$$y = \frac { 1 } { \sqrt { 7 - 6 x - x ^ { 2 } } }$$
$x=-\frac{\sqrt{16y^{2}-1}}{y}-3$
$x=\frac{\sqrt{16y^{2}-1}}{y}-3\text{, }y\geq \frac{1}{4}$
$y=\frac{1}{\sqrt{\left(1-x\right)\left(x+7\right)}}$
$x>-7\text{ and }x<1$
