Finding a function with a set of points I have a function graph and I need to find the matching function. Here is the graph :
The graph
And here are three points from the graph :

*

*(-2;0)

*(0;-6)

*(3;0)

Note: As you can see, this is not a linear function
Does anyone know how to achieve that?
I was thinking of interpolation, the problem is that I don't find the formula...
 A: Quick and simple answer:$$
y = x^2 -x -6
$$
A: Solving a problem like this, at least by hand, usually involves figuring out the general family of functions that yours belongs to.  As the previous answer states, this is a quadratic.  However, being able to see that takes experience and there is in general no way to know.  Some hints that we have that indicate that it might be a quadratic:

*

*It sounds like a pre-calculus course question, and they study polynomials extensively

*The graph is obviously nonlinear, and if we are searching for polynomials we want to try the least complex one we can first.  The least complicated nonlinear polynomial is the quadratic.

*You were given only three points.  A quadratic has the general formula $ax^2 + bx + c = y$.  There are three unknown parameters, so with three points the best you can do is identify those three parameters.

So, how do you do it? Ultimately, you want to turn this into a set of linear equations if you can.  Plug in each point to get three equations:
$$0 = a(-2)^2 + b(-2) + c$$
$$-6 = a(0)^2 + b(0) + c$$
$$0 = a(-3)^2 + b(3) + c$$
From these three equations, you have three unknowns, and you can solve for each of them.  This is a general technique that can sometimes work if your equation is nice (that is to say, if it nicely fits into this technique of producing linear equations): plug in the points, get the equations, and solve them simultaneously.  The REALLY hard part about this is determining what kind of function you're solving for.
I should mention that this problem has a nice property: you were given the two zeros of the parabola and a point on it, which might be the vertex (I'm not sure, I didn't look that closely), so this particular problem is much simpler to solve.  However, the method I show above works in general, no matter which points you're given.
