Subtracting 1 from inside the square root will move the graph one to the right of the graph of $\displaystyle\sqrt{{x}}\usepackage{cmbright}\Huge\$ , so the x-values will start at 1. However, because $\displaystyle{f{{\left({1}\right)}}}={1}-\sqrt{{{1}-{1}}}={1}-\sqrt{{0}}={1}\usepackage{cmbright}\Huge\$ , the graph will begin at $\displaystyle{\left({1},{1}\right)}\usepackage{cmbright}\Huge\$ . Also, the negative sign in front of the radical will result in a graph that decreases without bound. Therefore, the correct answer is: Yes, because it will begin at (1, 1) and decrease without bound.