Question 801038
In general...

{{{x[min]=-b/2a=-6/(2*1)=-3}}}
{{{y[min]=f(x[min])=(-3)^2+6(-3)+9=0}}}

v(-3,0)


but there is an easier way if we use the fact that it is a perfect square...

{{{y=x^2+6x+9}}}
{{{y=(x+3)^2 }}}

so that {{{h=-3}}}, {{{k=0}}}