Because the low point is $\displaystyle{\left(\frac{\pi}{{10}},-{3}\right)}\usepackage{cmbright}\Huge\$ , the depth of the valley is 3. This is the amplitude of the function that was graphed. The amplitude of $\displaystyle{y}=-{3}{\sin{{5}}}{x}\usepackage{cmbright}\Huge\$ is also 3. The point $\displaystyle{\left(\frac{\pi}{{10}},-{3}\right)}\usepackage{cmbright}\Huge\$ satisfies $\displaystyle{y}=-{3}{\sin{{5}}}{x}\usepackage{cmbright}\Huge\$ .

The length of one cycle is $\displaystyle\frac{{{2}\pi}}{{5}}\usepackage{cmbright}\Huge\$ . This is equal to the period of the function that was graphed. The period of $\displaystyle{y}=-{3}{\sin{{5}}}{x}\usepackage{cmbright}\Huge\$ is also $\displaystyle\frac{{{2}\pi}}{{5}}\usepackage{cmbright}\Huge\$ . So this could be the function that was graphed.