# Is it possible to find out $x^2$ parabola and function from 3 given points?

I am programming a ball falling down from a cliff and bouncing back. The physics can be ignored and I want to use a simple $y = ax^2$ parabola to draw the falling ball.

I have given two points, the edge of the cliff at $C(-0.9; 0.8)$ and the point where the ball hits the bottom $B(0.1; -1.05)$. Due to its symmetry we know there is another point at $A(-1.9; -1.05)$. So that are 3 points I could work with. $C$ is the vertex.

I've tried this approach but my parabola is not as exact as I need it to estimate for example the intersects with the x- or y-axis.

As the legs of the parabola are down, all I can tell is that $a$ is negative:

$$y = ax^2 + bx +c; a < 0$$

I tried drawing it in Excel which allows to add an polynomial trendline. This is the best approximation I could do so far (dotted line).

But I need the function. Is it somehow possible to calculate the equation somehow?

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Yes, we can find the equation for the parabola with the given information.

In this problem there are three unknowns, the coefficients $a,b$ and $c$ so to find them we will need three pieces of information. In this case we know that the parabola must pass through the three given points.

Substitute the given points into the equation for the parabola. Then:

$$0.8 = a(-0.9)^2 + b(-0.9) + c,$$ $$-1.05 = a(0.1)^2 + b(0.1) + c,$$ $$-1.05 = a(-1.9)^2 + b(-1.9) + c.$$

This is a system of three equations for three unknowns which we can solve in the following way.

From the first equation we can write $c = 0.8 - a(-0.9)^2 - b(-0.9).$ This expression for $c$ can be substituted into the second and third equations. Then:

$$-1.85 = a\left[(0.1)^2 - (-0.9)^2\right] + b(-0.8),$$ $$-1.85 = a\left[(-1.9)^2-(-0.9)^2\right] + b(-2.8) + 0.8.$$

From the first of these expressions $b = -\left[-1.85 - a\left[(0.1)^2 - (-0.9)^2\right]\right/0.8$

Substitute this into the final expression and solve for $a$, then $b$, then $c$. In this case:

$$a = -1.850,$$ $$b = -3.330,$$ $$c = -0.6985.$$

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Ah, thanks that worked! – 5chdn May 29 '13 at 7:00

Linear equation is the way to go. If you work it all out you'll get formulas for coefficients below: $$a = \frac{y_2(x_3 - x_1) - y_1(x_3 - x_2) - y_3(x_2 - x_1)}{x_1^2(x_2 - x_3)-x_3^2(x_2-x_1)-x_2^2(x_1-x_3)}$$ $$b = \frac{y_2 - y_1 + a(x_1^2 - x_2^2)}{x_2 - x_1}$$ $$c = -ax_1^2 - bx_1 + y_1$$

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Solve the system of linear equations $$\begin{cases} ax_1^2 + bx_1 +c=y_1; \\ ax_2^2 + bx_2 +c=y_2; \\ ax_3^2 + bx_3 +c=y_3; \end{cases}$$ w.r.t. unknown variables $a,\ b,\ c.$

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