A minivan is purchased for $29,248. The value of the vehicle depreciates over time.

Describe the advantages and disadvantages of using a linear function to represent the depreciation of the car over time.
Describe the advantages and disadvantages of using an exponential function to represent the depreciation of the car over time.

The minivan depreciates $3,000 in the first year. Write either a linear or exponential function to represent the value of the car x years after it was sold.
Read another student’s post and comment on it by answering the following questions:

Do you agree or disagree with the advantages and disadvantages stated about the linear function and exponential function? Why do you agree or disagree?
Do you agree with the linear or exponential function created to represent the value of the car x years after it was sold? Why or why not?

Hey!

If a car is losing value at a constant rate, then a linear function would work better. For example, say that a car dealership is buying used cars. They will buy it for the asking price, but subtract 500 dollars for each year the car has been in existence. So, if you had a car from 2015 that cost you $2,000, and it's 2018, the dealership would buy it for 2000-(500 * the amount of years it has been since the car was made). Therefore, as 2018-2015=3, we have 2,000-500*3=2,000-1,500=500 as the value of the car after three years, and linear function (in the form of

, with y representing the end cost, b representing the starting cost, m representing the amount the car depreciates in value every year, and x being the amount of years) is 500=(-500*3+2000) in our situation here. Similarly, if the minivan depreciates 3,000 dollars in the first year, based on our knowledge and being forced to use linear function, we can assume that the minivan loses 3,000 dollars every year, making our equation y=(-3000*x)+29,248, with y representing the value of the car at the end, x representing the amount of years, and b representing the starting cost.

However, say that the car loses 50% of its value every year. This means that if you originally bought a car for 2,000 dollars in 2015, in 2016 it would be world 2,000/2=1,000 , 1,000/2=500 the year after that, and so on. We can try to put this into a linear function, but it's hard to put into one equation because we're not subtracting or adding to the cost of the car by a specific amount each year. If we had our equation as y=-0.5*x+2000, this wouldn't work because it's adding to the original amount, not multiplying. However, if we used an exponential equation such as y=b(m^a), with y representing the end cost, b representing the starting cost, m representing the amount multiplied per year, and a representing the amount of years. This works because we start with a value, and multiply it by an amount each year. Since 50%=0.5, we plug that into m to get y=2,000(0.5^a). Therefore, this works really well here. If the minivan were to depreciate by 3,000 every year, starting at $29,248, this means that we first have to find out what we multiply 29,248 by the first year to subtract 3,000. As, after one year, the value is 29,248-3,000=26,248, we have 26,248=29,248(m^1). Therefore, we can divide 29,248 by both sides to get around 0.9 as our answer for m. Thus, our equation is y=29,248(0.9^a). This type of equation would not work if we subtracted an amount every year because we're not multiplying by the same amount then. For example, if a toy was valued at 3$ and gained a value of one dollar every year, we could multiply the toy by 4/3 the first year to get 4, but the next year we wouldn't get 5.

Feel free to ask further questions! :)