Adding 1 on the inside of the square root will move the graph one to the left of the graph of $\displaystyle\sqrt{{x}}\usepackage{cmbright}\Huge\$ , so the x-values will start at $\displaystyle-{1}\usepackage{cmbright}\Huge\$ . 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 positive sign in front of the radical will result in a graph that increases without bound. Therefore, the correct answer is: No, because it will begin at $\displaystyle{\left(-{1},{1}\right)}\usepackage{cmbright}\Huge\$ and increase without bound.