We have to find the extreme point of the curve y = 3x - 6x^2.

To do that we find the first derivative of 3x - 6x^2 and equate it to zero. This is solved for x.

Now y = 3x - 6x^2

y' = 3 - 12x

3 - 12x = 0

=> 12x = 3

=> x = 1/4

At x = 1/4, y = 3*(1/4) - 6*(1/4)^2

=> 3/4 - 6/16

=> 3/4 -3/8

=> 3/8

Also at x = 1/4, y'' is -12 which is negative. So we have the point of maxima at x = 1/4

The extreme point is at** x = 1/4** and this is the **maximum point **with the **expression equal to 3/8**.

To determine the extreme point of the curve, we'll have to calculate the critical point of the function, that is the root of the first derivative of f(x).

f'(x) = (3x-6x^2)'

f'(x) = 3 - 12x

We'll put f'(x) = 0.

3 - 12x = 0

We'll divide by 3:

1 - 4x = 0

We'll subtract 1:

-4x = -1

x = 1/4

The extreme value of the function is:

f(1/4) = 3/4 - 6/16

f(1/4) = (12-6)/16

f(1/4) = 6/16

We'll simplify and we'll get;

f(1/4) = 3/8

**The extreme point of the curve is the vertex of the parabola and it is the maximum point of the function: (1/4 ; 3/8).**