Cubic Bezier point and tangent calculation

Equation

$$\mathbf P(u)=(1-u)^3\mathbf B_0+3u(1-u)^2\mathbf B_1+3u^2(1-u)\mathbf B_2+u^3\mathbf B_3$$

At u = 0.5

Weights: $0.125,\ 0.375,\ 0.375,\ 0.125$.

$$\begin{aligned} x&=0.125(0)+0.375(1)+0.375(4)+0.125(5)=0.375+1.5+0.625=2.5 \\ y&=0.125(0)+0.375(3)+0.375(4)+0.125(0)=1.125+1.5=2.625 \end{aligned}$$

At u = 0.25

Weights: $(0.75)^3=0.421875$; $3(0.25)(0.75)^2=0.421875$; $3(0.25)^2(0.75)=0.140625$; $(0.25)^3=0.015625$.

$$\begin{aligned} x&=0.421875(1)+0.140625(4)+0.015625(5)=0.421875+0.5625+0.078125=1.0625 \\ y&=0.421875(3)+0.140625(4)=1.265625+0.5625=1.828125 \end{aligned}$$

End tangents

$$\mathbf P'(0)=3(\mathbf B_1-\mathbf B_0)=3(1,3)=(3,\ 9),\qquad \mathbf P'(1)=3(\mathbf B_3-\mathbf B_2)=3(1,-4)=(3,\ -12)$$

De Casteljau check at u = 0.5

Repeated mid-points:

The result $(2.5,\ 2.625)$ equals the formula value.

Answer: P(0.25) = (1.0625, 1.828); P(0.5) = (2.5, 2.625); P'(0) = (3, 9); P'(1) = (3, -12).