body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

Recommended

Tests

An error ocurred, try again later!

eCourses /

Algebra 1 /

Chapter {{ article.chapter.number }}

{{ article.number }}.

Show less Show more

Lesson Settings & Tools

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount }}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount }}
	{{ 'ml-lesson-time-estimation' \| message }}

Image Credits

{{ item.file.title }}
{{ presentation }}

No file copyrights entries found

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})

and, from the previous result, it is also known that

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

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

SubstituteII

$y_{4} - y_{3} = m (x_{4} - x_{3})$ , $y_{3} - y_{1} = m (x_{3} - x_{1})$

y_{4} - y_{1} = m (x_{4} - x_{3}) + m (x_{3} - x_{1})

▼

Evaluate

FactorOut

Factor out $m$

y_{4} - y_{1} = m (x_{4} - x_{3} + x_{3} - x_{1})

SubTerms

Subtract terms

y_{4} - y_{1} = m (x_{4} - x_{1})

As shown,

(x_{4}, y_{4})

satisfies the equation as well. This process can be repeated with all the points, and all will satisfy the linear equation

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

And since every point satisfies the equation, every point is a solution. Therefore, every point of the data set lies on the line representing this linear equation.

It is possible to use the information presented above to set a criterion for identifying data that follows a linear relation. Data sets that have a constant rate of change between consecutive points can be described by linear relations, and the rate of change is the slope of the line.

Example

Identifying Linear Tendencies in Cryptocurrencies

Diego's brother is very interested in crypotcurrencies and is telling Diego about how their values changed in the first weeks of the last year. One of them is of special interest to him because its graph is a straight line. He tried to print some information to show Diego, but his printer broke before it could print the complete tables of values for the cryptocurrencies.

Diego's brother wants to identify the table in which the values have a linear relationship. For each table, the left column represents the week number and the right column represents the value of the cryptocurrency in dollars. Which of the three tables best describes a linear relation?

Hint

Linear relations grow by equal differences over equal intervals.

Solution

In order to identify the data set that follows a linear relation, it is useful to remember that linear relations have a constant rate of change. Therefore, they grow by equal differences over equal intervals.

Linear j Relation ⇕ Constant Rate of Change

Let

x

represent the week number and

y

represent the value of the cryptocurrencies. The differences between consecutive

x -

and

y -

values will be calculated for each table. This will first be done for Table A.

Since the relation represented in Table A increases by equal differences in equal intervals, the data can be modeled with a linear relation. Now, the same procedure will be applied to other tables.

Since the differences between consecutive $y -$ values are not constant for equal intervals, Table B and Table C do not satisfy the condition. As such, they do not represent linear relations.

Pop Quiz

Identifying Linear Relations From Tables of Values

The following table shows the relationship between two variables, $x$ and $y .$ Look at the data carefully and determine whether the relationship between $x$ and $y$ can be described by linear equation.

Discussion

Linear Functions and Intercepts

In a previous example, it was shown how substituting a specific $x -$ value into a linear equation in two variables determined an associated $y -$ value. The set of all the points obtained in this way represent a line in the coordinate plane. This idea is depicted below by showing some points for a particular linear equation.

Linear equation with assiciated table of values and graph

This type of relation in which an input is assigned only one specific output is known as a function. To each input, typically represented by the independent variable

x,

the function assigns an output represented as

f (x) .

In the example above

y

depends on

x .

This can be denoted explicitly by using function notation.

Equation y = 2 x + 1 l a l a l a l a Function f (x) = 2 x + 1

In general, functions can assign an output value to the input value in many different ways and their graphs can be very different. A particular function whose graph is a line is known as a linear function. Therefore, a linear equation can represent a linear function.

Concept

Linear Function

A linear function is a function with a constant rate of change. A linear function can be represented by a linear equation in two variables. Graphically, a linear function is a nonvertical line.

Using that line, the rate of change can be determined by finding the horizontal change

Δ x

and the vertical change

Δ y

between any two points on the line. Any function whose graph is not a straight line cannot be linear.

When working with linear functions, there are two points of special interest. These are the points where the line crosses the axes, known as intercepts.

Concept

Intercept

The $x -$ intercept of a line is the $x -$ coordinate of the point where the line crosses the $x -$ axis. The $y -$ intercept of a line is the $y -$ coordinate of the point where the line crosses the $y -$ axis. The $y -$ intercept of an equation is also known as its initial value.

