Suppose a Bezier curve $P(t)$ is defined by the following four control points in the $xy$-plane:

$P_0=(-2,0),\ P_1=(-2,4),\ P_2=(2,4),\ P_3=(2,0)$

Then which of the following statements are correct?

A. Bezier curve $P(t)$ has degree $3$.

B. $P\left(\frac{1}{2}\right)=(0,3)$.

C. Bezier curve $P(t)$ may extend outside the convex hull of its control points.

Since there are $4$ control points, the Bezier curve is cubic.

So, degree $=4-1=3$. Hence, statement A is true.

For cubic Bezier curve,

$P(t)=(1-t)^3P_0+3t(1-t)^2P_1+3t^2(1-t)P_2+t^3P_3$

At $t=\frac{1}{2}$,

$P\left(\frac{1}{2}\right)=\frac{1}{8}P_0+\frac{3}{8}P_1+\frac{3}{8}P_2+\frac{1}{8}P_3$

$x=\frac{1}{8}(-2)+\frac{3}{8}(-2)+\frac{3}{8}(2)+\frac{1}{8}(2)=0$

$y=\frac{1}{8}(0)+\frac{3}{8}(4)+\frac{3}{8}(4)+\frac{1}{8}(0)=3$

So, $P\left(\frac{1}{2}\right)=(0,3)$. Hence, statement B is true.

Bezier curve always lies inside the convex hull of its control points.

So, statement C is false.