Determine the maximum of the function x^2-x-6=0.

To determine the maximum of the given function means to
find the vertex of a parabola. 

We'll write the function
as:

f(x) = a(x-h)^2 + k, where the vertex has the
coordinates v(h,k)

We'll re-write the given
function:

f(x) = 1(x^2 - 1x) - 6

We'll
complete the square:

x^2 -2*(1/2) x + (1/2)^2 = (x -
1/2)^2

So, we'll add and subtract the value
1/4:

f(x) = 1(x^2 - x + 1/4) - 1/4 -
6

We'll combine like terms outside the
brackets:

f(x) = (x - 1/2)^2 -
25/4

We'll compare the result with the standard form and
we'll get the coordinates of the vertex:

 (x - 1/2)^2 -
25/4 = a(x-h)^2 + k

h = 1/2

k
= -25/4

The coordinates of the vertex are:V
(1/2 ; -25/4)

Another way to determine the
local extreme of the function is to use the first derivative of the function (the vertex
is a local extreme).

f'(x) = 2x -
1

We'll determine the critical value of
x:

2x - 1 = 0

2x =
1

x = 1/2

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

f(1/2) = (1/2)^2 -
1/2 - 6

f(1/2) = 1/4 - 1/2 -
6

f(1/2) =
(1-2-24)/4

f(1/2) =
-25/4

We notice that the
vertex of the function is located in the 4th
quadrant.