So will give ! amy owns a graphic design store. she purchases a new printer to use in her store. the printer depreciates by a fixed rate of 14% per year. the function v = 2,400(1 – 0.14)t can be used to model the value of the printer in dollars after t years. question: part a: explain what the parameter 2,400 represents in the equation of the function. part b: what is the factor by which the printer depreciates each year? part c: amy also considered purchasing a printer that costs $4,000 and depreciates by 25% each year. which printer will have more value in 5 years? justify your answer by showing/explaining your work.

Given that the printer depreciates at the rate of 14% p.a. This has  been modeled by the function V=2400(1-0.14)^t, this follows an exponential form given by
y=a(b)^x
where:
V=y
a=2400
b=1-0.14
x=t
thus:
Part A: Explain what the parameter 2,400 represents in the equation of the function. 
The parameter 2400 represents the initial value of the printer at time t=0. This is the original value.

Part B: What is the factor by which the printer depreciates each year?
The factor of depreciation is 14% percent. This is the rate at which the printer depreciates and it accounts for the value of the printer at the end of every year.

Part C: Amy also considered purchasing a printer that costs $4,000 and depreciates by 25% each year. Which printer will have more value in 5 years? 
Value after 5 years of the $2400 printer that depreciates at 14% per year will be:
V(t)=2400(1-0.14)^5=$1,129.025

Value after 5 years of the $4000 printer that depreciates at 25% per year will be:
F(t)=4000(1-0.25)^t
F(5)=4000(1-0.25)^5=$949.22

The printer that costs $2400 will be more valuable compared to the printer that cost $4000

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

f(-2) = -3f(1) = 6 . . . estimated from f(x) = (x+2)^2-3, which matches points on the graph (-1,-2) and (0, 1)

so the change in y-value for function f is (6 )) = 9.

the table tells us

g(-2) = 1g(1) = 31

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).