# 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 ( -b/ 2a , (4ac - b^2)/4a)

Here the parabola is y = 3x^2 - 12x

So, we have a = 3 , b = -12 and c = 0

Substituting these values in the coordinaates of the vertex, we get

-b/2a = 12 / 6 = 2

(4ac - b^2)/4a = 0 - 12^2 / 12 = -12.

**The required vertex is ( 2 , -12).**

We'll recall the coordinates of the vertex of the function, that is a parabola:

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