What is the vertex of this quadratic function?

f ( x ) = ( x + 2 ) ( x - 6 ) 

Real solutions:

f = -15, x = -12

f = -15, x = 1

f = -8, x = -6

f = -8, x = 2

f = -5, x = -4

f = -5, x = 3

f = -3, x = -3

f = -3, x = 4

f = 0, x = -2

f = 0, x = 6

f = 7, x = -1

f = 7, x = 12

f(x) = (x + 2)(x - 6)

We need to put that into vertex form:

f(x) = a(x - h)² + k

which puts the vertex at point (h, k).

So first, I'll call it "y" instead of f(x) and then expand the right side:

y = (x + 2)(x - 6)

y = x² - 4x - 12

We need the right side to be in the form of (x² + bx) to prepare to complete the square so let's add 12 to both sides:

y + 12 = x² - 4x

Now we'll complete the square by adding 4 to both sides:

y + 16 = x² - 4x + 4

Factor:

y + 16 = (x - 2)²

Then solve for y again:

y = (x - 2)² - 16

back in function form:

f(x) = (x - 2)² - 16

The vertex is (2, -16)