but this must be the point (0, 0). From this we calculate that (1/4) + (1/2)y' + (1/4) = 0 and thus y' = −1 and

The acid test is to see if **B**(1/4) satisfies y = x^{2}.
**B**(1/4) = (9/16)(−1, 1) + (3/8)(0, −1) + (1/16)(1, 1)
= (−9/16 + 1/16, 9/16 − 3/8 + 1/16) = (−1/2, 1/4).
Hmm, perhaps it is a parabola.

The general point on this Bézier curve is:
**B**(t) = (1 − t)^{2}(−1, 1) + 2(1 − t)t(0, −1) + t^{2}(1, 1)

= (− (1 − 2t + t^{2}) + t^{2},
(1 − 2t + t^{2}) − 2(t − t^{2}) + t^{2})

= (2t − 1, 1 − 4t + 4t^{2}).
This is indeed exactly our parabola!
Some affine transformation can adapt this calculation to any 2nd degree Bézier curve.

The claim about the cubic remains unjustified. The third order Bézier curve is defined by 8 numbers and defines a curve with two endpoints. The general cubic requires 9 independent reals to define. It may still be that every Bézier is a cubic. This matters because I know efficient ways to plot conic sections and cubics with adds in the inner loop.

I settle this by a Scheme program (and see this) which computes 10 points on a third order Bézier curve and evaluates the determinant to see if all the points are on some 3rd degree curve.

The determinant is 0; ten points on the Bézier curve all lie on the same cubic curve. I am flummoxed. I wish I could prove this.

Indeed it seems that any parametrically defined plane curve defined by nth degree polynomials is a nth degree curve.
This is an elementary algebraic proposition, but much too messy for me.
The code corroborates the theorem for degrees 1 thru 7.
Here is the conjecture.
Affine transformations are not powerful enough to derive this result.

random notes;
A 19th century theorem settles the question.