Thank you for the opportunity to help you with your question!

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

Please let me know if you need any clarification. I'm always happy to answer your questions.