# 4.8: Linear Function as a Model

**At Grade**Created by: CK-12

## Applications of Linear Functions

**Introduction**

In this lesson you will learn that linear relationships are often used to model real-life situations. To do this, two data values related to the real-life situation must be present in the problem. These two data values in context will give you the information necessary to create a graph and an equation to model the real-life situation. The graph and the equation will be more meaningful if the axis is labelled according to the items in the problem and if variables representing these items are used in the equation. When the data values have been represented graphically and the equation of the line has been determined, questions relating to the real-life situation can be presented and answered.

**Objectives**

The lesson objectives for Linear Function as a Model:

- Understanding how to determine the equation of a line that models a real-life situation
- Understanding the meaning of the slope of a line as it applies to real-life situations
- Understanding how to use the equation to answer problems related to the real-life situation

**Introduction**

In this lesson, you will learn that linear functions can be applied to real-life problems. When equations and graphs are used to model real-life situations, the domain of the graph is often \begin{align*}x \epsilon N\end{align*}

**Watch This**

Khan Academy Basic Linear Function

**Guidance**

Joe’s Warehouse has banquet facilities to accommodate a maximum of 250 people. When the manager quotes a price for a banquet she is including the cost of renting the room plus the cost of the meal. A banquet for 70 people costs $1300. For 120 people, the price is $2200.

(a) Plot a graph of cost versus the number of people.

(b) From the graph, estimate the cost of a banquet for 150 people.

(c) Determine the slope of the line. What quantity does the slope of the line represent?

(d) Write an equation to model this real-life situation.

(a) On the \begin{align*}x-\end{align*}

(b) The approximate cost of a banquet for 150 people is $2700.

(c) The two data points (70, 1300) and (120, 2200) will be used to calculate the slope of the line.

\begin{align*}m &= \frac{y_2-y_1}{x_2-x_1}\\
m &= \frac{2200-1300}{120-70}\\
m &= \frac{900}{50}\\
m &= \frac{18}{1}\end{align*}

The slope represents the cost of the banquet for each person. The cost is $18 per person.

When a linear function is used to model the real life situation, the equation can be written in the form or in the form \begin{align*}y=mx+b\end{align*}

(d) \begin{align*}y &= mx+b\\ {\color{red}1300} &= {\color{red}18}({\color{red}70})+b\\ 1300 &= 1260+b\\ 1300 {\color{red}-1260} &= 1260 {\color{red}-1260}+b\\ 40 &=b\end{align*}

The \begin{align*}y-\end{align*}intercept is (0, 40)

The equation to model the real-life situation is \begin{align*}y=18x+40\end{align*}. The variables should be changed to match the labels on the axes. The equation that best models the situation is \begin{align*}c=18n+40\end{align*} where ‘\begin{align*}c\end{align*}’ represents the cost and ‘\begin{align*}n\end{align*}’ represents the number of people.

**Example A**

A cab company charges $2.00 for the first 0.6 miles and $0.50 for each additional 0.2 miles.

(a) Draw the graph of cost versus distance.

(b) Determine the equations that model this situation.

(c) What is the cost to travel 16 miles by cab?

**This example will demonstrate a real-life situation that cannot be modeled with just one equation.**

(a) On the \begin{align*}x-\end{align*}axis is the distance in miles and on the \begin{align*}y-\end{align*}axis is the cost in dollars. The first graph from \begin{align*}A\end{align*} to \begin{align*}B\end{align*} extends horizontally across the distance from 0 to 0.6 miles. The cost is constant at $2.00. The equation for this constant function is \begin{align*}y=2.00\end{align*} or \begin{align*}c=2.00\end{align*}. The second graph from \begin{align*}B\end{align*} to \begin{align*}C\end{align*} and upward is not constant.

(b) The equation that models the second graph can be determined by using the data points (0.6, 2.00) and (1, 3.00)

\begin{align*}m &= \frac{y_2-y_1}{x_2-x_1} && \text{Use the data points to calculate the slope.}\\ m &= \frac{3.00-2.00}{1-0.6}\\ m &= \frac{3.00-2.00}{1-0.6}\\ m &= \frac{1.00}{0.4}\\ m &= 2.5\\ \\ y-y_1 &= m(x-x_1) && \text{Use the slope and one point to determine the equation.}\\ y- {\color{red}2} &= {\color{red}2.5} (x- {\color{red}0.6})\\ y-2 &= 2.5x-1.5\\ y-2 {\color{red}+2} &= 2.5x-1.5 {\color{red}+2}\\ y &= 2.5x+0.5\\ c &= 2.5d+0.5\end{align*}

