Problem 83 - Algebra for College Students

Question

Solve each problem. The rising base price \(P\) (in dollars) for a new Ford \(\mathrm{F} 150\) can be modeled by the function \(P=793 n+15,950,\) where \(n\) is the number of years since 2000. a) What will be the base price for a new Ford F150 in \(2009 ?\) b) By what amount is the price increasing annually? c) Graph the equation for \(0 \leq n \leq 10\)

Step-by-Step Solution

Answer

a) \(23,087 in 2009. b) By \)793 yearly. c) Plot points for 0 ≤ n ≤ 10 and connect.

Step 1: Identify the Values Given in the Problem

Identify the variables and values from the problem. Here, the base price is given by the function \[ P = 793n + 15,950 \] and the number of years since 2000, represented by \( n \).

Step 2: Calculate Base Price in 2009

To calculate the base price in 2009, determine the value of \( n \) when the year is 2009: \[ n = 2009 - 2000 = 9 \]Substitute \( n = 9 \) into the equation: \[ P = 793(9) + 15,950 \]Calculate the result: \[ P = 7,137 + 15,950 \]\[ P = 23,087 \]Therefore, the base price in 2009 is \$23,087.

Step 3: Determine the Annual Price Increase

The annual increase in price is represented by the coefficient of \( n \) in the equation, which is \( 793 \). This means the price increases by \$793 each year.

Step 4: Graph the Equation for 0 ≤ n ≤ 10

To graph the equation, list values for \(n\) from 0 to 10. Calculate corresponding \( P \) values:- When \( n = 0 \): \[ P = 793(0) + 15,950 = 15,950 \]- When \( n = 1 \): \[ P = 793(1) + 15,950 = 16,743 \]- When \( n = 2 \): \[ P = 793(2) + 15,950 = 17,536 \]- Continue up to \( n = 10 \):Plot these points on a graph with the x-axis representing \(n\) and the y-axis representing \(P\).

Step 5: Plotting Points and Connecting

Plot the ordered pairs: (0, 15,950), (1, 16,743), (2, 17,536), (3, 18,329), (4, 19,122), (5, 19,915), (6, 20,708), (7, 21,501), (8, 22,294), (9, 23,087), (10, 23,880).Connect the points with a straight line to visualize the increase in base price over time.

Key Concepts

Slope-Intercept Form

In mathematics, the slope-intercept form is a method to write the equation of a line. This form makes it easy to identify both the slope and the y-intercept of the line. A linear function in slope-intercept form is written as: \[ y = mx + b \] Here,

• \textbf{m} represents the slope of the line.
• \textbf{b} is the y-intercept, the point where the line crosses the y-axis.

In the given problem, the function that models the rising base price of the new Ford F150 is: \[ P = 793n + 15,950 \] Comparing this with the slope-intercept form, we see that

• The slope (\textbf{m}) of the line is 793. This tells us how much the price increases each year.
• The y-intercept (\textbf{b}) is 15,950. This is the base price of the car in the year 2000.

Understanding the slope-intercept form helps us quickly determine these key details when analyzing linear functions.

Graphing Linear Equations

Graphing linear equations involves plotting points on a coordinate plane and connecting them to form a straight line. Let's revisit the equation we are dealing with: \[ P = 793n + 15,950 \]To graph this equation for values of \textbf{n} from 0 to 10, we follow these steps:

1. Choose values for \textbf{n}, such as 0, 1, 2, 3, ..., 10.
2. Calculate the corresponding values of \textbf{P} (price):
   For example, when \textbf{n} = 0, \textbf{P} = 15,950; when \textbf{n} = 1, \textbf{P} = 16,743; and so on.
3. Plot these points on the graph with \textbf{n} on the x-axis and \textbf{P} on the y-axis. Some points to plot are:
   • (0, 15,950)
   • (1, 16,743)
   • (2, 17,536)
   • (3, 18,329)
   • (4, 19,122)
   • (5, 19,915)
   • (6, 20,708)
   • (7, 21,501)
   • (8, 22,294)
   • (9, 23,087)
   • (10, 23,880)
4. Connect the points with a straight line. This line shows the trend of increasing prices over the years.

Graphing the equation visually represents how the base price increases annually, which can be very helpful for understanding the relationship.

Solving for Variables

Solving for variables is a fundamental skill in algebra and involves isolating the variable on one side of the equation. In our problem, we needed to find the price in the year 2009. Here, the variable \textbf{n} represents years since 2000. To solve for a specific year, follow these steps:

1. Determine the value of \textbf{n} for the desired year. For example, for 2009: \[ n = 2009 - 2000 = 9\] This means \textbf{n} is 9.
2. Substitute \textbf{n} back into the equation: \[ P = 793(9) + 15,950\]
3. Calculate the result. First, multiply 793 by 9: \[ 793 \times 9 = 7,137\] Add this to 15,950: \[ 7,137 + 15,950 = 23,087 \] So, the base price in 2009 is \$23,087.

Solving for variables allows us to predict outcomes and understand relationships within data. This skill is crucial for tackling many math problems and real-world scenarios effectively.