When talking about functions, the

x -

intercepts are the zeros of the function. Sometimes, only one coordinate of these points is referenced. For example, if the

x -

intercept lies at

(a, 0),

it can be said that the

x -

intercept is at

x = a .

The same is true for the

y -

intercept. A relation can have several intercepts. A function can have multiple

x -

intercepts, but it can only have one

y -

intercept.

Example

Interpreting and Finding the Intercepts of a Linear Function in Different Scenarios

Diego wants to make sure he has enough data available on his smartphone to stream a series while taking a trip with his family. He visits a local branch of his service provider and updates to the Unlimited Plan, which costs an initial service fee and a special price per gigabyte of data used. The graph represents the costs of the service plan.

Linear function representing the service provider plan

a What is the

y -

intercept of this linear function? Write the answer as a single number.

What does the

y -

intercept represent?

b After updating his service plan, Diego does some research online about his smartphone model. He finds the following linear function.

y = - 10 x + 100

In this function,

y

is the percentage of charge remaining in the battery after

x

number of hours using the phone for streaming videos. What is the

x -

intercept for this linear function? Write the answer as a single number.

What does the

x -

intercept represent?

Hint

a The

y -

intercept is the

y -

value corresponding to

x = 0 .

b The

x -

intercept is the

x -

value corresponding to

y = 0 .

Solution

a Recall that the

y -

intercept is the

y -

value corresponding to

x = 0 .

From the given graph, it is possible to see that the line intersects the

y -

axis at the point

(0, 4) .

Therefore, the $y -$ intercept is $4 .$ Since this is the cost of the service when $0$ gigabytes of data are used, $x = 0,$ it can be concluded that $$ 4$ is the initial service fee.

b In this case it is necessary to find the

x -

intercept of the given function. This is the

x -

value corresponding to

y = 0 .

Therefore, to find the

x -

intercept,

y = 0

will be substituted into the linear equation and the equation will be solved for

x .

y = - 10 x + 100

Substitute

$y = 0$

0 = - 10 x + 100

▼

Solve for

x

SubEqn

$LHS - 100 = RHS - 100$

- 100 = - 10 x

DivEqn

$LHS / (- 10) = RHS / (- 10)$

10 = x

RearrangeEqn

Rearrange equation

x = 10

It has been found that the

x -

intercept is

10 .

The given linear equation relates the number of hours of streaming

x

to the percentage of battery charge left on the phone

y .

This means that the

x -

intercept — the number of hours after which

y = 0

— represents the time it takes for the battery to drain completely.

Pop Quiz

Practice Identifying the Intercepts

The following applet alternates between a graph and a linear equation, both of which represent linear functions. Identify the indicated intercept of the given function and write the answer as a single number.

Interactive graph showing a table or expression

Discussion

Average Rate of Change

The rate of change of a nonlinear function is not constant — it may even change from point to point. To measure the change for this type of function, an average rate of change is defined by averaging the total change of the function over a specific interval. The following graph illustrates how the average rate of change depends on the interval considered.

The average rate of change can be thought of as an approximation for the rate of change of the nonlinear function. This approximation is done using the slope of a linear function passing through the endpoints of a specific interval.

Linear function 2*(x-1)+1 intersecting (x-1)^2+1 at the points (x1, y,1) and (x2, y2). 2 Legends denoting which function is linear and which is nonlinear are present.

The slope of the linear function can be calculated by using the points common to both functions,

(x_{1}, y_{1})

and

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

slope = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

The

y -

values that correspond to

x_{1}

and

x_{2}

are also the nonlinear function's values

f (x_{1})

and

f (x_{2}),

respectively. Then, the slope formula can be rewritten and identified as the average rate of change of the function

f (x)

over the interval

[x_{1}, x_{2}] .

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

Example

Modeling the Takeoff and Landing Speeds of an Airplane Using Average Rates of Change

Diego decides to fly back home after visiting his family in his hometown. He is very excited as the airplane readies for takeoff. The plane gets into position at the beginning of a long, straight runaway surface and starts accelerating. Diego enjoys the view through the window.

a On the screen in front of his seat, Diego can see statistics of the flight in progress. He selects the option to display the information about the takeoff as a table of values, which shows the distance covered by the airplane at a specific time period during the takeoff process. He obtains the following information.

