What is vertex of x^2-x-6=0?

There are 3 ways, at least, 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 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

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

We'll
compare the result with the standard form:

 (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 is to use the first derivative of
the function, since 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