Sketch the parametric curve by plotting points.

$x=3t+1,y=t,t\in [-2,5]$

$x=3t+1,y=t,t\in [-2,5]$

A collection of quantities is defined as a function of one or more independent variables called parameters in a parametric equation.

Consider the parametric curves $x=1-2t,y=5-3t,t\in \mathrm{ℝ}$

The goal is to draw the parametric curve.

To draw the graph for the parametric equations assume $t=-2,-1,0,1,2$

Find the values of $x,y$by substituting different $t$values in the parametric equations.

The point $(x,y)$When $t=-2$is,

$$\begin{gathered} (x,y)=(1-2t,5-3t) \\ (x,y)=(1-2·-2,5-3·-2)[\text{since by substituting}t=-2] \\ (x,y)=(1+4,5+6) \\ (x,y)=(5,11)\text{simplify} \end{gathered}$$

The point $(x,y)$When $t=-1$is,

$$\begin{gathered} (x,y)=(1-2t,5-3t) \\ (x,y)=(1-2·-1,5-3·-1) \\ (x,y)=(1+2,5+3)[\text{since by substituting}t=-1] \\ (x,y)=(3,8)\text{simplify} \end{gathered}$$

The point $(x,y)$When $t=0$is,

$$\begin{gathered} (x,y)=(1-2t,5-3t) \\ (x,y)=(1-2·0,5-3·0)\text{[since by substituting}t=0\left] \\ \right(x,y)=(1,5)\text{simplify} \end{gathered}$$

The point $(x,y)$wither $t=1$is,

$$\begin{gathered} (x,y)=(1-2t,5-3t) \\ (x,y)=(1-2·1,5-3·1)[\text{since by substituting}t=1] \\ (x,y)=(1-2,5-3) \\ (x,y)=(-1,2)\text{simplify} \end{gathered}$$

The point $(x,y)$ When $t=2$ is,

The tabular representation of the points is as follows,

The graphical representation is shown below,

Therefore, the solution is a required graph.