Therefore, the equations that model this situation are:

\begin{align*}c=\begin{Bmatrix} 2.00 & 0<d \le 0.6 \\ 2.5d+0.5 & d>0.6 \end{Bmatrix}\end{align*}

(c) The cost to travel 16 miles in the cab is:

The distance is greater than 0.6 miles. The cost must be calculated using the equation \begin{align*}c=2.5d+0.5\end{align*}. Substitute 16 in for ‘\begin{align*}d\end{align*}’.

\begin{align*} & c = 2.5d+0.5\\ & c = 2.5 ({\color{red}16})+0.5\\ & c = 2.5({\color{red}16})+0.5\\ & \boxed{c = \$ 40.50}\end{align*}

**Example B**

When a 40 gram mass was suspended from a coil spring, the length of the spring was 24 inches. When an 80gram mass was suspended from the same coil spring, the length of the spring was 36 inches.

(a) Plot a graph of length versus mass.

(b) From the graph, estimate the length of the spring for a mass of 70 grams.

(c) Determine an equation that models this situation. Write the equation in slope-intercept form.

(d) Use the equation to determine the length of the spring for a mass of 60 grams.

(e) What is the \begin{align*}y-\end{align*}intercept? What does the \begin{align*}y-\end{align*}intercept represent?

(a) On the \begin{align*}x-\end{align*}axis is the mass in grams and on the \begin{align*}y-\end{align*}axis is the length of the spring in inches.

(b) The length of the coil spring for a mass of 70 grams is approximately 33 inches.

(c) The equation of the line can be determined by using the two data values (40, 24) and (80, 36).

\begin{align*}m &= \frac{y_2-y_1}{x_2-x_1}\\ m &= \frac{36-24}{80-40}\\ m &= \frac{12}{40}\\\ m &= \frac{3}{10}\\ \\ y &= mx+b\\ {\color{red}24} &= {\color{red}\frac{3}{10}}({\color{red}40})+b\\ 24 &= \frac{3}{\cancel{10}} \left(\overset{{\color{red}4}}{\cancel{40}}\right)+b\\ 24 &= 12+b\\ 24 {\color{red}-12} &= 12 {\color{red}-12}+b\\ 12 &= b\end{align*}

The \begin{align*}y-\end{align*}intercept is (0, 12). The equation that models the situation is

\begin{align*}y=\frac{3}{10}x+12\end{align*}

\begin{align*}\boxed{l=\frac{3}{10} m+12}\end{align*}

where ‘\begin{align*}l\end{align*}’ is the length of the spring in inches and ‘\begin{align*}m\end{align*}’ is the mass in grams.

(d) \begin{align*}& l = \frac{3}{10}m+12 && \text{Use the equation and substitute} \ 60 \ \text{in for} \ `m'.\\ & l = \frac{3}{10}({\color{red}60})+12\\ & l = \frac{3}{\cancel{10}} \left(\overset{{\color{red}6}}{\cancel{60}}\right)+12\\ & l = 18+12\\ & \boxed{l = 30 \ inches}\end{align*}

(e) The \begin{align*}y-\end{align*}intercept is (0, 12). The \begin{align*}y-\end{align*}intercept represents the length of the coil spring before a mass was suspended from it. The length of the coil spring was 12 inches.

**Example C**

Juan drove from his mother’s home to his sister’s home. After driving for 20 minutes he was 62 miles away from his sister’s home and after driving for 32 minutes he was only 38 miles away. The time driving and the distance away from his sister’s home form a linear relationship.

(a) What is the independent variable? What is the dependent variable?

(b) What are the two data values?

(c) Draw a graph to represent this problem. Label the axis appropriately.

(d) Write an equation expressing distance in terms of time driving.

(e) What is the slope and what is its meaning in this problem?

(f) What is the time-intercept and what does it represent?

(g) What is the distance-intercept and what does it represent?

(h) How far is Juan from his sister’s home after he had been driving for 35 minutes?

(a) The independent variable is the time driving. The dependent variable is the distance.

(b) The two data values are (20, 62) and (32, 38).

(c) On the \begin{align*}x-\end{align*}axis is the time in minutes and on the \begin{align*}y-\end{align*}axis is the distance in miles.

