# Vertex of the function .Determine the vertex of the function f(x)=3x^2-12x

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For a parabola y = ax^2 + bx + c, the vertex is given by the coordinates: (-b/2a , -(b^2 - 4ac)/4a)

The equation of the parabola given is : y = 3x^2-12x

substituting the values, the vertex is

(12/6 , -(144 - 0)/12)

=> (2 , -12)

**The vertex of y = 3x^2-12x lies at (2 , -12)**

We know that the coordinates of a vertex of a parabola are:

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

a,b,c are the coefficients of the quadratic:

y = ax^2 + bx + c

Comparing, we'll identify the coefficients:

a = 3, b = -12 , c = 0

We'll calculate the x coordinate of the vertex:

xV = -(-12)/2*3

xV = 12/6

xV = 2

We'll calculate the y coordinate of the vertex:

yV = -delta/4a

delta = b^2 - 4ac

delta = 144 - 4*3*0

delta = 144

yV = -144/4*3

yV = -144/12

yV = -12

**The requested coordinates of the vertex of the parabola are: V(2 , -12).**