Past exam of the computer science course of the University of Oxford / 2015 / Geometric modelling / 2 / e / Solution

Rotating the geometry through $180^{\circ }$ sends every point $p$ to $-p$. To start the extension at the origin, also reverse the parameter:Then $E(0)=(0,0)=F(1)$ and $E(1)=(2,0)$. Merely using $-F(t)$ would trace the same rotated image toward the origin, instead of away from the join.

The reversed and rotated de Boor points are $-D_2,-D_1,-D_0$, namely $(0,1),(0,-1),(4,1)$. The first two are already $D_1,D_2$, so there is only one new de Boor point, $\boxed{D_3=(4,1)}$. Using the same uniform basis matrix,Equivalently, the Bézier curve handles are $C=(0,0)$, $-B=(0,-1)$ and $-A=(2,0)$, which gives the same polynomials immediately.