(d) (20, 62) and (32, 38) are the coordinates that will be used to calculate the slope of the line.

\begin{align*}m &= \frac{y_2-y_1}{x_2-x_1}\\ m &= \frac{38-62}{32-20}\\ m &= \frac{-24}{12}\\ m &= -2\\ \\ y &= mx+b\\ {\color{red}62} &= {\color{red}-2} ({\color{red}20})+b\\ 62 &= -40+b\\ 62 {\color{red}+40} &= -40 {\color{red}+40}+b\\ 102 &= b \qquad \text{The} \ y- \text{intercept is} \ (0, 102)\end{align*}

\begin{align*}y &= mx+b\\ y &= -2x+102\\ d &= -2t+102\end{align*}

(e) The slope is \begin{align*}-2=\frac{-2}{1}=\frac{-2(miles)}{1(minute)}\end{align*}. The slope means that for each minute of driving, the distance that Juan has to drive to his sister’s home is reduced by 2 miles.

(f) The time-intercept is actually the \begin{align*}x-\end{align*}intercept. This value is:

\begin{align*}d &= -2t+102 && \text{Set} \ d=0 \ \text{and solve for} \ `t'.\\ {\color{red}0} &= -2t+102\\ 0 {\color{red}+2t} &= -2t {\color{red}+2t}+102\\ 2t &= 102\\ \frac{2t}{{\color{red}2}} &= \frac{102}{{\color{red}2}}\\ \frac{\cancel{2}t}{\cancel{2}} &= \frac{102}{2}\\ t &= 51 \ minutes\end{align*}

The time-intercept is 51 minutes and this represents the time it took Juan to drive from his mother’s home to his sister’s home.

(g) The distance-intercept is the \begin{align*}y-\end{align*}intercept. This value has been calculated as (0, 102). The distance-intercept represents the distance between his mother’s home and his sister’s home. The distance is 102 miles.

(h)\begin{align*}d &= -2t+102 && \text{Substitute} \ 35 \ \text{into the equation for} \ `t' \ \text{and solve for} \ `d'.\\ d &= -2 ({\color{red}35})+102\\ d &= -70+102\\ d &= 32 \ miles\end{align*}

After driving for 35 minutes, Juan is 32 miles from his sister’s home.

**Guided Practice**

1. Some college students, who plan on becoming math teachers, decide to set up a tutoring service for high school math students. One student was charged $25 for 3 hours of tutoring. Another student was charged $55 for 7 hours of tutoring. The relationship between the cost and time is linear.

(a) What is the independent variable?

(b) What is the dependent variable?

(c) What are two data values for this relationship?

(d) Draw a graph of cost versus time.

(e) Determine an equation to model the situation.

(f) What is the significance of the slope?

(g) What is the cost-intercept? What does the cost-intercept represent?

2. A Glace Bay developer has produced a new handheld computer called the ** Blueberry**. He sold 10 computers in one location for $1950 and 15 in another for $2850. The number of computers and cost forms a linear relationship

(a) State the dependent and independent variables.

(b) Sketch a graph.

(c) Find an equation expressing cost in terms of the number of computers.

(d) State the slope of the line and tell what the slope means to the problem.

(e) State the cost-intercept and tell what it means to this problem.

(f) Using your equation, calculate the number of computers you could get for $6000.

3. Handy Andy sells one quart can of paint thinner for $7.65 and a two quart can for $13.95. Assume there is a linear relationship between the volume of paint thinner and the price.

(a) What is the independent variable?

(b) What is the dependent variable?

(c) Write two data values for this relationship.

(d) Draw a graph to represent this relationship.

(e) What is the slope of the line?

(f) What does the slope represent in this problem?

(g) Write an equation to model this problem.

(h) What is the cost-intercept?

(i) What does the cost-intercept represent in this problem?

(j) How much would you pay for 6 quarts of paint thinner?

**Answers**

1. (a) The cost for tutoring depends upon the amount of time. The independent variable is the time.

(b) The dependent variable is the cost.

(c) Two data values for this relationship are (3, 25) and (7, 55).

(d) On the \begin{align*}x-\end{align*}axis is the time in hours and on the \begin{align*}y-\end{align*}axis is the cost in dollars.

