Interpreting Graphs: Linear Relation and Equations Insights

This lesson will introduce a way to compare and quantify how fast one quantity changes with respect to another. This will allow a more in-depth study of the relationship between two variables and the classification of those variables according to their type of dependency. The ideas covered in this lesson are the foundation of many areas of modern mathematics.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Variables
Table of Values
Linear Equations in Two Variables
Solution of an Equation
Coordinate Pair
Coordinate Plane
Line

Here are a few practice exercises before getting started with this lesson.

a Which of the following equations is a linear equation in two variables?

b Consider the linear equation in two variables

y = - 3 x + 5 .

Which of the following coordinate pairs is a solution to the equation?

c Take a look at the tables of values shown below.

Which of the tables represents the linear equation in two variables

y = 2 x + 4 ?

Challenge

Analyzing Dominika’s Speeds on Her Way to and From School

Dominika is curious about the amount of time she spends going to and from school. One day, she decides to measure the time it takes her to get to school on foot and compare it to the amount of time it takes her to arrive home by bus. She drew the following graph for her trip to school.

Graph representing Dominika's walk to school

a What was Dominika's average walking speed on the way to school?

b At

2 : 00 PM,

the school day was over and Dominika was waiting for the bus. After

9

minutes, the bus finally arrived, and she arrived back home at

2 : 12 PM .

Draw a graph similar to the one given in Part A, showing the distance from the school this time, and find Dominika's average speed when returning home.

c Was Dominika faster than the school bus? Explain.

Discussion

What Is a Rate of Change?

In order to understand the relationship between two quantities and make predictions about their behavior, it is important to be able to compare and quantify how one changes with respect to the other. These ideas define the concept of rate of change.

Concept

Rate of Change

The rate of change ROC is a ratio used to compare how a variable changes in relation to another variable. It is determined by dividing the change in the output variable $y$ by the change in the input variable $x .$ For any ordered pairs $(x_{1}, y_{1})$ and $(x_{2}, y_{2}),$ the rate of change is calculated using the following formula.

Rate of Change Rate of Change = \frac{Change in y}{Change in x} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

The Greek letter $Δ$ (Delta) is commonly used to represent a difference. This leads to an alternative way of writing the formula for ROC.

$Rate of Change = \frac{Δ y}{Δ x}$

Extra

Positive, Negative, and Zero Rate of Change

Depending on the relationship between the variables, the rate of change can be positive, negative, or zero.

Extra

Rate of Change Units

The units of a rate of change are the ratio of output units to the input units, meaning that they are derived units. Interpreting the rate of change depends on the context.

Output Units	Input Units	Rate of Change Units	Possible Interpretation
meters	seconds	$\frac{meters}{second}$	Car speed over a certain period of time
bacteria	hours	$\frac{bacteria}{hour}$	Growth rate of bacteria in an experiment
U.S. dollars	hours	$\frac{U . S . dollars}{hour}$	Worker's hourly wage

As mentioned above, the rate of change is a quantity that compares the change in the output variable to the input variable. These are also known as the dependent and independent variables, respectively.

Concept

Independent and Dependent Variables

In the context of functions, the input is often referred to as the independent variable because it can be chosen arbitrarily from the domain. Conversely, the output is called the dependent variable because its value depends on the value of the independent variable. For instance, if the price of oranges is

$ 2.50

per pound, the total cost is determined by the product of the unit price and the weight in pounds.

Cost y = Unit Price 2.50 \times Weight x

As shown, the total cost of oranges depends on how many pounds of fruit are purchased. Therefore, the cost of oranges

y

is the

dependent

variable

and the number of pounds purchased

x

is the

independent

variable .

Example

Exploring the Rate of Change in Different Scenarios

Diego is taking the bus back to his hometown. During the trip, he sees that there are signs by the side of the road that indicate the distance from the city.

He keeps track of some of the signs and notes the time taken to reach them. He puts the information together using a table of values.

Time (minutes)	Distance (kilometers)
$0$	$0$
$10$	$15$
$20$	$30$
$30$	$45$
$40$	$60$
$50$	$75$
$60$	$90$

a Considering the variables used in this context, what is the interpretation of the rate of change?

What is the rate of change between

10

and

20

minutes?

