Formula of parabola from two points and the $y$ coordinate of the vertex

Question

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:

Answer 1

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

Answer 2

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 $p$.

Answer 3

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 $$?=11+3\sqrt{2}$$