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.
y = 3x^2-4x+2.
to find the location of the vertex.
y = 3x^2-4x+2 .
y/3 = x^2 -(4/3)x+2
y/3 = (x-2/3)^2 +2-(2/3)^2
y/3 = (x-2/3)^2 + (18-4)/9
y/3 - 14/9 = (x-2/3)^2
[y -(14/3)]/ 3 = (x-2/3)^2 .... (1).
This is parabbola of the standard form X^2 = 4aY whose vertex is (X, Y) = (0 , 0).
So the parabola at (1) has the vertex at (x,y) = (2/3 14/3) at the 1st quadrant.