sketch the parametric curve by plotting points.

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

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

Consider the parametric curve $x=3t+1,y=t$at $t\in [-2,5]$.

The objective is to sketch the parametric curve.

To draw the graph for the parametric equations assume $t=-2,0,2,3,5$.

Substitute different t values in the parametric equations and find the values of x, y.

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

$(x,y)=(3t+1,t)$

$$\begin{gathered} (x,y)=\left(3\right(-2)+1,-2)[\sin cebysubstitutingt=-2] \\ (x,y)=(-6+1,-2) \\ (x,y)=(-5,-2)\{simplify\} \end{gathered}$$

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

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

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

$$\begin{gathered} (x,y)=(3t+1,t) \\ (x,y)=\left(3\right(2)+1,2)[\text{since by substituting}t=2] \\ (x,y)=(6+1,2) \\ (x,y)=(7,2)\text{simplify} \end{gathered}$$

The point$(x,y)$When $t=3$is,

$$\begin{gathered} (x,y)=(3t+1,t) \\ (x,y)=\left(3\right(3)+1,3)[\text{since by substituting}t=3] \\ (x,y)=(9+1,3)(x,y)=(10,3)\text{simplify} \end{gathered}$$

The point $(x,y)$When $t=5$is,

$$\begin{gathered} (x,y)=(3t+1,t) \\ (x,y)=\left(3\right(5)+1,5)[\text{since by substituting}t=5] \\ (x,y)=(15+1,5) \\ (x,y)=(16,5)\text{simplify} \end{gathered}$$

The tabular representation is as follows:

The graphical representation is