sketch the parametric curve by eliminating the parameter

$x=t+2,y=e^t,t\in \mathrm{ℝ}$

The parametric curve is$x=t+2,y=e^t,t\in \mathrm{ℝ}$

$x=t+2.$Consider the parametric equations $x=t+2,y=e^{\text{'}},t\in \mathrm{ℝ}$

The objective is to sketch the parametric curve by eliminating the parameter.

Take the parametric curve $x=t+2.$

Add $-2$on both sides of the equation.

$$\begin{gathered} x-2=t+2-2 \\ x-2=t+2-2 \end{gathered}$$

Thus.

$x-2=t$

Now substitute $t=x-2$in the parametric equation $y=e^{\text{'}}$.

Then,

$y=e^{x-2}[\text{since}t=x-2]$

Thus, the required equation after eliminating the parameter $t$is localid="1653375089492" $y=e^{r-2}$

To draw the graph of the equation assume width="122">x=-2,-1,0,1,2.

Substitute width="51">x=-2in the equation localid="1653375092667" $y=e^{x-2}.$

Then,

localid="1653375095753" $$\begin{gathered} y=e^{-2-2} \\ y=e^{-4} \\ y=0.018 \\ (x,y)=(-2,0.018) \end{gathered}$$

Substitute width="51">x=-1in the equation width="63">y=ex-2.

Then,

width="131">y=e-1-2y=e-3y=0.049(x,y)=(-1,0.049)

Substitute $x=0$in the equation $y=e^{x-2}$.

Then,

$$\begin{gathered} y=e^{0-2} \\ y=e^{-2} \\ y=0.13 \\ (x,y)=(0,0.13) \end{gathered}$$

Substitute $x=1$in the equation $y=e^{x-2}$Then,

$$\begin{gathered} y=e^{1-2} \\ y=e^{-1} \\ y=0.36 \\ (x,y)=(1,0.36) \end{gathered}$$

Substitute $x=2$in the equation$y=e^{x-2}$. Then,

$$\begin{gathered} y=e^{2-2} \\ y=e^0 \\ y=1 \\ (x,y)=(2,1) \end{gathered}$$

The graphical representation by using the points

is as follows,

$(-2,0.018)(-1,0.049)(0,0.13)(1,0.36)(2,1)$

Therefore, the required equation after eliminating the parameter is$y=e^{x-2}$