Problem 76 - Introductory Algebra for College Students

Question

A new car worth \(\$ 24,000\) is depreciating in value by \(\$ 3000\) per year. The mathematical model $$y=-3000 x+24,000$$ describes the car's value, \(y,\) in dollars, after \(x\) years. a. Find the \(x\)-intercept. Describe what this means in terms of the car's value. b. Find the \(y\)-intercept. Describe what this means in terms of the car's value. c. Use the intercepts to graph the linear equation. Because \(x\) and \(y\) must be nonnegative (why?), limit your graph to quadrant I and its boundaries. d. Use your graph to estimate the car's value after five years.

Step-by-Step Solution

Answer

The \(x\)-intercept of the given equation is 8, which means the car's value will hit 0 in 8 years. The \(y\)-intercept is \$24,000, implying the worth of the car at the time of purchase was \$24,000. According to the graph created using these intercepts, the car's value after five years is approximately \$15,000.

Step 1: Find the x-Intercept

To find the \(x\)-intercept, set \(y = 0\) in the given equation and solve for \(x\). By doing this, you would get \(x=8\). So, the \(x\)-intercept is \(8\). This means the value of the car will be 0 after 8 years.

Step 2: Find the y-Intercept

To find the \(y\)-intercept, set \(x = 0\) in the given equation and solve for \(y\). By doing this, you would get \(y=24000\). So, the \(y\)-intercept is \$24000. This implies the car was worth \$24000 at the time of purchase.

Step 3: Draw the graph

Plot the \(x\)-intercept (8,0) and \(y\)-intercept (0,24000) on the graph. Connect these points with a straight line that starts at the \(y\)-intercept and goes through the \(x\)-intercept.

Step 4: Estimate the Car's Value after Five Years

To estimate the car's value after five years, look for \(x=5\) on the graph and find the corresponding \(y\)-value. This value is approximately \$15,000.

Key Concepts

Depreciation

Depreciation refers to the decrease in value of an asset over time. In the case of a car, it loses value as it gets older and used more. This decline in value can be calculated using a linear equation, which is often the case for assets that lose value at a constant rate. For instance, when a car is purchased for \\(24,000 and then depreciates by \\)3,000 every year, we can model this as a straightforward linear equation which helps in estimating the car's future worth.
The given equation, \( y = -3000x + 24000 \), provides this depreciation model. Here, the slope \(-3000\) indicates the depreciation of \\(3,000 per year. This means that every year, the car’s value drops by \\)3,000. The negative sign of \( -3000 \) signifies this decrease.
Understanding depreciation not only helps in predicting future monetary value but is crucial for making informed decisions about buying or selling cars, especially after a certain number of years.

x-intercept

The \(x\)-intercept of a linear equation provides the point where the graph intersects the x-axis. At this point, the value of \( y \) is zero. Finding the \(x\)-intercept involves setting \( y = 0 \) and solving for \( x \).
For our car depreciation equation \( y = -3000x + 24000 \), the \(x\)-intercept is calculated as:

• Set \(y = 0: 0 = -3000x + 24000\)
• Solve for \(x: x = 8\)

This means, after 8 years, the car's value will reach zero, showing it holds no market worth.
Understanding the \(x\)-intercept provides crucial insight, especially in financial planning, illustrating when an asset like a car will entirely lose its purchase value.

y-intercept

The \(y\)-intercept is the point where the graph intersects the y-axis, and it shows the initial value of the asset when \(x = 0\). For the car's value equation, this is where we determine the initial market value of the car when it was brand new.
In our example of car depreciation:

• Set \(x = 0\) in \(y = -3000x + 24000\).
• You get \(y = 24000\).

Thus, the \(y\)-intercept is \(\\(24,000\). This indicates that when the car was first purchased, its value was \\)24,000.
Grasping the \(y\)-intercept helps individuals understand the starting point in the depreciation process, crucial when evaluating the worth of an asset over time.

Graphing in Quadrant I

Graphing the linear depreciation equation in Quadrant I involves plotting only the positive sections of the axes since neither the car's age nor its monetary value can be negative.
To graph \( y = -3000x + 24000 \) properly:

• Plot points at the \(y\)-intercept (0, 24000) and \(x\)-intercept (8, 0).
• Connect these points with a line within Quadrant I.

This portion of the graph helps visualize the decline in value over time.
Additionally, Quadrant I graphing confirms that both elements, age (in years) and value (in dollars), remain within realistic, positive measurements. These graphs are immensely useful for visualizing how the car's value decreases yearly, making the data more accessible and easily interpretable.