For the first equation we obtain $b=\frac23-\frac{39}{3}a$, which we plug in the second equation to obtain $$\frac{9a}{4}+\frac{1}{9a}+3=0$$ Solving this equations we obtain 
\begin{align}
a&=\frac29(-3-2\sqrt{2})\\
a&=\frac29(-3+2\sqrt{2})
\end{align}
and consequently you can find two values for $b$ and two values for $c$. Hence there are two values for the unknown ($-b/2a$) you are after $$?=11-3\sqrt{2}$$ and $$?=11+3\sqrt{2}$$
