Definition of Bezier Curve
$\textbf{Bezier Curve}$ is used to draw smooth curve along points on a path A Bezier Curve goes through points called anchor points and the shape between anchor points is defined by so call control points $\textbf{Bezier Curve}$ is used to draw smooth curve along points on a path A Bezier Curve goes through points called anchor points and the shape between anchor points is defined by so call control points

$\textbf{Movement between two points using vectors}$ A point $\mathbf{Q}$ that moves from $\mathbf{P_1}$ to $\mathbf{P_2}$ over the time $0 \leq t \leq 1$ can be described using position vectors. Let $\mathbf{O}$ be some fixd point, then the $\textbf{position vector}$ $\mathbf{OQ}$ points at $\mathbf{Q}$. Let $\mathbf{P_1 P_2}$ be the vector between $\mathbf{P_1}$ and $\mathbf{P_2}$, then $\mathbf{OQ}$ is described by $\overrightarrow{OQ} = \overrightarrow{O P_1} + t\overrightarrow{P_1 P_2} \quad t \in [0, 1]$ to find the coordinates of $\overrightarrow{P_1 P_2}$, position vector are used,
$\overrightarrow{P_1 P_2} = \overrightarrow{O P_2} - \overrightarrow{O P_1}$ We get
\begin{align*} \overrightarrow{OQ} &= \overrightarrow{O P_1} + t\overrightarrow{P_1 P_2} \\ &= \overrightarrow{O P_1} + t (\overrightarrow{O P_2} - \overrightarrow{O P_1}) \\ &= (1-t)\overrightarrow{O P_1} + t \overrightarrow{O P_2}\\ \end{align*} the position vector of a point has the same coordinates as the point. For this reason, you don't need to distinguish between points and vectors when doing computer graphic. The point $\mathbf{Q}$ can be written as $\mathbf{Q} = (1-t) \mathbf{P_1} + t \mathbf{P_2} \quad t \in [0, 1]$ $\textbf{Parametric equation of Bezier Curve}$ $\textit{Linear Bezier Curve}$: Two points are needed. Both are anchor points, as $\mathbf{Q_0}$ moves along the line $\mathbf{P_0}$ and $\mathbf{P_1}$ it traces out a linear Bezier curve. Let t be a parameter, then the linear Bezier curve can be written as parameter curve $\mathbf{Q_0} = (1-t) \mathbf{P_0} + t \mathbf{P_1} \quad t \in [0, 1]$ $\textbf{Quadratic Bezier curve}$: Three points $P_0, P_1, P_2$ are needed. $P_0, P_2$ are anchor points, $P_1$ is a control point. The $\mathbf{Q}$-points move along lines between the $\mathbf{P}$-points. As $R_0$ moves along a line $Q_0$ and $Q_1$, it traces out a quadratic Bezier curve, the three movements are described by\\ \begin{equation} \begin{aligned} \mathbf{Q_0} &= (1-t) \mathbf{P_0} + t\mathbf{P_1} \\ \mathbf{Q_1} &= (1-t) \mathbf{P_1} + t\mathbf{P_2} \\ \mathbf{R_0} &= (1-t) \mathbf{Q_0} + t\mathbf{Q_1} \\ \mathbf{R_0} &= (1-t)^2 \mathbf{P_0} + 2(1-t)t \mathbf{P_1} + t^2 \mathbf{P_2} \quad t \in [0, 1] \nonumber \end{aligned} \end{equation} Just use $\mathbf{P}$ point to describe the movement $R_0$ we get that
$\mathbf{R_0} = (1-t)^2 \mathbf{P_0} + 2(1-t)t \mathbf{P_1} + t^2 \mathbf{P_2} \quad t \in [0, 1]$ $\textbf{Cubic Bezier Curve}$: for points $P_0$, $P_1$, $P_2$ $P_3$ are needed. $P_0$ and $P_3$ are anchor points, the other two are control points. Just use $\mathbf{P}$-points to describe $\mathbf{T_0}$ we get that \begin{equation} \begin{aligned} \mathbf{R_0} &= (1-t) \mathbf{Q_0} + t\mathbf{Q_1} \\ \mathbf{R_1} &= (1-t) \mathbf{Q_1} + t\mathbf{Q_2} \\ \mathbf{S_0} &= (1-t) \mathbf{R_0} + t\mathbf{R_1} \\ \mathbf{S_0} &= (1-t) ((1-t) \mathbf{Q_0} + t\mathbf{Q_1}) + t ((1-t) \mathbf{Q_1} + t\mathbf{Q_2}) \\ \mathbf{S_0} &= (1-t)^3 \mathbf{P_0} + 3(1-t)^2 t\mathbf{P_1} + 3(1-t) t^{2}\mathbf{P_3} + t^{3} \mathbf{P_4} \nonumber \end{aligned} \end{equation}

$\textbf{Quartic Bezier Curve}$: Five points $P_0$, $P_1$, $P_2$, $P_3$, $P_4$ are needed, $P_0$ and $P_4$ are anchor points, the others are control points. Just use $\mathbf{P}$-points to describe $\mathbf{T_0}$ we get that $\mathbf{T_0} = (1-t)^4 \mathbf{P_0} + 4(1-t)^3 t\mathbf{P_1} + 6(1-t)^2 t^{2}\mathbf{P_2} + 4(1-t) t^3 \mathbf{P_3} + t^4 \mathbf{P_4} \\$ $\textit{Binomial Expansion}$: $[t + (1-t)]^n = \sum_{k=0}^{n} \binom{n}{k} t^k (1-t)^{n-k} \\$ $\textit{Bezier Curve can be defined as following}$: \begin{equation} \begin{aligned} \textbf{C}(t) &= \sum_{k=0}^{n} \textbf{B}_{n,k}(t) \mathbf{P}_k \\ \textbf{B}_{n,k}(t) &= \binom{n}{k} t^k (1-t)^{n-k} \\ 1 &= \textbf{B}_{4,0}(t) + \textbf{B}_{4,1}(t) + \textbf{B}_{4,2}(t) + \textbf{B}_{4,3}(t) + \textbf{B}_{4,4}(t) \quad \text{where } n = 4 \quad \because [t+(1-t)]^4 = 1 \\ \end{aligned} \end{equation} $\textit{Plot } \textbf{B}_{4,0}(t) \,, \textbf{B}_{4,1}(t) \,, \textbf{B}_{4,2}(t) \,, \textbf{B}_{4,3}(t) \,, \textbf{B}_{4,4}(t)$ 