Question

Find an expression for the function whose graph is on the given curve. The bottom half of the parabola x+(y-3)^2=0

Answer 1

The bottom half of the parabola is going to occur when we subtract whatever square root, instead of adding it, so we have
\[\begin{align}
&x+(y-3)^2=0\\
&(y-3)^2=-x\\
&y-3=-\sqrt{-x}\\
&y=3-\sqrt{-x}
\end{align}\]
If you're doing this for someone fussy, you should probably restrict the domain to \((-\infty,0]\).

Answer 2

Thanks so much

Answer 3

My pleasure.