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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

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