TITLE: Is it possible to find out $x^2$ parabola and function from 3 given points?
QUESTION [4 upvotes]: 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?

REPLY [4 votes]: Linear equation is the way to go.  If you work it all out you'll get formulas for coefficients below: 
$$
a = \frac{y_2(x_3 - x_1) - y_1(x_3 - x_2) - y_3(x_2 - x_1)}{x_1^2(x_2 - x_3)-x_3^2(x_2-x_1)-x_2^2(x_1-x_3)}
$$
$$
b = \frac{y_2 - y_1 + a(x_1^2 - x_2^2)}{x_2 - x_1}
$$
$$
c = -ax_1^2 - bx_1 + y_1
$$
{"set_name": "stack_exchange", "score": 4, "question_id": 405451}