There is a significant value in the difference in the value of the car after nine years
The table of values is given as:
Years 1 2 3
Car 1 17,390 16,346.60 15,365.80
Car 2 17,500 16,500 15,500
The values of car 1 have a common ratio of 0.94; this is calculated by dividing the terms of the function
The common ratio implies that the function of car 1 is a geometric function.
The values of car 2 have a common difference of -1000.
The common difference implies that the function of car 2 is a linear function.
Car 1
In (a), we have:
Common ratio, r = 0.94
The initial value is given as:
a = 18,500
So, the function of car 1 is
f(x) = 18500 * 0.94^x
Car 2
In (a), we have:
Common difference, d = -1000
The initial value is given as:
a = 18,500
So, the function of car 2 is
f(x) = 18500 - 1000x
This means that
x = 9
So, we have:
Car 1
f(9) = 18500 * 0.94^9
f(9) = 10600
Car 2
f(9) = 18500 - 1000 *9
f(9) = 9500
10600 is greater than 9500
Hence, there is a significant value in the difference in the value of the car after nine years
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