Answer
$\left\{\begin{array}{l}
x=t,\\
y=t^{2}+2t
\end{array}\right.\qquad t\leq -1$
Work Step by Step
The parabola $y=ax^{2}+bx+c$
is symmetric to the line$\quad x=-\displaystyle \frac{b}{2a}=-1.$
So the left half of the parabola is obtained when $x \leq -1$
Defining a parameter $t=x$, we have
$\left\{\begin{array}{l}
x=t,\\
y=t^{2}+2t
\end{array}\right.\qquad t\leq -1$
as a possible parametrization.
