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

What is vertex of the parabola whose equation is f(x) = x^2-x-6?

There are two ways to find the vertex of a parabola.  

1.  Completing the square to get it into the form:

f(x) = a(x-h)² + k and then the vertes is (h,k)

f(x) = 1x² - 1x - 6

Factor the coefficient of x² out of the first two terms:

f(x) = 1(x² - 1x) - 6

Take half the coefficient of x and square it

(1/2)×(1) = 1/2, then (1/2)² = 1/4

Add it then subtract it in the right side of the parentheses:

f(x) = 1(x² - 1x + 1/4 - 1/4) - 6

Change the parentheses to brackets:

f(x) = 1[x² - 1x + 1/4 - 1/4] - 6

Factor the trinomial formed by the first three terms inside the bracket

f(x) = 1[(x - 1/2)(x - 1/2) - 1/4] - 6

Write the factorization as the square of a binomial:

f(x) = 1[(x - 1/2)² - 1/4] - 6

Remove the brackets using the distributive law, leaving the 

parentheses intact

f(x) = 1(x - 1/2)² - 1/4 - 6

Combine the two number terms on the right

f(x) = 1(x - 1/2)² - 25/4

Compare to

f(x) = a(x - h)² + k, so

a=1, h=1/2, k=-25/4

So vertex = (h,k) = (1/2,-25/4)

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Second way:  Use the vertex formula 

Vertex formula for f(x) = ax² + bx + c:

Vertex is (h,k) where:

h = -b/(2a)

k = f(h)

f(x) = x² - x - 6

f(x) = 1x² - 1x - 6

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

h = -b/(2a) = -(-1)/(2·1) = 1/2

k = h² - h - 6 = (1/2)² - (1/2) - 6 = 1/4 - 1/2 - 6 = -25/4

So vertex = (h,k) = (1/2, -25,4)

Edwin