f(x)=3x^2 -2x +4
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3x^2 -2x +4
x vertex = -b\2a = 4\6 = 2\3
so y vertex is 3(2\3)^2 -2(2\3) + 4 = (4)
so the vertex is (2\3 , 4)
3( x^2 -2\3 x + (4\9) ) + 4 - 4\3
3(x-2\3) ^2 - 5\3
when x = 2\3 that is gonna makes the first part to be zero which is the minimum point for the equation and there is no maximum because it just have to have one since it is a second degree equation
axis of symmetry is x = 2\3
and thank you
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