Eliminate the parameters to find a Cartesian equation of the curve and sketch the curve. $x = e^t – 1, y = e^{2t}$.

Eliminate the parameters to find a Cartesian equation of the curve and sketch the curve.

$x = e^t – 1, y = e^{2t}$.

My attempt:

$x = e^{t} - 1$

$x + 1 = e^{t}$

$\ln(x+1) = t$

so

$y = e^{2t} = e^{2\ln(x+1)} = (x+1)^2$ [fixed mistake]

I think I eliminated the parameters now how would I sketch this?

I made a table

\begin{array}{|c|c|c|c|} \hline t& 0 & 1 & 2 & 3 & 4 \\ \hline x & 0& e^{1}-1 & e^{2} - 1 & e^{3}-1 & e^{4}-1\\ \hline y & 1 & e^{2} & e^{4} & e^{6} & e^{8}\\ \hline \end{array}

If I graph above with respect to x and y, then would this be correct?

As you were told in the comments, there is an error in your computations and the answer is $y=e^{2\ln(x+1)}=\left(e^{\ln(x+1)}\right)^2=(x+1)^2$. But there is an even shorter path to that conclusion: since $x+1=e^t$, then $y=e^{2t}=(e^t)^2=(x+1)^2$.