sketch the parametric curve by plotting points. $x=3+1,y=2t,t\in \lfloor 0,s\rfloor$

The parametric curves at $x=3+1,y=2t,t\in \lfloor 0,s\rfloor$.

The objective is to sketch the parametric curve.

To draw the graph for the parametric equations, assume $t=0,2,4,6,8$.

Substitute different $t$values in the parametric equations and find the values of localid="1653374216385" $x,y$

The point localid="1653374219177" $(x,y)$When localid="1653374221862" $t=0$is,

localid="1653374224361" $$\begin{gathered} (x,y)=(3t+1,2t) \\ (x,y)=(3·0+1,2·0) \\ (x,y)=(1,0) \end{gathered}$$

The point localid="1653374229338" $(x,y)$When localid="1653374227049" $t=2$is,

localid="1653374232057" $$\begin{gathered} (x,y)=(3+1,2t) \\ (x,y)=(3·2+1,2·2) \\ (x,y)=(6+1,4) \\ (x,y)=(7,4) \end{gathered}$$

The point $(x,y)$When localid="1653374235752" $t=4$is,

localid="1653374239359" $$\begin{gathered} (x,y)=(3+1,2t) \\ (x,y)=\left(3\right(4)+1,2·4) \\ (x,y)=(12+1,8) \\ (x,y)=(13,8) \end{gathered}$$

The point localid="1653374253432" $(x,y)$When localid="1653374257560" $t=6$is,

localid="1653374261724" $$\begin{gathered} (x,y)=\left(3\right(6)+1,2-6) \\ (x,y)=(18+1,12) \\ (x,y)=(19,12) \end{gathered}$$

The point $\left(x_{v}y\right)$When $t=8$is,

$$\begin{gathered} (x,y)=(3t+1,2t) \\ (x,y)=\left(3\right(8)+1,2·8) \\ (x,y)=(24+1,16) \\ (x,y)=(25,16) \end{gathered}$$

The tabular representation of the points is as follows.