What is the rate of change between

0

and

60

minutes?

b When he arrived at his hometown, Diego found a magazine containing statistics for different cars. The following graph shows information about the performance of a new electric car, indicating the speed reached during the first few seconds of the car accelerating from zero.

Graph of a line showing three points: (0,0),(2,50), and (4,100)

Find the rate of change between

(0, 0)

and

(2, 50) .

Find the rate of change between

(2, 50)

and

(4, 100) .

c The magazine also included a linear equation in two variables describing the performance of a hybrid car. This time it refers to the amount of kilometers the car can travel depending on the amount of gallons of gas.

y = 50 x + 30

The

y -

variable represents the distance in kilometers that the car can travel and the

x -

variable the gas in gallons. Find the rate of change between

x = 1

and

x = 3 .

Find the rate of change between

x = 4

and

x = 7 .

Find the rate of change between

x = 1

and

x = 7 .

Hint

a In this case, the independent variable is the time, which is measured in seconds. On the other hand, the dependent variable is the distance, which is measured in meters.

b In a coordinate pair, the first coordinate represents the value of the independent variable, while the second coordinate represents the value of the dependent variable.

c A linear equation can be represented by a line on the coordinate plane. When the

x -

coordinate of a point on the line is known, its

y -

coordinate can be found by substituting the

x -

value into the equation of the line.

Solution

a First, the units of the rate of change will be determined and interpreted. Then, the rate of change will be calculated for the given interval.

Interpreting the Rate of Change

The rate of change is defined as the ratio of the change in the dependent variable to the change of the independent variable.

Rate of Change = \frac{Δ y}{Δ x}

Note that the distance covered by the car depends on the time elapsed. On the other hand, time passes by without relation to the the car and can take any non-negative value without restrictions. Therefore, the dependent variable in this case is the distance, and time is independent. Let

t

represent the time and

d

the distance.

Rate of Change = \frac{Δ d}{Δ t} = \frac{d _{2} - d _{1}}{t _{2} - t _{1}}

Recall that the unit of the rate of change is the ratio of the unit of the dependent variable to the unit of the independent variable.

Rate of Change Units = \frac{kilometers}{minute}

Therefore, the unit of the rate of change is

\frac{km}{min},

which quantifies kilometers traveled per minute. The rate of change can be interpreted as the speed of the bus.

\underline{Interpretation for the Rate of Change} The speed of the car during a specific period of time

Finding the Rate of Change Between Different Times

Now, by using the rate of change formula together with the information from the table, the rate of change of the car between

10

and

20

minutes will be calculated. Notice that the corresponding distances are

15

and

30

kilometers, respectively.

Rate of Change = \frac{d _{2} - d _{1}}{t _{2} - t _{1}}

SubstitutePoints

Substitute $(10, 15)$ & $(20, 30)$

Rate of Change = \frac{3 0 - 1 5}{2 0 - 1 0}

▼

Evaluate

SubTerms

Subtract terms

Rate of Change = \frac{1 5}{1 0}

CalcQuot

Calculate quotient

Rate of Change = 1.5

The rate of change between

10

and

20

minutes is

1.5 \frac{km}{min} .

By following the same procedure it is possible to find the rate of change between the other points of interest. The following table summarizes these results for the rate of change corresponding to the required time intervals.

Rate of Change $= \frac{d _{2} - d _{1}}{t _{2} - t _{1}}$
Times	Substitute	Evaluate
$10 - 20$ min	$\frac{3 0 km - 1 5 km}{2 0 min - 1 0 min}$	$1.5 \frac{km}{min}$
$0 - 60$ min	$\frac{9 0 km - 0 km}{6 0 min - 0 min}$	$1.5 \frac{km}{min}$

b Similarly to Part A, the rate of change formula will be used to find the rate of change of the graph. In this case, the variables are the speed

s

and the time

t .

Rate of Change = \frac{Δ s}{Δ t}

Notice that in a coordinate pair, the first coordinate represents the value of the independent variable, while the second coordinate corresponds to the dependent variable. Using this information, the rate of change between the points

(0, 0)

and

(2, 50)

will be calculated for the given graph.

Rate of Change = \frac{s _{2} - s _{1}}{t _{2} - t _{1}}

SubstitutePoints

Substitute $(0, 0)$ & $(2, 50)$

Rate of Change = \frac{5 0 - 0}{2 - 0}

▼

Evaluate

SubTerm

Subtract term

Rate of Change = \frac{5 0}{2}

CalcQuot

Calculate quotient

Rate of Change = 25

The acceleration in the first

2

seconds is

25 \frac{km}{hr \cdot s} .

