Question 111223
In general, for {{{ax^2+bx+c}}}, vertex at (h, k), {{{h=-b/(2a)}}}, and {{{k=ah^2+bh+c}}}.
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So, for your particular problem, b = -1 and a = 3.  That makes {{{h=(-(-1))/(2*3)=1/6}}}, and
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{{{k=3(1/6)^2-(1/6)-1}}}
{{{k=3/36-1/6-1}}}
{{{k=3/36-6/36-36/36}}}
{{{k=-39/36}}}
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Putting the vertex squarely on the point (1/6,-39/36)
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Send me a note if there is any part of that you don't understand.
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John