Time (seconds)	Distance (meters)
$0$	$0$
$3$	$18$
$6$	$72$
$9$	$162$
$12$	$288$
$15$	$450$
$18$	$648$
$21$	$882$

According to this table, which of the following options best approximates the speed of the plane at takeoff?

b The trip has been completed and the plane is finally landing. Diego is curious about the statistics for the landing process. This time he chooses to display the data using a graph. The graph shows the distance covered after the airplane came in contact with the runaway and started decelerating.

What is the average speed during the first

28

seconds of the landing?

What is the average speed during the time interval

t = 24

seconds to

t = 28

seconds?

Hint

a The rate of change represents the speed of the plane, which is not constant. Therefore, the takeoff speed can be approximated by averaging over an interval as specific to the takeoff as possible.

b Identify the intervals of interest and use the formula for the average rate of change.

Solution

a In this case, the rate of change represents the speed of the plane. Since the plane is accelerating, its speed is increasing and the rate of change from the table of values is not constant. Then, the best that can be done is to approximate the speed by using the average rate of change formula.

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

Note that since distance covered depends on the time value but time has no restriction, distance

d

is the dependent variable and time

t

is the independent variable. Now, to best approximate the speed at takeoff, the interval used for finding the average rate of change should be as short and as close to takeoff as possible.

By doing this, the calculation will not average the takeoff speed with the speeds at the beginning and the middle of the motion, and a better approximation for the speed at time of takeoff alone will be obtained. As indicated by the arrows on the table above, the coordinate pairs to be used are

(18, 648)

and

(21, 882) .

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

SubstitutePoints

Substitute $(18, 648)$ & $(21, 882)$

Average Rate of Change = \frac{8 8 2 - 6 4 8}{2 1 - 1 8}

▼

Evaluate

SubTerms

Subtract terms

Average Rate of Change = \frac{2 3 4}{3}

CalcQuot

Calculate quotient

Average Rate of Change = 78

It was found that in the interval of time just before takeoff, the average speed of the plane was

78 \frac{m}{s} .

b For this exercise, the average speed during the first

28

seconds of the landing process must be found. To find it, the interval to be used is

t = 0

seconds to

t = 28

seconds. The average speed can be found by using these values together with their corresponding distances.

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

SubstitutePoints

Substitute $(0, 0)$ & $(28, 1176)$

Average Rate of Change = \frac{1 1 7 6 - 0}{2 8 - 0}

▼

Evaluate

SubTerms

Subtract terms

Average Rate of Change = \frac{1 1 7 6}{2 8}

CalcQuot

Calculate quotient

Average Rate of Change = 42

The average speed during the first

28

seconds of the landing process is

42 \frac{m}{s} .

The average speed during the smaller interval,

t = 24

seconds to

t = 28

seconds, can be found in the same way.

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

SubstitutePoints

Substitute $(24, 1152)$ & $(28, 1176)$

Average Rate of Change = \frac{1 1 7 6 - 1 1 5 2}{2 8 - 2 4}

▼

Evaluate

SubTerms

Subtract terms

Average Rate of Change = \frac{2 4}{4}

CalcQuot

Calculate quotient

Average Rate of Change = 6

During this period, the average speed of the plane was just

6 \frac{m}{s},

as this calculation included only a short period of time closer to when the plane stopped moving. It does not include the data corresponding to times when the plane had a greater speed, as the average speed for the first

28

seconds of the landing process does.

Pop Quiz

Practice Calculating the Average Rate of Change

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

Alternating table of values / graphs for nonlinear data

Discussion

Rate of Change for Graphs Comprised of Different Lines

It is important to mention that the graphs of some functions can be comprised of straight lines with different slopes. The average rate of change is used for these functions as well, since the graph is not a single straight line and the rate of change is not constant.

With this information in mind, consider again the challenge presented at the beginning of this lesson.

Example

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

Answer

120

meters per minute

b Graph:

Average Speed: $100$ meters per minute

c No, see solution.

Hint

a Use the formula for the average rate of change. Use the information from the graph to identify the appropriate interval to be used in the calculation.

b Use the formula for the average rate of change. Use the information from the graph to identify the appropriate interval to be used in the calculation.

c Think carefully about what is being averaged in Part B.

Solution

a In this case, the dependent variable represents the distance covered in meters, while the independent variable represents the time in minutes required to cover that distance. The units of the rate of change are the ratio of the units of the dependent variable to the units of the independent variable.

