= Solution
Rotating the geometry through $180^\circ$ sends every <point> $p$ to $-p$. To start the extension at the origin, also reverse the parameter:
$$
E(t)=-F(1-t),\qquad
\boxed{E(t)=(2t^2,\,-2t+2t^2),\quad 0\le t\le1.}
$$
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>,
$$
\boxed{E(t)=\frac12[t^2\ t\ 1]
\begin{pmatrix}1&-2&1\\-2&2&0\\1&1&0\end{pmatrix}
\begin{pmatrix}0&1\\0&-1\\4&1\end{pmatrix}.}
$$
Equivalently, the <Bézier curve> handles are $C=(0,0)$, $-B=(0,-1)$ and $-A=(2,0)$, which gives the same <polynomials> immediately.
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