Question

Baxter is thinking about buying a car for $18,500. 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) 17,390 16,346.60 15,365.80
Car 2 (value in dollars) 17,500 16,500 15,500

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 9 years. Will there be any significant difference in the value of either car after 9 years? Explain your answer, and show the value of each car after 9 years. (4 points)

Answer 1

@Vocaloid PLease help!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 2

So then A is linear right

Answer 3

hm. so B is clearly linear but I just double checked A and A isn't linear

Answer 4

since the differences in A aren't fixed

Answer 5

So the answer to part a is not linear

Answer 6

Car 1 is nonlinear (exponential) and Car 2 is linear

Answer 7

So then it is exponential

Answer 8

make sure to specify which car is which.

Answer 9

Okay whats next

Answer 10

car 2 is easier to write an equation for

it's: value = (rate of change)x + initial value

Answer 11

replace "rate of change" with the amount it decreases each year, and "intial value" with 18500

Answer 12

We on part b right?

Answer 13

yes

Answer 14

okay

Answer 15

I read what you typed but im confused again.

Answer 16

how much does car 2 decrease per year?

Answer 17

1,000

Answer 18

good

take

f(x) = (rate of change)x + initial value

replace "rate of change" with -1000 and "initial value" with 18500

don't do anything else, let me know what you get

Answer 19

Done!

Answer 20

what did you get?

Answer 21

f(x) = (-1000)x + 18500

Answer 22

good, that's car 2

car 1 is a little harder so bear with me

Answer 23

Okay awesome! I will

Answer 24

if we take the year 1/year 0 value, year 2/year 1 value, year 3/year 2 value, we can see that the car value decreases by 6% each year

Answer 25

in other words:

value = (value of the previous year)*(0.94)

Answer 26

so we can express this as:

f(x) = initial value(0.94)^x

Answer 27

Wait what is the initial value

Answer 28

same as before

Answer 29

18500

Answer 30

okay

Answer 31

f(x) = 18500(0.94)^x
is your eq. for car 2

and that's it for b

Answer 32

OKyy

Answer 33

for part c, take both functions and calculate f(x) when x = 9

Answer 34

car 1 f(x) = (-1000)x + 18500

car 2 f(x) = 18500(0.94)^x

let x = 9 and find car 1 and car 2

Answer 35

So replace x with 9?

Answer 36

yes

Answer 37

Okay hold on

Answer 38

What is the site you used for math???

Answer 39

I mean, I used wolframalpha but any online calculator could do fine

Answer 40

I got what i needed lol. Hold on im almost done

Answer 41

For car 1 is it f(9) = 10600.40384122

Answer 42

good, round that to the nearest hundreth since we're talking about money

Answer 43

so $10600.40 for car 1

what about car 2?

Answer 44

Okay hold on

Answer 45

f(9) = 9500

Answer 46

good

so to answer the question, car 1 is worth more than car 2 by how much?

Answer 47

1100.4

Answer 48

good, so there ~is~ a significant difference and Baxter should buy car 1

that's it for c

Answer 49

Okay

Answer 50

Thank you!!