Uniform cubic B-spline segment points

Matrix form

For one uniform cubic segment ($0\le u\le1$):

$$\mathbf P(u)=\frac16\begin{bmatrix}u^3&u^2&u&1\end{bmatrix} \begin{bmatrix}-1&3&-3&1\\3&-6&3&0\\-3&0&3&0\\1&4&1&0\end{bmatrix} \begin{bmatrix}\mathbf P_0\\\mathbf P_1\\\mathbf P_2\\\mathbf P_3\end{bmatrix}$$

This is equivalent to the blending functions

$$\begin{aligned} N_0&=\tfrac16(1-u)^3 \\ N_1&=\tfrac16(3u^3-6u^2+4) \\ N_2&=\tfrac16(-3u^3+3u^2+3u+1) \\ N_3&=\tfrac16u^3 \end{aligned}$$

Weights

Points

At $u=0$:

$$x=\frac{1+4(2)+5}{6}=\frac{14}{6}=2.333,\qquad y=\frac{1+4(4)+5}{6}=\frac{22}{6}=3.667$$

At $u=0.5$:

$$\begin{aligned} x&=0.0208(1)+0.4792(2)+0.4792(5)+0.0208(7)=3.521 \\ y&=0.0208(1)+0.4792(4)+0.4792(5)+0.0208(2)=4.375 \end{aligned}$$

At $u=1$:

$$x=\frac{2+4(5)+7}{6}=4.833,\qquad y=\frac{4+4(5)+2}{6}=4.333$$

Tangents

$\mathbf P'(0)=\tfrac12(\mathbf P_2-\mathbf P_0)=\tfrac12(4,4)=(2,\ 2)$

$\mathbf P'(1)=\tfrac12(\mathbf P_3-\mathbf P_1)=\tfrac12(5,-2)=(2.5,\ -1)$

The curve does not touch the control points; it starts at $(2.333, 3.667)$, a weighted average of $\mathbf P_0,\mathbf P_1,\mathbf P_2$.

Answer: P(0) = (2.333, 3.667); P(0.5) = (3.521, 4.375); P(1) = (4.833, 4.333); tangents (2, 2) at u = 0 and (2.5, -1) at u = 1.