Calculate the extreme values of x^2 + 6x - 6 .

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).