Baxter is thinking about buying a car. 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) 18,500, 17,390, 16,346.60
Car 2 (value in dollars) 18,500, 17,500, 16,500

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. (2 points)

Part B: Write one function for each car to describe the value of the car f(x), in dollars, after x years. (4 points)

Part C: Baxter wants to purchase a car that would have the greatest value in 10 years. Will there be any significant difference in the value of either car after 10 years? Explain your answer, and show the value of each car after 10 years.

Question

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Respuesta :

semsee45 semsee45 · Answer 1 · 2022-02-24T16:59:03+01:00

Answer:

A) car 1: exponential

car 2: linear

B) car 1: [tex]f(x)=18500(0.94)^{x-1}[/tex]

car 2: [tex]f(x)=-1000x+19500[/tex]

C) Yes, there will be a significant difference of $1,100.40 which is approx 10% of the value of the cars after 10 years.

Step-by-step explanation:

Part A

Car 1: exponential as the projected value is not decreasing by the same amount each year

Car 2: linear as the projected value decreases by the same amount each year ($1000)

Part B

Car 1

General form of exponential function: [tex]y=ab^x[/tex]

a = initial value = 18500

b = growth factor = [tex]\dfrac{17390}{18500}=0.94[/tex]

[tex]\implies f(x)=18500(0.94)^{x-1}[/tex]

NB We have to change x to "x-1" since we are told in the table that after 1 year the car's value is 18500.

Car 2

General form of linear function: [tex]y=mx+b[/tex]

[tex]m=\dfrac{\textsf{change in} \ y}{\textsf{change in} \ x}=\dfrac{17500-18500}{2-1}=-1000[/tex]

[tex]y-y_1=m(x-x_1)[/tex]

[tex]\implies y-17500=-1000(x-2)[/tex]

[tex]\implies f(x)=-1000x+19500[/tex]

Part C

Car 1 after 10 years: [tex]\implies f(10)=18500(0.94)^{10-1}=\$10,600.40[/tex]

Car 2 after 10 years: [tex]\implies f(10)=-1000(10)+19500=\$9,500[/tex]

Difference = 10600.40 - 9500 = 1100.40

Yes, there will be a significant difference of $1,100.40 which is approx 10% of the value of the cars after 10 years.