Answer
The slope of the tangent line at $t=0$ is 4. This is equal to the slope of the segment $\overline{P_0 P_1}$.
Work Step by Step
Since
$P_0=\left(a_0,b_0\right)=(1,4)$,
$P_1=\left(a_1,b_1\right)=(3,12)$,
$P_2=\left(a_2,b_2\right)=(6,15)$,
$P_3=\left(a_3,b_3\right)=(7,4)$.
Substituting these values into Eq. (9) and Eq. (10) we obtain the parametric equations
$x(t)=1+6t+3t^2-3t^3$,
$y(t)=4+24t-15t^2-9t^3$.
Using Eq. (8) the slope of the tangent line is given by
$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{-27 t^2-30 t+24}{-9 t^2+6 t+6}$.
So,
$\frac{dy}{dx}|_{t=0 }=\frac{24}{6}=4$.
The slope of the segment $\overline{P_0 P_1}$ is $\frac{12-4}{3-1}=\frac{8}{2}=4$.
Thus, the slope of the tangent line at t=0 is equal to the slope of the segment $\overline{P_0 P_1}$.
