**Problem:** Identify the shape of the parametric curve

*x* = *e*^{t} − 1,

*y* = *e*^{2t} − 2*e*^{t}*t* real.

**Solution:** There is no point in using some theorem, because it would only
(with a bit of luck) guarantee a local existence, while we want a global and
concrete description. We could also try the standard procedure for drawing
parametric curves (see
Parametric functions in
Derivative - Theory - Graphing, or
Parametric functions in
Derivative - Methods Survey - Graphing), but that would only give us a
sketch, while the question says "identify". The only procedure that would
actually precisely identify the shape is changing the parametric description
either into a function, or at least to some implicit equation that we
recognize.

This is done by eliminating *t* from the description. Can we express
*t* from one of the two equations above? The first equation looks quite
inviting, so we can try it. We notice that we do not really need *t*
itself, since we will want to substitute for *t* into the formula for
*y*, but there this *t* is always as *e*^{t},
so it is enough to express this exponential from the first equation.

The equation
*y* = *x*^{2} − 1*x*-axis at *x* and *y*?

The expression
*x* = *e*^{t} − 1*t*
going through all real numbers has range
*e*^{t} > 0*x* > −1.

How about the other coordinate? The expression
*y* = *e*^{2t} − 2*e*^{t}*z* = *e*^{t}*y* as
*y* = *z*^{2} − 2*z*,*y* = *z*(*z* − 2).*z* > 0?

The conclusion is that the given parametric equations describe a piece of the
parabola
*y* = *x*^{2} − 1*x* > −1.