= Solution
For a fixed $t\in[0,1]$, <De Casteljau's algorithm> first takes the same fractional position along both adjacent sides of the <control polygon>:
$$
L(t)=(1-t)A+tB=(-2+2t,t),\qquad
R(t)=(1-t)B+tC=(0,1-t).
$$
It then applies another <linear interpolation> with that same $t$:
$$
F(t)=(1-t)L(t)+tR(t)
=(1-t)^2A+2t(1-t)B+t^2C.
$$
This is the degree-two <Bernstein basis> representation of a quadratic <Bézier curve>. Its <coordinate> <polynomials> are
$$
\boxed{F_x(t)=-2(1-t)^2=-2+4t-2t^2,\qquad
F_y(t)=2t(1-t)=2t-2t^2.}
$$
The degree is exactly two, since its second <derivative> $(-4,-4)$ is nonzero. As a check on the construction, at $t=1/2$ the two intermediate <points> are $(-1,1/2)$ and $(0,1/2)$ and their <midpoint> is $F(1/2)=(-1/2,1/2)$.
Back to article page
