First, this graph passes through the origin and has the hill and valley shape of a sine function. Second, because $\displaystyle{a}={2}\usepackage{cmbright}\Huge\$ in the equation, the amplitude is 2. Finally, because $\displaystyle{b}=\frac{{1}}{{2}}\usepackage{cmbright}\Huge\$ , the period of this function is $\displaystyle\frac{{{2}\pi}}{{\frac{{1}}{{2}}}}=\frac{{{2}\pi}}{{1}}\cdot\frac{{2}}{{1}}={4}\pi\usepackage{cmbright}\Huge\$ . This graph has one full cycle on the interval $\displaystyle{\left[{0},{4}\pi\right]}\usepackage{cmbright}\Huge\$ , so it has the correct period also.