Sketch the parametric curve by plotting points.

$x=2t-1,y=3t+5,t\in \mathrm{ℝ}$

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

The objective is to sketch the parametric curve.

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

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

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

$$\begin{gathered} (x,y)=(2t-1,3t+5) \\ (x,y)=(2·-2-1,3·-2+5) \\ (x,y)=(-4-1,-6+5) \\ (x,y)=(-5,-1) \end{gathered}$$

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

$$\begin{gathered} (x,y)=(2t-1,3t+5) \\ (x,y)=(2·-1-1,3·-1+5) \\ (x,y)=(-2-1,-3+5) \\ (x,y)=(-3,2) \end{gathered}$$

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

$$\begin{gathered} (x,y)=(2t-1,3t+5) \\ (x,y)=(2·0-1,3·0+5) \\ (x,y)=(-1,5) \end{gathered}$$

The point $(x,y)$When $t=\mathrm{I}$is,

$$\begin{gathered} (x,y)=(2t-1,3t+5) \\ (x,y)=(2·1-1,3·1+5) \\ (x,y)=(2-1,3+5) \\ (x,y)=(1,8) \end{gathered}$$

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

$$\begin{gathered} (x,y)=(2t-1,3t+5) \\ (x,y)=(2·2-1,3·2+5) \\ (x,y)=(4-1,6+5) \\ (x,y)=(3,11) \end{gathered}$$

The following is a tabular depiction of the point.

Therefore, the graph is