jmc

First, complete square as follows: $$y=3x^2+6x+9=3\left(x^2+2x\right)+9.$$ To complete the square, we need to add ${\left(\frac{2}{2}\right)}^2=1$ after the $2x.$ So we have $$y+3=3\left(x^2+2x+1\right)+9.$$ This gives $$y=3{\left(x+1\right)}^2+6.$$ Now, since ${\left(x+1\right)}^2\ge 0,$ the minimum value is when the squared term is equal to $0.$ So the minimum value is $$y=3{\left(x+1\right)}^2+6=3\cdot 0+6=\boxed{6}.$$