(e) Use the two data values (3, 25) and (7, 55) to calculate the slope of the line.

\begin{align*}m &= \frac{y_2-y_1}{x_2-x_1}\\ m &= \frac{55-25}{7-3}\\ m &= \frac{30}{4}\\ m &= \frac{15}{2}\end{align*}

Determine the \begin{align*}y-\end{align*}intercept of the graph.

\begin{align*}y&=mx+b\\ {\color{red}25}&= {\color{red}\frac{15}{2}}( {\color{red}3})+b && \text{Use the slope and one of the data values to determine the value of} \ `b'.\\ 25 &= \frac{45}{2}+b\\ 25 {\color{red}-\frac{45}{2}} &= \frac{45}{2} {\color{red}-\frac{45}{2}}+b\\ {\color{red}\frac{50}{2}}-\frac{45}{2} &= b\\ \frac{5}{2} &= b\end{align*}

The \begin{align*}y-\end{align*}intercept is \begin{align*}\frac{5}{2}\end{align*}

The equation to model the relationship is \begin{align*}y=\frac{15}{2}x+\frac{5}{2}\end{align*}. To match the variables of the equation with the graph the equation is \begin{align*}\boxed{c=\frac{15}{2}t+\frac{5}{2}}\end{align*}. The relationship is cost in dollars versus time in hours. The equation could also be written as \begin{align*}\boxed{c=7.50t+2.50}\end{align*}

(f) The slope of \begin{align*}\frac{15}{2}\end{align*} means that it costs $15.00 for 2 hours of tutoring. If the slope is expressed as a decimal, it means that it costs $7.50 for 1 hour of tutoring.

(g) The cost-intercept is the \begin{align*}y-\end{align*}intercept. The \begin{align*}y-\end{align*}intercept is (0, 2.50). This value could represent the cost of having a scheduled time or the cost that must be paid for cancelling the appointment. In a problem like this, the \begin{align*}y-\end{align*}intercept must be represent a meaningful quantity for the problem.

2. (a) The number of dollars in sales from the computers depends upon the number of computers sold. The dependent variable is the dollars in sales and the independent variable is the number of computers sold.

(b) On the \begin{align*}x-\end{align*}axis is the number of computers and on the \begin{align*}y-\end{align*}axis is the cost of the computers.

(c) Use the data values (10, 1950) and (15, 2850) to calculate the slope of the line.

\begin{align*}m &= \frac{y_2-y_1}{x_2-x_1}\\ m &= \frac{2850-1950}{15-10}\\ m &= \frac{900}{5}\\ m &= 180\end{align*}

Determine the \begin{align*}y-\end{align*}intercept of the graph.

\begin{align*}y &= mx+b\\ 1950 &= 180(10)+b\\ 1950 &= 1800+b\\ 1950-1800 &= 1800-1800+b\\ 150 &= b\end{align*}

The \begin{align*}y-\end{align*}intercept is (0, 150).

The equation of the line that models the relationship is

\begin{align*}\boxed{y=180x+150}\end{align*}

To make the equation match the variables of the graph the equation is

\begin{align*}\boxed{c=180n+150}\end{align*}

(d) The slope is \begin{align*}\frac{180}{1}\end{align*}. This means that the cost of one computer is $180.00

(e) The cost intercept is the \begin{align*}y-\end{align*}intercept. The \begin{align*}y-\end{align*}intercept is (0, 150). This could represent the cost of renting the location where the sales are being made or perhaps the salary for the sales person.

(f) \begin{align*}c &= 180n+150\\ 6000 &= 180n+150\\ 6000-150 &= 180n+150-150\\ 5850 &= 180n\\ \frac{5850}{180} &= \frac{180n}{180}\\ \frac{5850}{180} &= \frac{\cancel{180} n}{\cancel{180}}\\ 32.5 &= n\end{align*}

With $6000 you could get **32** computers.

3. (a) The independent variable is the volume of paint thinner.

(b) The dependent variable is the cost of the paint thinner.

(c) Two data values are (1, 7.65) and (2, 13.95).

(d) On the \begin{align*}x-\end{align*}axis is the volume in quarts and on the \begin{align*}y-\end{align*}axis is the cost in dollars.

(e) Use the two data values (1, 7.65) and (2, 13.95) to calculate the slope of the line.

\begin{align*}m &= \frac{y_2-y_1}{x_2-x_1}\\ m &= \frac{13.95-7.65}{2-1}\\ m &= \frac{6.30}{1}\end{align*}

(f) The slope represents the cost of one quart of paint thinner. The cost is $6.30.

(g) \begin{align*}y &= mx+b\\ 7.65 &= 6.30(1)+b\\ 7.65 &= 6.30+b\\ 7.65-6.30 &= 6.30-6.30+b\\ 1.35 &= b\end{align*}

The \begin{align*}y-\end{align*}intercept is (0, 1.35). The equation to model the relationship is

\begin{align*}y=6.30x+1.35\end{align*}

The equation that matches the variables of the graph is \begin{align*}\boxed{c=6.30v+1.35}\end{align*}

(h) The cost-intercept is (0, 1.35).

