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

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### 4 Answers

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)

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.

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)

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).