Find max or min by completing the square, need help!?

y = -3x² + 4x + 2
y = -√3√3x² + 2 * √3x * ⅔√3 - (⅔√3)² + (⅔√3)² + 2
y = -(√3√3x² - 2 * √3x * ⅔√3 + (⅔√3)²) + (⅔√3)² + 2
y = -(√3√3x² + 2 * √3x * -⅔√3 + (⅔√3)²) + (⅔√3)² + 2
y = -(√3x - ⅔√3)² + (⅔√3)² + 2

For maximum value of y, the term -(√3x - ⅔√3)² should be maximum.

The maximum value of -(√3x - ⅔√3)² is 0, because it is a negative value.

√3x - ⅔√3 = 0
√3x = ⅔√3
x = 2/3
y = -3(2/3)² + 4(2/3) + 2
= -4/3 + 8/3 + 6/3
= 10/3

Complete the square means reduce it to a first order ie (x+a)^2 where a can be positive or negative.

so your equation right now cannot be easily factored. We must manipulate it to make it so.

First we need to get rid of the -3 in front of x^2, this will make your life so much easier when solving these problems, so we get:

Y/-3 = x^2 -4/3x -2/3

Now since (x+a)^2 = x^2 + 2ax + a^2 we require for our specific problem that:

2ax = -4/3x and therefore a = -4/6 or a = -2/3.

However a^2 = 4/9 and in our equation we have -2/3. So we must add something to both sides to get what we want.

to make -2/3 into 4/9 we must add 10/9 to each side as -2/3 = -6/9 and -6/9 +10/9 = 4/9, now our equation becomes

Y/-3 + 10/9 = x^2 -4/3x +10/9

our right hand side will always reduce to (x+a)^2

Y/-3 + 10/9 = (x-2/3)^2

simplifying we get:

Y = ((x-2/3)^2 - 10/9)/-3

Now we must maximize the numerator in order to maximize the function. Don't forget the negative sign in the denominator though. For right now we can forget about the 1/3.

So we get -((x-2/3)^2 - 10/9) which reduces to ((10/9-(x-2/3)^2), since (x-2/3)^2 cannot be negative, the lowest it can be is zero. This is so when x = 2/3 as (2/3-2/3)^2 = 0.

Now we simply plug x = 2/3 into our equation for Y to get the max of the function:

Y = (0 - 10/9)/-3 = 10/9/3 = 10/3.

Once you know how to compute derivatives the preceding can be done in two lines.

y = - 3x^2 + 4x + 2
y = -3 (x^2 - 4/3 x - 2/3)
complete the square: x^2 - 4/3 x + 4/9 is a square
y = -3 (x^2 - 4/3 x + 4/9 - 4/9 - 2/3)
y = -3 ( (x-2/3)^2 - 10/9)
y = -3 (x-2/3)^2 + 10/3

anything squared is zero or greater, so the max or min occurs when the squared term is zero.
this occurs at x = 2/3
y = -3(0)^2 + 10/3 = 10/3

for any other value of x:
(x-2/3)^2 > 0
-3 (x-2/3)^2 < 0
-3(0)^2 + 10/3 < 10/3
y < 10/3
therefore the point is a maximum