Assuming you mean f(x)=3x²-6x+4, then:
f(x)=3(x²-2x)+4,
f(x)=3(x²-2x+1-1)+4,
f(x)=3((x-1)²-1)+4,
f(x)=3(x-1)²-3+4=3(x-1)²+1,
f(x)-1=3(x-1)² is usually written:
y-1=3(x-1)², which has vertex (1,1) and this is the vertex form of f(x). So the idea is to complete the square so that you have (x-h)² as a component, where h is a constant to be found (in this case, h=1) which is the x-coordinate of the vertex. By rewriting the quadratic you then find another constant k (in this case k=1 also) which is the y-coordinate of the vertex.