whereas the tangent vector for end $P_{1}$ is defined by line joining $P_{1}$ and another point $P_{3}(6,0)$

Solution:

$P_{0}=\left[\begin{array}{ll}{1} & {3}\end{array}\right]$

$P_{1}=\left[\begin{array}{ll}{7} & {2}\end{array}\right]$

$P_{2}=\left[\begin{array}{ll}{10} & {8}\end{array}\right]$

$P_{3}=\left[\begin{array}{ll}{6} & {0}\end{array}\right]$

Tangent vector,

$P_{0}^{1}=\left[P_{2}-P_{0}\right]$

$P_{0}^{1}=[9 \quad 5]$

$P_{1}^{1}=\left[P_{3}-P_{1}\right]$

$P_{1}^{1}=[-1 \quad -2]$

$P(t)=P_{0}\left[2 t^{3}-3 t^{2}+1\right]+P_{1}\left[-2 t^{3}+3 t^{2}\right]+P_{0}^{1}\left[t^{3}-2 t^{2}+t\right]+P_{1}^{1}\left[t^{3}-t^{2}\right]$

For X -direction

$P_{x}=1\left[2 …