To determine the coordinates of the vertex of the parable, we'll use the formula:

V (-b/2a ; -delta/4a)

We'll identify the coordinates a,b,c, of the expression of the function:

a = 3 , b = -4 , c = 2

Now, we'll calculate xV:

xV = 4/6

We'll divide by 2:

xV = 2/3

yV = -delta/4a

yV = (4ac-b^2)/4a

yV = (24-16)/12

yV = 8/12

We'll divide by 4:

yV = 2/3

The coordinates of the vertex of the parable are V(2/3 , 2/3) and they are located on the bisecting line of the first quadrant.** **

f(x) = 3x^2-4x+2.

Let y = 3x^2-4x+2

y/3 = x^2-4/3x+2/3

= (x-2/3)^2 -(2/3)^2 +2/3

=(x-2/3)^2 +(-4+6)/9

=(x-2/3)^2 +2/9

(y/3-2/9) = (x-2/3)^2

(1/3)(y-2/3) = (x-2/3)^2 which is of the standard parabola of the type: X^2= 4aY, with X= 0 an Y = 0as vertex

Comparing the two parabolas,

a = (1/3)/4 = 1/12.

x-2/3 = 0 and y-2/3 = 0 gives the coordinates of vertex .

(x,y) = (2/3,2/3) are the coordinates of the vertex.