= Solution
Choose a quadratic uniform <B-spline> segment on a unit <spline knot> interval. This choice is useful because adjacent pieces can later share <de Boor points> at simple, equally spaced <spline knots>. Its three active basis pieces on local $t\in[0,1]$ are
$$
b_0(t)=\frac{(1-t)^2}{2},\qquad
b_1(t)=\frac{1+2t-2t^2}{2},\qquad
b_2(t)=\frac{t^2}{2}.
$$
They follow from the <Cox-de Boor recurrence> for degree two and consecutive integer <spline knots>. If the three <de Boor points> are $D_0,D_1,D_2$, then the equivalent <Bézier curve> handles are
$$
A=\frac{D_0+D_1}{2},\qquad B=D_1,\qquad
C=\frac{D_1+D_2}{2}.
$$
One can verify this conversion simply by substituting these expressions into the <Bernstein basis> formula and collecting each $D_i$. Solving gives
$$
\boxed{D_0=2A-B=(-4,-1),\qquad D_1=B=(0,1),\qquad D_2=2C-B=(0,-1).}
$$
The desired <matrix> representation is therefore
$$
\boxed{F(t)=\frac12[t^2\ t\ 1]
\begin{pmatrix}1&-2&1\\-2&2&0\\1&1&0\end{pmatrix}
\begin{pmatrix}-4&-1\\0&1\\0&-1\end{pmatrix}.}
$$
The left row is again the monomial row, the middle <matrix> contains the monomial <coefficients> of $b_0,b_1,b_2$, and the right <matrix> lists the <de Boor points>. Unlike the <Bézier curve> basis <matrix>, the constant row averages two handles. A clamped <B-spline> with repeated end <spline knots> could instead reuse $A,B,C$ directly; the simple uniform convention above specifies which representation is being used and supplies the shared handles required for the extension.
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