#### Answer

a)
$y=(x+1)^{2}$
$x \gt -1$
b)
As $t$ increases, $x$ also increases.
So the arrow should be pointing in the direction in which x increases.
There should be a hole at $(-1,0)$.

#### Work Step by Step

a)
Given that $x=e^{t}-1$
$x+1=e^{t}$
Also given that $y=e^{2t}$
we can write as $y=(e^{t})^{2}$
Use the first equation to replace $e^{t}$ with $x+1$
$y=(x+1)^{2}$
(After graphing)
Domain: $x \gt -1$
Range: $y \gt 0$
b)
we know that
$y=(x+1)^{2}$
where
$x \gt -1$
Given that $x=e^{t}-1$
As $t$ increases, $x$ also increases.
So the arrow should be pointing in the direction in which x increases.
There should be a hole at $(-1,0)$.