$$y= { \left(x+1 \right) }^{ 2 } +1$$
$y=\left(x+1\right)^{2}+1$
$x=-\left(\sqrt{y-1}+1\right)$
$x=\sqrt{y-1}-1$
$x=-\left(\sqrt{y-1}+1\right)$
$x=\sqrt{y-1}-1\text{, }y\geq 1$
