Locate the vertex of y=3x^2-x+3.

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aliddi's profile pic

aliddi | High School Teacher | (Level 1) Adjunct Educator

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Given the equation: y = 3x^2 - x + 3:

To find the vertex of a parabola:

1. Establish the values for a, b, and c in the given equation:

a = 3, b = -1, c = 3

2. Solve for the x-coordinate of the vertex by using the formula x = -b/2a and simplify:

x = -(-1)/2(3)

x = 1/6

3. Substitute the value for x into the original equation and simplify: y = 3(1/6)^2 - (1/6) + 3

y = 3(1/36) - (1/6) + 3     (square)

y = (3/36) - (1/6) + 3     (multiply)

y = (1/12) - (2/12) + (36/12)     (like terms)

y = 35/12     (add/subtract)

4.  Write as a coordinate (x, y):

(1/6, 35/12)

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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y= 3x^2 - x + 3

Where a = 3     b = -1     and c = 3

Then the vertix xoordinates are:

Vx = -b/2a = 1/2*3 = 1/6

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

    = -(1-4*3*3)/4*3

     = -(-35)/12 = 35/12

Then Vertix V is:

  V(1/6,  35/12)   which is located in the first quadrant.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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The given parabola is y = 3x^2-x+3

We rite this in the standard form  X^2 = 4aY, whose focus is at (X,Y) = (0,0) and focus is at (0,a).

y= 3x^2-x+3.

Divide by 3.

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

y/3 = x^2-x/3+1

y/3 = (x-1/6)^2 - (1/6)^2+1

y/3 = (x-1/6)^2 +35/36

y/3 -35/36= (x-1/6)^2.

{y-(35/12)}/3 = (x-1/6)^(1/2).

If we compare this to X^2 = 4aY, the vertex is at:

X = (x-1/6) = 0 and

Y = (y-(35/12)/3 = 0.

So the focus is  at (x,y) = (1/6 , 35/12)

 

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To find out where it is located the vertex of the parable y, we'll have to establish the quadrant where the coordinates of the vertex of the graph of y are located.

We know that the coordinates of the parabola vertex are:

 V(-b/2a;-delta/4a), where a,b,c are the coefficients of the  function and delta=b^2 -4*a*c.

y=f(x)=3x^2-  x + 3

We'll identify the coefficients:

a=3, b=-1, c=3, 2a=6, 4a=12

delta=(-1)^2 -4*3*3=1-36=-35

V(-b/2a;-delta/4a)=V(-(-1)/6;-(-35)/12)

V(-(-1)/6;-(-35)/12) = V(1/6 ; 35/12)

Because the coordinates are both positive, the vertex is located in the first quadrant: V(1/6 ; 35/12).

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