We have the curve y=x^2 - 2x + ln e^2. Now we need to find the extreme point of the curve. For this, we first find the first derivative of the curve.

y = x^2 - 2x + ln e^2

=> y = x^2 - 2x + 2

y' = 2x - 2

Equate this to 0

=> 2x - 2 = 0

=> 2x = 2

=> x = 1

At x = 1

y = 1^2 - 2*1 + 2

=> y = 1 - 2 + 2

=> y = 1

Also y'' = 2 which is positive at x = 1, therefore this is the point with the minimum value.

Here the curve has the minimum value.** **

**We derive the extreme point of y = x^2 - 2x + ln e^2 as ( 1 , 1).**

To find the extreme value of y = x^2 - 2x + ln e^2.

Solution:

y(x) = x^2 -2x + 2, as lne^2 = 2.

y(x) = (x-1)^2 -1+2.

y(x) = (x-1)^2+1.

Since (x-1)^2 is a perfect square, (x-1)^2 > = 0. y(x) > = y(1) = 1. So y(x is least when x= 1.

y(x) = (x-1)^2-1 +lne^2 is > = 1 for all x.

**Therefore y(x) = x^2-2x+ln e^2 = 1 is the minimum of y (x), when x= 1.**

If we want to determine the local extreme of the behavior curve, we'll use the first derivative of the function (the vertex is a local extreme).

f'(x) = (x^2-2x+ln e^2)'

f'(x) = 2x - 2 + 0

f'(x) = 2x - 2

We'll determine the critical value of x, that is the root of the first derivative.

2x - 2 = 0

We'll add 2 both sides:

2x = 2

We'll divide by 2:

x = 2/2

x = 1

Now, we'll calculate the y coordinate of the local extreme:

f(1) = (1)^2 - 2 + 2ln e, but ln e = 1

f(1) = 1 - 2 + 2

We'll eliminate like terms:

**f(1) = 1**

**The coordinates of the local extreme of the behavior curve are (1 ; 1).**

**We remark that the coordinates are located on the bisectrix** **of the 1st quadrant.**