Question 3 (essay worth 40 points) (04.07 hc) autumn is thinking about buying a car for $38,000. the table below shows the projected value of two different cars for three years. number of years 1 2 3 car 1 (value in dollars) 32,000 26,000 20,000 car 2 (value in dollars) 32,300 27,455 23,336.75 part a: what type of function, linear or exponential, can be used to describe the value of each of the cars after a fixed number of years? explain your answer. part b: write one function for each car to describe the value of the car f(x), in dollars, after x years. part c: autumn wants to purchase a car that would have the greatest value in 5 years. will there be any significant difference in the value of either car after 5 years? explain your answer, and show the value of each car after 5 years.

Step-by-step explanation:

Consider the provided information

Part A)  

For Car 1: The dollars value is 32000, 26000, 20000, if we observe we can see that the common difference of 6000. Each year the value decreases by 6000.

Therefore, A linear function can be used to describe the value of each of the cars after a fixed number of years.

For Car 1: The dollars value is 32300, 27455, 23,336.75  which has a common ratio of  0.85 . The common ratio of  and if we compare the next term we get the same ratio .

Therefore, A exponential function can be used to describe the value of each of the cars after a fixed number of years.

Hence, linear function is suitable for Car 1 and exponential function is suitable  for Car 2.

Part B)

The linear function is given by

Where m is common difference, y is value after x year, x is number of years and c is the buying cost of car.

For Car 1, function decrease by 6000 each year that means m=-6000 and The buying cost of car is 38000, There for c = 380000. So the equation is .

The exponential function is given by

Where, y is the value after x years , x is the number of years , a is the value after 1 yr and r is the common ratio.

For Car 2 we have a=38000 and r= 0.85 , so the equation is .

The function for Car 1 is: and the function for car 2 is: .

Part C)

To find the highest value after 5 years substitute x = 5 in both the functions as shown:

Car 1:

Substitute x = 5 in above function.

After 5 yr Car 1 value is 8000.

For car 2:

Now, substitute x = 5 in above function.

After 5 yr Car 2 value is 16861 approximately.

Therefore, the value of Car 2 will higher after 5 years. The value of car 1 can reach to zero or below because of constant change but the value of Car 2 never gets to a zero.

wheres graph

step-by-step explanation:

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