You probably made a mistake with the calculation of the period. The amplitude of this graph and $\displaystyle{y}=\frac{{3}}{{2}}{\sin{{\left(\frac{{1}}{{2}}{x}\right)}}}\usepackage{cmbright}\Huge\$ is $\displaystyle\frac{{3}}{{2}}\usepackage{cmbright}\Huge\$ . However, this graph has two cycles in the interval $\displaystyle{\left[{0},{2}\pi\right]}\usepackage{cmbright}\Huge\$ , so a period of $\displaystyle\pi\usepackage{cmbright}\Huge\$ . But $\displaystyle{y}=\frac{{3}}{{2}}{\sin{{\left(\frac{{1}}{{2}}{x}\right)}}}\usepackage{cmbright}\Huge\$ has a period of $\displaystyle{4}\pi\usepackage{cmbright}\Huge\$ . The correct answer is the graph with one hill over the interval $\displaystyle{\left[{0},{2}\pi\right]}\usepackage{cmbright}\Huge\$ .