1. ## Parametrics

These are normally really easy for me and this one appears to be incredibly simple but I can't seem to get the correct answer.

Given $x = e^t - 6$ and $y = e^{2t}$

Eliminate the parameter to find a Cartesian equation of the curve.

Isolate t:

$x+6=e^t$

$t=\ln {(x+6)}$

Plug it into the y equation:

$y=e^{2\ln {(x+6)}}$ , $x> {-6}$

2. Error in one of your steps:

$e^{t} = x + 6$
$\ln e^{t} = \ln (x + 6)$
$t = \ln (x + 6)$

6 is inside the logarithm as well.

3. That was just a typo by me. It still isn't the right answer and I have no idea why.

4. Oooh. You just need to simplify a bit more. Remember your logarithm rules? You can bring the 2 up:

$y = e^{\ln(x+6)^{2}}$

And hopefully you'll see how the e and the ln disappears

Or you could've realized:
$y = e^{2t} = \left(e^{t}\right)^{2} = (x + 6)^{2}$

5. I don't know how I missed that.

6. Hello, topher0805!

Given: . $\begin{array}{cccc}x& = & e^t - 6 & {\color{blue}[1]}\\ y &= &e^{2t} & {\color{blue}[2]}\end{array}$

Eliminate the parameter to find a Cartesian equation of the curve.

From [1], we have: . $e^t \:=\:x+6\quad\Rightarrow\quad e^{2t} \:=\x+6)^2" alt="e^t \:=\:x+6\quad\Rightarrow\quad e^{2t} \:=\x+6)^2" />

Substitute into [2]: . $y \:=\x+6)^2" alt="y \:=\x+6)^2" />

This is an up-opening parabola with vertex (-6, 0).

However, note the domains of $x\text{ and }y\!:\;\;\begin{Bmatrix} x &\in &(\text{-}6,\,\infty) \\ y &\in &(0,\,\infty) \end{Bmatrix}$
Code:
                     |*
|
|
*
|
*|
* |
*   |
--------o------+-----
-6      |

The graph is only the "right half" of the parabola.