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:


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:
$$6=a(5+b)^2+10,$$
$$8=a(8+b)^2+10$$
