Where is the location of the vertex of y = 3x^2 - 4x + 2 ?

To determine the location of the vertex, we'll have to
calculate the coordinates of the vertex  of  the
parable.

We'll apply the
formula:

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

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

a = 3 , b = -4 , c =
2

Now, we'll calculate the
coordinate xV:

xV = 4/6

We'll
divide by 2:

xV =
2/3

We'll calculate the coordinate
yV:

yV = -delta/4a

delta = b^2
- 4ac

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

yV =
(24-16)/12

yV = 8/12

We'll
divide by 4:

yV =
2/3

Since both coordinates are positive, the
vertex V(2/3 , 2/3) is located in the first
quadrant.

Furthermore, the coordinates of the vertex are
equal, V(2/3 , 2/3), so they are located on the bisecting line of the first
quadrant.