Is it possible to work backwards from completing the square

What I mean is, I'm basically trying to model a piece of data with a quadratic function. I know that, if you have the equation to start of with, you can you completing the square to find out the turning point on the graph, and the points where the line cuts the x-axis. However, I was curious, for my coursework, is it possible to work backwards from completing the square, so you know you're turning point and where the points cuts the x-axis, and at the end of working backwards, you'll have the full quadratic equation.

Is this theory possible?

Thankyou

Sure. The vertex (turning point) and any other point are sufficient to define a quadratic function.

However, not every quadratic function crosses the x-axis at all. So you won't always have that information to start from. And note that since you only need two points, the third has to be consistent with them (specifically, the x-coordinate of the vertex is halfway between the x-intercepts).

In completing the square you found the vertex by rearranging the equation into this form:
(y - y0) = A(x - x0)^2
If you know the vertex (x0, y0), you can write an equation in this form.
We'll rearrange it later, but first we need to solve for A.

Any point (x,y) on the curve has to satisfy the equation. This includes your intercept point, which we'll call (c, 0).
(0 - y0) = A(c - x0)^2
A = - y0 / (c - x0)^2
For example, a vertex of (2, -1) and point (5,0) gives A = 1 / 9.

In fact, the point did not have to be an intercept. For a vertex (x0,y0) and any point (x1, y1) you could find
A = (y1 - y0) / (x1 - x0)^2

Now that you have A, you can rearrange the vertex-form equation into standard form.
(y - y0) = A(x - x0)^2
y = Ax^2 - (2Ax0) x + (Ax0^2 + y0)

Yes, you would need to apply other concepts line multiply the entire equation by a constant, make sure what you do us done to both sides of the equation, you can't just add anything back in, you have to make sure you otherwise find the axis points a second time