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

Because the rate of change is quantifying how much distance is covered per minute, it can be interpreted as Dominika's speed. Note that the rate of change is not constant. For this reason, the average speed can then be found by using the the average rate of change formula.

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

Because the average speed during the whole trip is desired, the interval of interest is

t_{1} = 0

t_{2} = 10

minutes. From the graph, the corresponding distances at times

t_{1}

and

t_{2}

can be found as follows.

t_{1} = 10 t_{2} = 10 \Rightarrow d_{1} = 0 \Rightarrow d_{2} = 1200

The average speed will now be calculated.

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

SubstitutePoints

Substitute $(0, 0)$ & $(10, 1200)$

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

▼

Evaluate

SubTerms

Subtract terms

Average Rate of Change = \frac{1 2 0 0}{1 0}

CalcQuot

Calculate quotient

Average Rate of Change = 120

Therefore, Dominika's average speed during the her walk to school was

120

meters per minute. Note that this speed does not correspond to the rate of change of any of the lines on the graph. Instead, this is the rate of change of the line passing through the initial and final points.

b To draw the graph, remember that Dominika waited for the bus for the first

9

minutes. During this period of time, her position did not change and she was constantly

0

meters away from school.

After this time, she got on the bus and arrived back home at $t = 12$ minutes. Recall that the distance between her home and her school is $1200$ meters.

Now that the graph is complete, the average speed for the return trip will be calculated. This time, the interval of interest is

t_{1} = 0

t_{2} = 12

minutes. From the graph, the corresponding distances at times

t_{1}

and

t_{2}

can be found as follows.

t_{1} = 10 t_{2} = 12 \Rightarrow d_{1} = 0 \Rightarrow d_{2} = 1200

The average speed will now be calculated.

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

SubstitutePoints

Substitute $(0, 0)$ & $(12, 1200)$

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

▼

Evaluate

SubTerms

Subtract terms

Average Rate of Change = \frac{1 2 0 0}{1 2}

CalcQuot

Calculate quotient

Average Rate of Change = 100

The average speed on the trip home was

100

meters per minute. Once more, note that this speed is the rate of change of the line passing through the initial and final points of the graph.

c The fact that Dominika's average speed on her way to school was greater than the average speed when returning home may seem confusing since she went to school on foot and returned by bus. How can this be possible?

Average speed going to school on foot 120 \frac{m}{min} Average speed returning from school by bus 100 \frac{m}{min}

This does not mean that Dominika's average speed was greater than the bus's. For the calculation of the return trip, the

9

minutes that Dominika waited — and therefore did not move — were averaged together with the motion of the bus. The following graph shows the

waiting

period

and the

bus

ride

in different colors.

The average speed of the bus can be found by repeating the same calculation but considering only the

last

3

minutes

of the return trip — that is, using only the interval from

t_{1} = 9

t_{2} = 12 .

t_{1} = 19 t_{2} = 12 \Rightarrow d_{1} = 0 \Rightarrow d_{2} = 1200

Now the calculation will be repeated using these values.

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

SubstitutePoints

Substitute $(9, 0)$ & $(12, 1200)$

Average Rate of Change = \frac{1 2 0 0 - 0}{1 2 - 9}

▼

Evaluate

SubTerms

Subtract terms

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

CalcQuot

Calculate quotient

Average Rate of Change = 400

This result shows that the actual average speed of the bus is

400

meters per minute. This value corresponds to the rate of change of the

last

segment

in the previous graph. It is also considerably faster than Dominika's average walking speed of

120

meters per minute.

Closure

Summary and Further Applications for the Rate of Change

In this lesson, the concepts of $x -$ and $y -$ intercepts, as well as linear functions, were introduced. However, it is important to note that the concept of a function is more general and that there are also nonlinear functions.

Examples of different graphs showing linear and nonlinear functions

Other main ideas of the lesson were the rate of change and the average rate of change. These basic concepts lead to the definition of instantaneous rate of change. Unlike the average rate of change, which is an approximation for an interval, the instantaneous rate of change gives the exact value of the rate of change at a specific point.

Examples of different graphs showing average rate of change and instantaneous rate of change

The definition of an instantaneous rate of change has many different applications in real-life situations. It is used to calculate the thrust of rockets, the exact velocity of projectiles, electrical current in circuits, and to design optimization processes for software to run faster or maximize the profits of a business, among other things.

Loading content

{{ article.displayTitle }}

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

Hint

Solution

Hint

Solution

Hint

Solution

Answer

Hint

Solution