Question 38295
{{{y=3x^2-12x+16}}} original
{{{y-16=3x^2-12x}}} add constant
{{{(y/3)-(16/3)=x^2-4x}}} divide by three
{{{(y/3)-(16/3)+(12/3)=(x-2)^2}}} complete the square
{{{(y/3)-(4/3)=(x-2)^2}}} combine
{{{(y/3)=(x-2)^2+(4/3)}}} add
{{{y=3(x-2)^2+4}}} multiply by three
Use stantard form {{{y=a(x-h)^2+k}}} where the vertex is (h,k).
Vertex is (2,4).