Find the vertex of the given quadratic function. f(x)=(x-8)(x-4) a.) (4,8) b.) (6,-4) c.) (-6,4) d.) (-4,8) hlp !

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

Work Shown:

Step 1 is to expand out (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

So (x-8)(x-4) turns into 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 see that a = 1, b = -12, c = 32

Use the values of a & 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

Then 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

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

So we get the same result using either expression

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

Since h = 6 and k = -4, the vertex is (h,k) = (6,-4). So that's why the answer is choice B.

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