If f(x) = x2 − 6x + 19, complete the square and determine the minimum or maximum value of the function.
a) f(x) = (x + 3)2 + 28 and f(x) has a maximum value f(3) = 28.
b) f(x) = (x + 3)2 + 28 and f(x) has a minimum value f(3) = 28.
c) f(x) = (x − 3)2 + 10 and f(x) has a maximum value f(3) = 10.
d) f(x) = (x − 3)2 + 10 and f(x) has a minimum value f(3) = 10. eliminate

D

Step-by-step explanation:

(Using the formula in my attachment)

x^2 − 6x + 19

(x - 3)^2 = x^2 - 6x + 9

9 - e = 19

e = -10

(x - 3)^2 - e

(x - 3)^2 - - 10

(x - 3)^2 + 10

when x - 3 = 0, x = 3

Subsitute 3 into the equation

(3 - 3)^2 + 10

0 + 10

Minimum (3,10) in this case f(3) = 10

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!