Correct. $\displaystyle{y}={3}{{x}}^{{2}}-{6}{x}={3}{\left({{x}}^{{2}}-{2}{x}\right)}\usepackage{cmbright}\Huge\$ . Completing the square, you found $\displaystyle{y}+{3}{\left({1}\right)}={3}{\left({{x}}^{{2}}-{2}{x}+{1}\right)}\usepackage{cmbright}\Huge\$ . This can be factored as $\displaystyle{y}={3}{{\left({x}-{1}\right)}}^{{2}}-{3}\usepackage{cmbright}\Huge\$ . Note that the coordinates for the vertex would be (1, -3).