If this were an inverse variation function, its equation would be of the form $\displaystyle{y}=\frac{{k}}{{x}}\usepackage{cmbright}\Huge\$ . Since $\displaystyle{y}={8}\usepackage{cmbright}\Huge\$ when $\displaystyle{x}={1}\usepackage{cmbright}\Huge\$ (according to the table), k would have to be 8. But if $\displaystyle{k}={8}\usepackage{cmbright}\Huge\$ , then $\displaystyle{y}=\frac{{8}}{{x}}\usepackage{cmbright}\Huge\$ which equals 4 when $\displaystyle{x}={2}\usepackage{cmbright}\Huge\$ . Since this doesn't match the table, this is not an inverse variation function.