The general vertex form is

#color(white)("XXX")y=color(green)(m)(x-color(red)(a))^2+color(blue)(b)# with vertex at #(color(red)(a),color(blue)(b))#

Given

#color(white)("XXX")y=3x^2-12x+9#

Extract the #color(green)(m)# factor (see the general vertex form above)

#color(white)("XXX")y=color(green)(3)(x^2-4x)+9#

Since in its general form a squared binomial

#color(white)("XXX")(x-a)^2=x^2-2ax+a^2#

If #x^2-4x# are the first two terms of a squared binomial,

the third term needs to be #+4#

If we add a #+4# to complete the square inside the parentheses we will need to subtract #3xx(+4)# outside the parentheses to keep the same value.

#color(white)("XXX")y=color(green)(3)(x^2-4xcolor(brown)(+4))+9color(brown)(-)(color(green)(3))*(color(brown)(+4))#

Re-writing as a squared binomial and simplifying the constant term

#color(white)("XXX")y=color(green)(3)(x-color(red)(2))^2+color(blue)(""(-3))#

which is the vertex form for a parabola with vertex at #(color(red)(2),color(blue)(-3))#

Here's a graph of the original equation for verification purposes:

graph{3x^2-12x+9 [-1.637, 7.132, -4.24, 0.142]}