Problem 25 - College Algebra

Question

A new car worth \(\$ 24,000\) is depreciating in value by \(\$ 3000\) per year. a. Write a formula that models the car's value, \(y,\) in dollars, after \(x\) years. b. Use the formula from part (a) to determine after how many years the car's value will be \(\$ 9000\). c. Graph the formula from part (a) in the first quadrant of a rectangular coordinate system. Then show your solution to part (b) on the graph.

Step-by-Step Solution

Answer

a. The formula for the car's value \(y\), in dollars, after \(x\) years is \(y = -3000x + 24000\). b. The car's value will be \$9000 after 5 years. c. The graph of the formula should show a line decreasing at a rate of \$3000/year, starting from a y-intercept of \$24000, with \(x = 5\) marked to indicate when the car's value will be \$9000.

Step 1: Create a Mathematical Model

In the equation, \(y = mx + b\), \(m\) is the slope (rate of change) and \(b\) is the y-intercept (initial value). So here, \(m = -3000\) (as the value is decreasing) and \(b = 24000\) (as that's the initial value of the car). Thus, the equation that models the car's value, \(y\), in dollars, after \(x\) years is \(y = -3000x + 24000\).

Step 2: Calculate the Year When the Car's Value Will Be $9000

We want to find the year \(x\) when the car's value \(y\) will be \$9000. So, we plug \(y = 9000\) into our formula and solve for \(x\): \[9000 = -3000x + 24000\Rightarrow x = (24000-9000)/3000 \Rightarrow x = 5 \] Thus, the car's value will be \$9000 after 5 years.

Step 3: Graph the Formula

The graph should be a line with a negative slope, starting from a y-intercept of \$24000 and decreasing at a rate of \$3000/year. \(x = 5\) should be marked to indicate when the car's value will be \$9000.

Key Concepts

Understanding Linear Equations

A linear equation forms a straight line when graphed on a coordinate plane. These kinds of equations are written in the format:

• \( y = mx + b \)
• where \( m \) represents the slope and \( b \) is the y-intercept.

The elegance of a linear equation is in its simplicity. It tells us that there is a consistent rate of change, making it extremely useful in predicting future outcomes. In the context of this exercise, a linear equation is used to model the decreasing value of a car over time.
By understanding each component, \( m \) as the depreciation rate and \( b \) as the starting value, we can capture how the car loses value every year.

Explaining Slope

In a linear equation, the slope is a crucial element that defines the direction and steepness of a line. The slope is represented by \( m \) in the equation \( y = mx + b \). It measures how much \( y \) changes for a unit change in \( x \).
In this particular exercise, the slope is \( -3000 \). - This negative value indicates a decrease in the car's value.- Specifically, the car's value decreases by \$3000 each year.- Since this decrease is consistent, this value remains constant in the equation.Understanding slope helps us gauge how quickly or slowly a variable changes relative to another, which is invaluable in financial modeling.

Deciphering Y-intercept

The y-intercept is another key component of a linear equation represented as \( b \) in \( y = mx + b \).
It signifies the point where the line crosses the y-axis, which is also when \( x = 0 \).

• In the case of this exercise, the y-intercept is \$24,000.
• This is the initial value of the car when no time has passed.

The y-intercept provides a starting reference point for evaluating how a variable changes over time. In financial contexts like depreciation, it represents the initial worth or capital before any influences of the slope come into play.
This makes it crucial to understand the base value before adjustments are made.

Graphing Linear Equations

Graphing is a powerful visual tool that translates an equation into a visual format. It helps to see the relationship between variables vividly. To graph a linear equation:

• Identify the y-intercept (\\(24,000 in this case).
• From this point, use the slope (\(-3000\)) to make consistent steps, moving one unit along the x-axis (years) and \( -3000 \) units along the y-axis (value).

The graph in this context will depict a descending line, showing how car value decreases over time. Plotting specific points helps find values of interest, such as identifying that the car will be worth \\)9000 at \( x = 5 \) years.
This complete, linear representation on a graph solidifies the understanding of how y, the car's value, depends on x, the time passed.