By following the same procedure, the rate of change for the other pair of points can be found. The following table summarizes the results for the rate of change of the graph between the specified pair of points.

Rate of Change $= \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Points	Substitute	Evaluate
$(0, 0), (2, 50)$	$\frac{5 0 - 0}{2 - 0}$	$25$
$(2, 50), (4, 100)$	$\frac{1 0 0 - 5 0}{4 - 2}$	$25$

c In this case, before using the rate of change formula, the corresponding

x -

and

y -

values must first be found. This can be done by substituting the

x -

value of interest into the linear equation and evaluating the resulting expression on the right-hand side. This will be done for

x = 1

first to illustrate the procedure.

y = 50 x + 30

Substitute

$x = 1$

y = 50 (1) + 30

▼

Evaluate

Multiply

y = 50 + 30

AddTerms

Add terms

y = 80

It was found that the

y -

value corresponding to

x = 1

y = 80 .

Therefore, the coordinate pair

(1, 80)

is a point on the graph of the line representing the given linear equation. By repeating this process, the specific points of interest can be found.

Linear Equation: $y = 2 x + 3$
$x$	Substitute	Point
$x = 1$	$y y = 50 (1) + 30 = 80$	$(1, 80)$
$x = 3$	$y y = 50 (3) + 30 = 180$	$(3, 180)$
$x = 4$	$y y = 50 (4) + 30 = 230$	$(4, 230)$
$x = 7$	$y y = 50 (7) + 30 = 380$	$(7, 380)$

Now that the points are known, all that is left to do is to repeat the same procedure as in Part B. The following table summarizes the results for the pair of points of interest.

Rate of Change $= \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Points	Substitute	Evaluate
$(1, 80), (3, 180)$	$\frac{1 8 0 - 8 0}{3 - 1}$	$50$
$(4, 230), (7, 380)$	$\frac{3 8 0 - 2 3 0}{7 - 4}$	$50$
$(1, 80), (7, 380)$	$\frac{3 8 0 - 8 0}{7 - 1}$	$50$

It is interesting to note that the rate of change is the same between all the pairs of points used. This occurred in Parts A and B as well.

Pop Quiz

Practice Calculating the Rate of Change

The following applet shows different representations for the data of two variables. Practice finding the rate of change between the indicated values. If the answer is not an integer number, round it to two decimal places.

Interactive graph showing a table/ expression, or a graph

Discussion

Linear Relations and Constant Rates of Change

Rule

Constant Rate of Change

Any line in the coordinate plane has a constant rate of change between any pair of its points. This can be checked by moving the points in the following applet.

Interactive graph showing a linear function the rate of change between movable points

The opposite of this statement also holds true. If the rate of change between consecutive pairs of points of a data set is constant, then these points follow a linear relation and they all lie on the same line in the coordinate plane. The following applet illustrates this.

Interactive graph showing a table of values with constant rate of change and their graph

Proof

Algebraic Proof

Pairs of Points in a Line Have a Constant Rate of Change

Any linear equation in two variables can written in the following form where

y

is the dependent variable,

x

is the independent variable, and

m

and

b

can be any real numbers.

y = m x + b

Now, since this is a linear equation in two variables, all of its solutions lie on a line in the coordinate plane. In what follows, two arbitrary points

(x_{1}, y_{1})

and

(x_{2}, y_{2})

will be used, and the rate of change between them will be found by using the rate of change formula.

Rate of Change = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Because these arbitrary points are solutions to the linear equation, they satisfy the equation. Therefore, it is possible to find an explicit expression for

y_{1}

and

y_{2}

in terms of

x_{1}

and

x_{2},

respectively.

y_{1} = m x_{1} + b y_{2} = m x_{2} + b

Now that the explicit form for

y_{1}

and

y_{2}

is known, the rate of change between these points can be calculated.

