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

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To find the vertex of x^2-x-6 = 0.

x^2-x-6 = 0 whose solutions are the intercepts of the the parabola y = x^2-x-6 .

So x^2-x-6 = (x-3)(x+2) = 0. So x= 3 and x= -2 are the intercepts of the parabola y = x^2-x-6 makes on x axis.

y = x^2-x-6 could be written as

y = x^2-x +(1/2)^2 -(1/2)^2-6

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

y +25/4 = (x-1/2)^2 is a parabola of the standard form (y-k) = (x-h)^2 whose vertex is (h,k).

Therefore (h,k) = (1/2 , -25/4) is the vertex of y = x^2-x-6.

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**