Problem 91 - Contemporary Precalculus

Question

A Honda Civic LX sedan is worth \(\$ 15,350 now and will be worth \)\$ 9910\( in four years. (a) Assuming linear depreciation, find the cquation that gives the value \)y\( of the car in year \)x$ (b) At what rate is the car depreciating? (c) Estimate the value of the car six years from now.

Step-by-Step Solution

Answer

Answer: The depreciation rate of the car is $-1360 per year, and its estimated value after six years is $7190.

Step 1: Finding the slope (depreciation rate)

The slope of the line is equal to the change in the value of the car divided by the change in time. Using the points (0, 15350) and (4, 9910), we can find the slope: m = (y2 - y1) / (x2 - x1) m = (9910 - 15350) / (4 - 0) m = -1360 The slope of the line is -1360, which is also the depreciation rate.

Step 2: Find the y-intercept (initial value of the car)

To find the y-intercept, we can use the point-slope form of the equation and plug in one of the given points (0, 15350): y - 15350 = -1360 * (x - 0) y = -1360 * x + 15350

Step 3: Write the equation for the value of the car in year x

Now that we have the slope and y-intercept, we can write the equation for the value of the car in year x: y = -1360x + 15350

Step 4: Find the depreciation rate per year

As we've calculated in Step 1, the depreciation rate is $-1360 per year.

Step 5: Estimate the value of the car six years from now

Using the equation from Step 3, we can estimate the value of the car after six years: y = -1360*6 + 15350 y = -8160 + 15350 y = 7190 The estimated value of the car six years from now is $7190.

Key Concepts

Depreciation Rate Calculation

Understanding how to calculate the depreciation rate of an asset is crucial for financial management and accounting. Depreciation represents the loss in value of an asset over time due to use and wear. In this exercise, we focus on linear depreciation, which implies that the asset loses the same fixed amount of value each year. To determine the yearly depreciation rate, you need to know the initial value of the asset, its value after a certain period, and the time span.
To find the depreciation rate, subtract the future value of the asset from its initial value to get the total depreciation over the given period. Then, divide this amount by the number of years to get the annual depreciation rate. For instance, the Honda Civic LX sedan lost \(15350 - 9910 = 5440\) in value over four years. Thus, the annual depreciation rate is \(\frac{5440}{4} = 1360\), which is negative because it denotes a decrease in value.

Point-Slope form

The point-slope form of a line is extremely helpful for writing the equation of a line when you have the slope and coordinates of a point on the line. In algebra, the point-slope formula is expressed as \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \(\left(x_1, y_1\right)\) are the coordinates of the known point. In our exercise, using the initial value of the car as the point \((0, 15350)\) and the depreciation rate of \(1360\) as the slope, we can set up the equation.
Substituting the known values into the formula gives us \(y - 15350 = -1360(x - 0)\). After simplifying the equation, we obtain the relationship for the value of the car over time: \(y = -1360x + 15350\). This formula can now be used to estimate the car's value at any point in time.

Linear Equations

Linear equations form the foundation of algebra and provide a way to describe the relationship between variables with a straight line when graphed. These equations typically look like \(y = mx + b\), where \(m\) is the slope of the line, and \(b\) is the y-intercept, or the point at which the line crosses the y-axis.
In the context of our exercise, the linear equation we derived represents the value of the Honda Civic LX sedan over time. The slope \(m = -1360\) indicates how quickly the car's value is decreasing each year, and the y-intercept \(b = 15350\) reflects the car's initial value. By observing the linear equation, one can make financial decisions regarding the asset, such as the best time to sell the car before it loses too much value.