Correct. If $\displaystyle{y}={\left({3}\right)}{{2}}^{{x}}\usepackage{cmbright}\Huge\$ , then $\displaystyle{y}={6}\usepackage{cmbright}\Huge\$ when $\displaystyle{x}={1}\usepackage{cmbright}\Huge\$ and $\displaystyle{y}={12}\usepackage{cmbright}\Huge\$ when $\displaystyle{x}={2}\usepackage{cmbright}\Huge\$ , so this function works.