Find the vertex of the given quadratic function.

f (x) = (x-8) (x-4)

A.

(4,8)

B.

(-4,8)

C.

(-6,4)

D.

(6,-4)

C. (6,-4)

Step-by-step explanation:

y = ax^2 + bx + c

1. expand (x-8) (x-4) using the FOIL rule or the box method or the distribution rule

(x-8) (x-4) = x (x-4) - 8 (x-4)

(x-8) (x-4) = x*x+x * (-4) - 8*x-8 * (-4)

(x-8) (x-4) = x^2-4x-8x+32

(x-8) (x-4) = x^2-12x+32

x^2-12x+32

x^2-12x+32 is the same as 1x^2 + (-12x) + 32 which is in the form ax^2+bx+c. We know that a = 1, b = - 12, c = 32

2. Use the values of a and b to find the value of h, which is the x coordinate of the vertex

h = - b / (2*a)

h = - (-12) / (2*1)

h = 12/2

h = 6

3. This is plugged back into the original function to find the y coordinate of the vertex. We can use either (x-8) (x-4) or x^2-12x+32 since they are equivalent expressions

k = y coordinate of vertex

k = f (h) = f (6) since h = 6

f (x) = (x-8) (x-4)

f (6) = (6-8) (6-4)

f (6) = (-2) (2)

f (6) = - 4

keep in mind/note that

f (x) = x^2-12x+32

f (6) = (6) ^2-12 (6) + 32

f (6) = 36-72+32

f (6) = - 36+32

f (6) = - 4

you get the same result using either expression

k = f (h) = f (6) = - 4

Because h = 6 and k = - 4, the vertex is (h, k) = (6,-4).

Choice B. The vertex is (6,-4)