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A de Boor point is a control point in a B-spline representation. For a degree-two uniform segment with three active de Boor points $D_0,D_1,D_2$, the equivalent Bézier curve handles are $(D_0+D_1)/2,D_1,(D_1+D_2)/2$. The next segment shares $D_1,D_2$ and introduces one new handle. This formula assumes simple equally spaced spline knots; clamped or repeated spline knots have different endpoint conversions.

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 $t\in [0,1]$ areThey 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 areOne can verify this conversion simply by substituting these expressions into the Bernstein basis formula and collecting each $D_i$. Solving givesThe desired matrix representation is thereforeThe 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.

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.

Assign one unit parameter interval to each piece. ThenThere is no ambiguity at $u=1$, since both pieces give the origin. For a single B-spline representation take degree two, the four de Boor pointsand the equally spaced spline knot vector $\tau =(-2,-1,0,1,2,3,4)$. With $N_{i,2}$ denoting the normalized degree-two B-spline on this vector, the complete valid parameter interval is $[{\tau }_2,{\tau }_4]=[0,2]$ andOn $[0,1]$ the active handles are $D_0,D_1,D_2$, and on $[1,2]$ they are $D_1,D_2,D_3$. Shared handles and a simple interior spline knot are what merge the two pieces into one uniform B-spline. The exterior spline knots support the basis; they do not extend the interval being drawn.

In particular, the requested horizontal pieces and their vertical counterparts are