Question

Cars in high demand that are also in low supply tend to retain their value better than other cars. The data in the table is for a car that won a resale value award.

Year : value(%)
1 : 84
3 : 64
5 : 44
a. Write a function to represent the change in the percentage of the car's value over time. Assume that the function is linear for the first 5 years.
b. According to the model, by what percent did the car's value drop the day it was purchased and driven off the lot?
c. Would the linear model be useful after 10 years? Explain why or why not.
d. Suppose months were used instead of years to write the function. How would the model change? What's the relationship of the new function to the
original function?

Answer 1

a. To represent the change in the percentage of the car's value over time, we can use a linear function. The equation that represents the change in the percentage of the car's value over time is y - 84 = -10(x - 1). b. The percent drop in value the day the car was purchased and driven off the lot is 94%. c. The linear model may not be useful after 10 years because the rate of depreciation may vary and other factors can affect the car's value.

Step-by-step explanation:

a. To represent the change in the percentage of the car's value over time, we can use a linear function. Since the function is linear for the first 5 years, we can use the two-point form of a linear function. Let's take the data points (1, 84) and (5, 44). Using these points, we can find the equation of the line:

y - y1 = m(x - x1)

where x1 = 1, y1 = 84, x = 5, y = 44. Solving for m, we get:

m = (y - y1) / (x - x1)

Substituting the values, we have:

m = (44 - 84) / (5 - 1) = -10

Substituting the values of x1, y1, and m into the equation, we get:

y - 84 = -10(x - 1)

This is the equation that represents the change in the percentage of the car's value over time.

b. To find the percent drop in value the day the car was purchased and driven off the lot, we need to find the value at time 0. Using the equation y - 84 = -10(x - 1), we substitute x = 0 and solve for y:

y - 84 = -10(0 - 1)

y - 84 = 10

y = 94

The percent drop in value is 94%.

c. The linear model may not be useful after 10 years because it assumes that the rate of change in the car's value is constant over time. In reality, the rate of depreciation may vary after a certain point. Other factors such as wear and tear, maintenance, and market conditions may also affect the car's value. Therefore, a linear model may not accurately represent the change in value after 10 years.

d. If months were used instead of years, the model would need to be adjusted to reflect the change in time unit. The new function would have a different slope based on the rate of change per month instead of per year. The relationship between the new function and the original function would still be linear, but with a different rate of change.