What is the vertex of the quadratic function f(x) = (x-6)(x + 2)?

(2,-16)

Step-by-step explanation:

One way to find the vertex is as follows:

First, we see that the x-intercepts are 6 and -2.

(to see this, we use the fact that the x-intercepts are found by letting y i.e. f(x) = 0. Then we solve each of the factors equal to 0. so i.e.

x-6 = 0

x = 6.

Similarly, x+2 = 0, so x= -2).

Now that we know the x-intercepts, we can use the symmetry property of a quadratic to see the x-coordinate of the vertex occurs mid-way between -2 and 6. So i.e at (6+-2)/2 = 2.

Lastly, we put x = 2 into the equation to find the y-coordinate:

f(2) = (2-6)(2+2) = -4 x 4 = -16.

So the vertex is (2,-16)

just needed some points but try socratic

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8