Hermite curve points and tangents

Formulae

$$\mathbf P(u)=F_1\mathbf P_0+F_2\mathbf P_1+F_3\mathbf P_0'+F_4\mathbf P_1'$$

with $F_1=2u^3-3u^2+1$, $F_2=-2u^3+3u^2$, $F_3=u^3-2u^2+u$, $F_4=u^3-u^2$. The tangent uses the derivatives $F_1'=6u^2-6u$, $F_2'=-6u^2+6u$, $F_3'=3u^2-4u+1$, $F_4'=3u^2-2u$.

Since $\mathbf P_0=(0,0)$:

$$\mathbf P(u)=F_2(4,2)+F_3(4,4)+F_4(4,-4)$$

At u = 0.5

$F_2=0.5$, $F_3=0.125-0.5+0.5=0.125$, $F_4=0.125-0.25=-0.125$.

$$\begin{aligned} x&=0.5(4)+0.125(4)-0.125(4)=2.0 \\ y&=0.5(2)+0.125(4)-0.125(-4)=1+0.5+0.5=2.0 \end{aligned}$$

At u = 0.25

$F_2=-0.03125+0.1875=0.15625$; $F_3=0.015625-0.125+0.25=0.140625$; $F_4=0.015625-0.0625=-0.046875$.

$x=0.15625(4)+0.140625(4)-0.046875(4)=0.625+0.5625-0.1875=1.000$

$y=0.15625(2)+0.140625(4)+0.046875(4)=0.3125+0.5625+0.1875=1.0625$

At u = 0.75

$F_2=-0.84375+1.6875=0.84375$; $F_3=0.421875-1.125+0.75=0.046875$; $F_4=0.421875-0.5625=-0.140625$.

$x=0.84375(4)+0.046875(4)-0.140625(4)=3.000$

$y=0.84375(2)+0.046875(4)+0.140625(4)=1.6875+0.1875+0.5625=2.4375$

Tangent vectors

Using $\mathbf P'(u)=F_2'(4,2)+F_3'(4,4)+F_4'(4,-4)$:

Example at $u=0.5$: $F_2'=1.5$, $F_3'=-0.25$, $F_4'=-0.25$, so $P'_y=1.5(2)+(-0.25)(4)+(-0.25)(-4)=3-1+1=3$ and $P'_x=4(1.5-0.25-0.25)=4$.

End conditions

At $u=0$: $F_1=1$, others 0, so $\mathbf P=(0,0)$ and $\mathbf P'=(4,4)$. At $u=1$: $F_2=1$, so $\mathbf P=(4,2)$ and $\mathbf P'=(4,-4)$. Both are satisfied.

Answer: P(0.25) = (1.00, 1.0625), P(0.5) = (2.00, 2.00), P(0.75) = (3.00, 2.4375); tangents (4, 4.25), (4, 3), (4, 0.25).