# X-axis coordinates of outer control points (only) for a Quadratic Bézier curve through 3 points

I am interested in the distance between the 2 outer (left & right: P0 & P2) control-points of a quadratic Bézier curve that goes through 3 data points. The curve's non-equidistant control points are different from the 3 data points (sampled concurrent x&y values from the curve): image.

Is there a shortcut to calculate the x-axis coordinates of these outer control points (P0 & P2) for my curve? I could then simply fit a parabola to my 3 points (ax^2+bx+c) and "cut" it off vertically at these 2 points on the X-axis, so I have a curve with the same shape as my old Bézier curve, but this time in a simple non-parametric/explicit y=f(x) form!

Or is there another way to calculate this distance?

Hardly an original idea I guess, but any help is mucho appreciado!!

But it sounds like you want to take an arbitrary quadratic Bezier curve and express it in the form $y = ax^2 + bx + c$. In general, this is not possible -- take for example the curve with control points $P_0 = (0,0)$, $P_1 = (2, 1)$, $P_2 = (0,2)$. You can construct an equation of the form $y = ax^2 + bx + c$ that is an approximation of the original Bezier curve, but, as my example shows, it will sometimes be a very poor approximation.
Any quadratic Bezier curve is a parabola, but in general, this parabola will be rotated so that its axis of symmetry is at some angle (not vertical). So, it is possible to get an equation of the form $ax^2 + bxy + cy^2 + dx + ey + f = 0$, but that's the best you can do.