Hello, topher0805!
Given: .$\displaystyle \begin{array}{cccc}x& = & e^t  6 & {\color{blue}[1]}\\ y &= &e^{2t} & {\color{blue}[2]}\end{array}$
Eliminate the parameter to find a Cartesian equation of the curve.
From [1], we have: .$\displaystyle e^t \:=\:x+6\quad\Rightarrow\quad e^{2t} \:=\x+6)^2$
Substitute into [2]: .$\displaystyle y \:=\x+6)^2$
This is an upopening parabola with vertex (6, 0).
However, note the domains of $\displaystyle x\text{ and }y\!:\;\;\begin{Bmatrix} x &\in &(\text{}6,\,\infty) \\ y &\in &(0,\,\infty) \end{Bmatrix}$ Code:
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6 
The graph is only the "right half" of the parabola.