Question 370262
Complete the square to put the equation into vertex form,{{{y=a(x-h)^2+k}}}
{{{f(x)=x^2-12x-3}}}
{{{f(x)=(x^2-12x+36)-3-36}}}
{{{f(x)=(x-2)^2-39}}}
The vertex is ({{{2}}},{{{-39}}}).
The vertex lies on the axis of symmetry {{{x=2}}}
Since {{{a=1>0}}} then the parabola opens upwards and the value at the vertex is the function minimum.
{{{y[min]=-39}}}