Question 317553
Convert to vertex form, {{{y=a(x-h)^2+k}}} by completing ths square. 
(h,k) is then the vertex.
{{{f(x)=3(x^2-2x)+6}}}
{{{f(x)=3(x^2-2x+1)+6-3}}}
{{{f(x)=3(x-1)^2+3}}}
The vertex is ({{{1}}},{{{3}}}).
The axis of symmetry contains the vertex and is {{{x=1}}}.
The coefficient of {{{x^2}}} term is positive so the parabola opens upwards and the vertex value is the minimum value.
{{{fmin=3}}}
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{{{drawing(300,300,-3,6,-1,8,grid(1),circle(1,3,.3),blue(line(1,-10,1,28)),graph(300,300,-3,6,-1,8, 3(x-1)^2+3,3x^2-6x+6))}}}