Convert Hermite data to Bezier cubic

Relation between the forms

A cubic Bezier curve with control points $\mathbf B_0..\mathbf B_3$ has end points $\mathbf P(0)=\mathbf B_0$, $\mathbf P(1)=\mathbf B_3$ and end tangents

$$\mathbf P'(0)=3(\mathbf B_1-\mathbf B_0),\qquad \mathbf P'(1)=3(\mathbf B_3-\mathbf B_2)$$

If this curve is to equal the Hermite curve, the end conditions must be identical:

$$\mathbf B_0=\mathbf P_0,\quad \mathbf B_3=\mathbf P_1,\quad \mathbf B_1=\mathbf P_0+\tfrac13\mathbf P_0',\quad \mathbf B_2=\mathbf P_1-\tfrac13\mathbf P_1'$$

Both are cubics with the same four end conditions, so they are the same curve.

Control points

$$\begin{aligned} \mathbf B_0&=(0,0) \\ \mathbf B_1&=(0,0)+\tfrac13(3,6)=(1,\ 2) \\ \mathbf B_2&=(6,0)-\tfrac13(3,-6)=(5,\ 2) \\ \mathbf B_3&=(6,0) \end{aligned}$$

Point at u = 0.5

Weights $0.125, 0.375, 0.375, 0.125$:

$$x=0.375(1)+0.375(5)+0.125(6)=3.0,\qquad y=0.375(2)+0.375(2)=1.5$$

Point at u = 0.25

Weights $0.421875,\ 0.421875,\ 0.140625,\ 0.015625$:

$$\begin{aligned} x&=0.421875(1)+0.140625(5)+0.015625(6)=1.2188 \\ y&=0.421875(2)+0.140625(2)=1.125 \end{aligned}$$

Answer: Bezier control points (0,0), (1,2), (5,2), (6,0); P(0.5) = (3.0, 1.5); P(0.25) = (1.219, 1.125).