(i) This could represent the cost of the can that holds the paint thinner.

(j) \begin{align*}c &= 6.30v+1.35\\ c &= 6.30(6)+1.35\\ c &= 37.80+1.35\\ c &= \$39.15\end{align*}

The cost of 6 quarts of paint thinner is $39.15.

**Summary**

In this lesson you have learned that real-life problems can be represented by graphs and linear equations. The slope and the \begin{align*}y-\end{align*}intercept both have significance that is reflected in the problem. The linear relationship can be modeled by a linear equation that reflects the variables of the graph. The equations can be written in either standard form or in \begin{align*}y-\end{align*}intercept form.

Data that is discrete is represented using a linear graph instead of a set of plotted points.

**Problem Set**

**Completely answer the following problems...**

- Players on the school soccer team are selling candles to raise money for an upcoming trip. Each player has 24 candles to sell. If a player sells 4 candles a profit of $30 is made. If he sells 12 candles a profit of $70 is made. The profit and the number of candles sold form a linear relation.
- State the dependent and the independent variables.
- What are the two data values for this relation?
- Draw a graph and label the axis.
- Determine an equation to model this situation.
- What is the slope and what does it mean in this problem?
- Find the profit-intercept and explain what it represents.
- Calculate the maximum profit that a player can make.
- Write a suitable domain and range.
- If a player makes a profit of $90, how many candles did he sell?
- Is this data continuous or discrete? Justify your answer.

- Jacob leaves his summer cottage and drives home. After driving for 5 hours, he is 112 km from home, and after 7 hours, he is 15 km from home. Assume that the distance from home and the number of hours driving form a linear relationship.
- State the dependent and the independent variables.
- What are the two data values for this relationship?
- Represent this linear relationship graphically.
- Determine the equation to model this situation.
- What is the slope and what does it represent?
- Find the distance-intercept and its real-life meaning in this problem.
- How long did it take Jacob to drive from his summer cottage to home?
- Write a suitable domain and range.
- How far was Jacob from home after driving 4 hours?
- How long had Jacob been driving when he was 209 km from home?

**Answers**

- (a) The profit a player makes depends upon the number of candles he sells. Independent Variables \begin{align*}(x) \rightarrow\end{align*} Candles (#) Dependent Variable \begin{align*}(y) \rightarrow\end{align*} Profit ($) (b) Data Values: (4, 30) and (12, 70) (c) Graph:

(d) \begin{align*}m &= \frac{y_2-y_1}{x_2-x_1}\\ m &= \frac{70-30}{12-4}\\ m &= \frac{40}{8}\\ m &= \frac{5}{1}\\ m &= 5\\ m &= \frac{5}{1} \quad \frac{5(dollars)}{1(candle)}\end{align*}

\begin{align*}y &= mx + b\\ 30 &= 5(4) + b\\ 30 &= 20 + b\\ 30 - 20 &= b\\ 10 &= b\end{align*}

\begin{align*}y &= mx+b\\ P &= 5C+10\end{align*}

(e) The profit made by selling one candle is $5.00.

(f) The profit-intercept is $10.00. This means that each player starts with a profit of $10.00 before selling any candles. This could be an incentive bonus.

(g) A player has 24 candles to sell.

\begin{align*}P &= 5C + 10\\ P &= 5(24) + 10\\ P &= 120 + 10\\ P &= \$130\end{align*}

The maximum profit is $130.00.

(h) \begin{align*}P &= 5C + 10\\ 90 &= 5C + 10\\ 90 {\color{red}- 10} &= 5C + 10 {\color{red}- 10}\\ 80 &= 5C\\ \frac{80}{5} &= \frac{5C}{5}\\ 16 &= C\end{align*}

If a player made a profit of $90.00, he/she would have sold 16 candles.

(j) This is discrete data. Profit is based on the number of candles sold. Since candles are single units, the value between each number is meaningless. Therefore the points should not be joined.

## Summary

In this lesson you have learned to apply the equation of a linear relationship to a real-life situation. Two data values from the problem were used to determine the equation of the line. The slope of the line represented a meaningful quantity of the problem. The \begin{align*}y-\end{align*}intercept also had significance for the real-life situation. The significance of the \begin{align*}y-\end{align*}intercept must represent a quantity that is meaningful for the problem.

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