Rate of Change = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

SubstituteII

$y_{1} = m x_{1} + b$ , $y_{2} = m x_{2} + b$

Rate of Change = \frac{m x _{2} + b - ( m x _{1} + b )}{x _{2} - x _{1}}

▼

Evaluate

Distr

Distribute $- 1$

Rate of Change = \frac{m x _{2} + b - m x _{1} - b}{x _{2} - x _{1}}

SubTerms

Subtract terms

Rate of Change = \frac{m x _{2} - m x _{1}}{x _{2} - x _{1}}

FactorOut

Factor out $m$

Rate of Change = \frac{m ( x _{2} - x _{1} )}{x _{2} - x _{1}}

MovePartNumLeft

$\frac{a \cdot b}{c} = a \cdot \frac{b}{c}$

Rate of Change = m \cdot \frac{x _{2} - x _{1}}{x _{2} - x _{1}}

QuotOne

$\frac{a}{a} = 1$

Rate of Change = m \cdot 1

MultByOne

$a \cdot 1 = a$

Rate of Change = m

As stated before,

m

is a real number and, therefore, a constant. It has been found that, no matter which two points on the line are used, the rate of change between them will always be a constant value. This constant value

m

is called the slope of the line.

Consecutive Points in a Data Set with Constant Rate of Change Lie on a Line

Consider a set of data points for which the rate of change between every pair of consecutive points is a constant value $m .$ Consider two consecutive points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}) .$

Rate of Change	Rewrite
$\frac{y _{2} - y _{1}}{x _{2} - x _{1}} = m$	$y_{2} - y_{1} = m (x_{2} - x_{1})$

Note that since the rate of change between every pair of consecutive points is constant, they all can be rewritten in a similar way.

Rate of Change	Rewrite
$\frac{y _{2} - y _{1}}{x _{2} - x _{1}} = m$	$y_{2} - y_{1} = m (x_{2} - x_{1})$
$\frac{y _{3} - y _{2}}{x _{3} - x _{2}} = m$	$y_{3} - y_{2} = m (x_{3} - x_{2})$
$⋮$	$⋮$
$\frac{y _{n} - y _{n - 1}}{x _{n} - x _{n - 1}} = m$	$y_{n} - y_{n - 1} = m (x_{n} - x_{n - 1})$

Next, it will be shown that every point from the data set satisfies the following linear equation in two variables.

y - y_{1} = m (x - x_{1})

This will be shown first for the third point from the data set,

(x_{3}, y_{3}) .

The expression on the right-hand side of the equation will be rewritten by adding and subtracting

y_{2}

to obtain an identity.

y_{3} - y_{1} = y_{3} - y_{1}

AddDiffZero

$a = a + y_{2} - y_{2}$

y_{3} - y_{1} = y_{3} - y_{2} + y_{2} - y_{1}

Since

(x_{3}, y_{3})

and

(x_{2}, y_{2}),

and

(x_{2}, y_{2})

and

(x_{1}, y_{1}),

are pairs of consecutive points, the differences

y_{3} - y_{2}

and

y_{2} - y_{1}

can be rewritten in terms of the constant rate of change

m

and the corresponding

x -

values.

y_{3} - y_{1} = y_{3} - y_{2} + y_{2} - y_{1}

SubstituteII

$y_{3} - y_{2} = m (x_{3} - x_{2})$ , $y_{2} - y_{1} = m (x_{2} - x_{1})$

y_{3} - y_{1} = m (x_{3} - x_{2}) + m (x_{2} - x_{1})

▼

Evaluate

FactorOut

Factor out $m$

y_{3} - y_{1} = m (x_{3} - x_{2} + x_{2} - x_{1})

AddTerms

Add terms

y_{3} - y_{1} = m (x_{3} - x_{1})

Therefore, the third point

(x_{3}, y_{3})

satisfies the linear equation

y - y_{1} = m (x - x_{1}) .

Now it will be shown that the fourth point of the data set,

(y_{4}, x_{4}),

also satisfies the equation.

y_{4} - y_{1} = y_{4} - y_{1}

AddDiffZero

$a = a + y_{3} - y_{3}$

y_{4} - y_{1} = y_{4} - y_{3} + y_{3} - y_{1}

Recall that it is known that

y_{4} - y_{3} = m (x_{4} - x_{3}

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Catch-Up and Review

Extra

Extra

Hint

Solution

Interpreting the Rate of Change

Finding the Rate of Change Between Different Times

Proof

Pairs of Points in a Line Have a Constant Rate of Change

Consecutive Points in a Data Set with Constant Rate of Change Lie on a Line