To find the extreme values of x^2+6x-6.

We write x^2+6x-6 as (x^2+6x+3^2) -3^2 -6 = 0. We added here 3^2 and subtracted.

Therefore x^2+6x-6 = (x+3)^2 -9-5

x^2+6x-6 = (x+3)^2 -15.

x^2+6x-6 > A perfect square (x+3)^2 -15.

Therefore x^2+6x -6 is always greater than -15, as (x+3)^2 > 0 for all x and is minimum when x+3 = 0 or x = -3.

Therefore x^2+6x-6 has the minimum (or extreme value) -15 when x = -3.

The extreme values of a function are the points where the value of the differential of the function is equal to zero. Here we have the function f(x) = x^2 + 6x - 6 .

Differentiating f(x) = x^2 + 6x - 6, we use the rule that the derivative of x^n is n*x^(n-1)

Therefore f'(x) = 2x + 6

Equating f'(x) to 0

=> 2x +6 = 0

=> x = -3

Now substituting x= -3 in f(x) = x^2 + 6x -6 = 9 - 18 -6 = -15

**Therefore the extreme value is -15.**

Since the expression of the function is a quadratic, we'll consider the vertex of the parable as the extreme point.

Since the coefficient of x^2 is positive, the vertex is a minimum point.

f(x) = x^2 + 6x - 6

The minimum value is V(-b/2a , -delta/4a)

We'll identify the coefficients a,b,c:

a = 1

b = 6

c = -6

delta = b^2 - 4ac

delta = 36 + 24

delta = 60

The vertex V has the following coordinates:

xV = -b/2a

xV = -6 / 2*1

**xV = -3**

yV = -60/4*1

**yV = -15**

**The minimum point of the function is V(-3 , -15).**

**Another method to determine the extreme point of a function, is the first derivative test.**

We'll calculate the first derivative of the function:

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

f'(x) = 2x + 6

Now, we'll compute the roots of f'(x):

2x + 6 = 0

We'll subtract 6 both sides:

2x = -6

x = -3

The roots of the first derivative represent the extreme points of the function.

We'll calculate the extreme point by substituting x by -3 in the original function:

f(-3) = (-3)^2 + 6(-3) - 6

f(-3) = 9 - 18 - 6

f(-3) = -15

**The extreme point has the coordinates: (-3 , -15).**