Finding the equation of a parabola containing $(2,4)$ and $(-4,4)$ Is it possible to write equation of a parabola that $(2,4)$ and $(-4,4)$ lie on it?
My effort:
$x=\frac{-b}{2a}=-1$, so $b=2a$
$y=ax^2+bx+c$
$4=4a+2a+c$ and $4=16a-4a+c$
But these give $a=b=0$.
 A: If $y=f(x)$ is a quadratic going through those points then
$$f(x)-4$$
is a quadratic going through the points $(2,0)$ and $(-4,0)$.  Therefore
$$f(x)-4=a(x-2)(x+4)$$
for some constant $a$, and
$$y=f(x)=a(x-2)(x+4)+4\ .$$
It is impossible to find the specific value of $a$ unless you have some more information.
A: Pretty sure you have a calculation error. Here is my attempt:
Plugging in the points $(2,4)$ and $(-4,4)$ into the general equation $y=ax^2+bx+c$, as you did, we get a system of 2 equations in 3 variables:
\begin{align}
(4)&=a(2)^2+b(2)+c\tag{1}\implies4=4a+2b+c\\
(4)&=a(-4)^2+b(-4)+c\tag{2}\implies4=16a-4b+c
\end{align}
Subtracting the 2nd equation from the first we get:
$$0=-12a+6b\implies b=2a$$
And plugging into the 1st equation we get:
$$4=8a+c\implies c=4-8a$$
Now we can choose the variable $a$ as we like and we will get a solution to the system. It doesn't matter what value we choose for $a$  (You can check by plugging in different solutions with different choices of $a$ into the system). Each choice will give us a valid solution to the system, and thus, as mentioned in the comments, there are many parabolas that pass through the desired points.
Plugging in a solution of the system to the general equation $y=ax^2+bx+c$ will give us a parabola passing through the desired points.
If I tried plugging in $b=2a$ to both the 1st and 2nd equations, as you did, we would just get the same equation twice, but this doesn't help us solve the system.
A: For $x=2,$ you should have $bx = 2a\cdot2 = 4a,$ but you wrote $2a.$ For $x=-4,$ you should have $bx = 2a\cdot-4 = -8a,$ but you wrote $-4a.$ If you correct these errors and combine the terms in $a$ in your equations, you'll find that your two equations are identical and $a$ is not determined by them.
A better approach is to simply pick a vertex point. You know what the $x$-coordinate of the vertex must be. The $y$-coordinate must not be $4,$ since the point $(-1,4)$ is collinear with $(2,4)$ and $(-4,4)$ and you would end up with a straight line $(a=0)$, not a parabola.
Once you have chosen a vertex point, write the equation of the parabola in vertex form. You will be able to write all of the constants except $a$ as explicit numbers.
You next will want to solve for $a,$ but you have three points on the parabola,
two of which actually are useful for that purpose.
Once you have found the vertex form you can convert it to the form $y=ax^2 + bx + c$ if needed.
