[tex]\displaystyle\\f(x)=3x^2-4x+2\\\\\text{Coordonatele varfului parabolei }~~V(x,~y)~~\text{sunt:}\\\\x=\frac{-b}{2a}=\frac{-(-4)}{2\cdot3}=\frac{4}{6}=\boxed{\bf\frac{2}{3}}\\\\y=\frac{-\Delta}{4a}=\frac{-(b^2-4ac)}{4a}=\frac{-(16-4\cdot3\cdot2)}{4\cdot3}= \frac{-(16-24)}{12}=\frac{8}{12}=\boxed{\bf\frac{2}{3}}\\\\\implies~~\boxed{\bf~~V\left(\frac{2}{3},~~\frac{2}{3}\right)}[/tex]
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