How to find position of object on a parabola? I'm making a computer game, where an object flies along a parabola curve. This object is 'thrown' by a 'robot' towards another robot.
I know the vertex of the parabola, and also it's two intersections with the x axis.
I also now the object's x position.
I need a way to find the object's y position on the parabola.
I'm not great in math, so I'd like the method to do this be as simple as possible. Please try to explain it in a way I'll be able to understand.
My parabola: 
Thanks a lot
 A: Ok, assuming that $y = ax^{2} + bx + c$:
The vertex is given by 
$$
y' = 0 = 2ax + b \implies x_{vertex} = -\frac{b}{2a}.
$$
Now we also have the intercepts of the x axis given by the quadratic formula,
$$
x = \frac{-b}{2a}\pm\frac{\sqrt{b^{2}-4ac}}{2a}.
$$
Now does y = x = 0? since that will give c. So from knowing the vertex value and the roots and the intercept of y at x = 0 we can define any point on y in terms of x. Alternatively if we set robot 1 to have x-cord of (x,y)=(0,0) then c = 0. From here you can compute what a, and b are from the vertex value and the points that the robots are placed. 
More explicitly:
$$
\mathrm{vertex} = -\frac{b}{2a}\\
\mathrm{robot\,1\,position} = \frac{-b}{2a} - \frac{\sqrt{b^{2}-4ac}}{2a}\\
\mathrm{robot\,2\, position} = \frac{-b}{2a} + \frac{\sqrt{b^{2}-4ac}}{2a}
$$
A: You are given the coordinate of the vertex $(h,k)$ and two distinct real roots, which by symmetry can be written as $h\pm r$  for some $r>0.$
Then the equation for the parabola is
$$y=-\frac{k}{r^2}(x-h)^2+k.$$

Another way to express this:
You are given distinct real roots $r_1,r_2$ and coordinate of vertex $\left(\frac{r_1+r_2}{2},k\right).$
Then the equation for the parabola is
$$y=-\frac{4k}{(r_1-r_2)^2}(x-r_1)(x-r_2).$$
