Quadratic in vertex form

We have
\[f(x)=y= -\frac{1}{2} (x-3)^2 -3   \Rightarrow   -\frac{1}{2} (x-3)^2=y+3\]
\[\Rightarrow (x-3)^2=-2(y+3)\]
\[\Rightarrow x-3=\pm\sqrt{-2(y+3)}\]
\[\Rightarrow x=3\pm\sqrt{-2(y+3)}.\]

Note that thif function is not invertable in general, but if we restrict the function $f(x)$ to $x \geq 3$ or $x\leq 3$, then it will be invertable, and the inverse will be obtained by switching the roles of $x$ and $y$ in the above equation:
\[ y=3+\sqrt{-2(x+3)}    \text{or}    y=3-\sqrt{-2(x+3)},\]
respectively.