# Find the vertex x^2-x-6=0

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### 2 Answers

Since the given equation represents the equation of a quadratic function `f(x) = ax^2 + bx + c` , you may find the vertex of parabola, that is the graph of quadratic function, using the following formulas:

`x = -b/(2a), y = (4ac - b^2)/(4a)`

You need to compare the given equation to the standard quadratic equation to identify a,b,c, such that:

`a = 1 , b = -1, c = -6`

Substituting the values of a,b,c in formulas yields:

`x = -(-1)/(2*1) => x = 1/2`

`y = (4*1*(-6) - (-1)^2)/(4*1) => y = -25/4 => y = -15 `

**Hence, evaluating the vertex of the parabola that is represented by the given quadratic function, yields **`x = 1/2, y = -25/4.`

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