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:

 A: Hints: I will first give a straight-forward method, and then give a cleverer method.
Straightforward method:
You know the formula for a parabola is $$y=ax^2+bx+c.$$
The idea now is just to plug in your points and solve the resulting system of equations. The nonvertex points are easy to deal with - they give you the equations
$$6=25a+5b+c,$$
$$8=64a+8b+c.$$
Now you need to deal with the vertex point. Recall the $x$ coordinate of the vertex is $-\frac{b}{2a}$, so we can plug this in to get the final equation we need:
$$10=\frac{b^2}{4a^2}a-\frac{b}{2a}b+c=-\frac{b^2}{4a}+c$$
Cleverer method:
This time we realize that we can write the parabola in a completed square form. That is we can write
$$y=a(x+b)^2+c$$
This is helpful because we know that the $y$-coordinate of the vertex corresponds to when the squared term $(x+b)^2=0$ - in other words the place where the parabola reaches an extremum. Hence we directly have $c=10$. Now we can plug in the other points as before and have an easier system to solve:
$$6=a(5+b)^2+10,$$
$$8=a(8+b)^2+10$$
A: 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 $$?=11+3\sqrt{2}$$
A: We are dealing with $P=\left\{ \langle x,y\rangle\mid y=q-a\left(x-p\right)^{2}\right\} $
where $(p,q)$ is the vertex. 
Here $q$ is known and the two points belonging to $P$ are giving us two equations in $a$ and $p$.
