\begin{equation}\begin{array}{l}{51-56 \text { Find an expression for the function whose graph is the }} \\ {\text { given curve. }}\end{array}\end{equation}The bottom half of the parabola $x+(y-1{)}^2=0$

Key Concepts: Parabolas Explanation: The given equation represents a downward facing parabola with vertex at $(0,1)$ and axis of symmetry being the line $x=0$. Since the bottom half of the parabola is needed, only the negative $y$ values on or below the vertex need to be considered. Step by Step Solution: Step 1. Substitute $y-1$ by $u$ to get $u^2=-x$. Step 2. The graph of the given curve can be obtained by translating the graph of $y=u^2$ one unit to the right followed by $1$ unit downward. So, $f(x)=(u+1{)}^2-1=(\sqrt{-x}+1{)}^2-1=x+2\sqrt{-x}$. Therefore, an expression for the function whose graph is the given curve is $f(x)=x+2\sqrt{-x}$. Final Answer: f(x) = x + 2\sqrt{-x}