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).

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1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.