Correct. For $\displaystyle{f{{\left({x}\right)}}}={a}{{x}}^{{2}}+{b}{x}+{c}\usepackage{cmbright}\Huge\$ , the vertex is: $\displaystyle{\left(-\frac{{b}}{{{2}{a}}},{f{{\left(-\frac{{b}}{{{2}{a}}}\right)}}}\right)}\usepackage{cmbright}\Huge\$

First find the x-coordinate by using the values $\displaystyle{a}=-{1}\usepackage{cmbright}\Huge\$ and $\displaystyle{b}={6}\usepackage{cmbright}\Huge\$ :

$\displaystyle-\frac{{b}}{{{2}{a}}}=-\frac{{6}}{{{2}{\left(-{1}\right)}}}=\frac{{-{6}}}{{-{{2}}}}={3}\usepackage{cmbright}\Huge\$

This tells you that the y-coordinate of the vertex is $\displaystyle{f{{\left({3}\right)}}}\usepackage{cmbright}\Huge\$ . Now:

$\displaystyle{f{{\left({3}\right)}}}=-{{3}}^{{2}}+{6}{\left({3}\right)}-{7}=-{9}+{18}-{7}={2}\usepackage{cmbright}\Huge\$