Chapter 7. Thinking Mathematically. Algebra: Graphs, Functions, and Linear Systems. Seventh Edition

Thinking Mathematically

Chapter 7

Algebra: Graphs,

Functions, and

Linear Systems

Objectives

1. Plot points in the rectangular coordinate system.

2. Graph equations in the rectangular coordinate system.

3. Use function notation.

4. Graph functions.

5. Use the vertical line test.

6. Obtain information about a function from its graph.

Points and Ordered Pairs

The horizontal number line is the x-axis. The vertical number line is the y-axis.

The point of intersection of these axes is their zero point, called the origin.

Negative numbers are shown to the left of and below the origin. The axes divide the plane into four quarters called “quadrants”.

Points and Ordered Pairs

Each point in the rectangular coordinate system

corresponds to an ordered pair of real numbers, (x, y). Look at the ordered pairs (−5, 3) and (3, −5).

The first number in each pair, called the

x-coordinate, denotes the distance and direction from the origin along the

x-axis.

The second number in each pair, called the y-coordinate, denotes the vertical distance and

direction along the x-axis or parallel to it.

The figure shows how we plot, or locate the points corresponding to the ordered pairs.

Example 1: Plotting Points in the

Rectangular Coordinate System

Plot the points: A(3, 5 ,) B(2, 4 , ) C

 

 

 

E and F

Solution: We move from the origin and plot the

point in the following way:

 3, 5 :

A  _{3 units left, 5 units up}

 2, 4 :

B 2 units right, 4 units down  5, 0 :

C 5 units right, 0 units up or down

 5, 3 :

Graphs of Equations

A relationship between two quantities can be expressed as an equation in two variables, such as

2

4 .

y =  x

A solution of an equation in two variables, x and y, is an ordered pair of real numbers with the following property:

When the x-coordinate is substituted for x and the y coordinate is substituted for y in the equation, we obtain a true statement.

The graph of an equation in two variables is the set of all points whose coordinates satisfy the equation.

Example 2: Graphing an Equation

Using the Point−Plotting Method

Graph y = 4  x2. Select integers for x, starting with −3

and ending with 3.

Solution: For each value of x, we find the

Example 2: Graphing an Equation

Using the Point−Plotting Method

Now plot the seven points and join them with a smooth curve.

Functions

If an equation in two variables (x and y) yields

precisely one value of y for each value of x, we say that y is a function of x.

The notation y  f x( ) _{indicates that the variable y is a} function of x. The notation f x( ) is read “f of x.”

Example 3: An Application Involving

Function Notation

Tailgaters beware: If your car is going 35 miles per hour on dry pavement, your required stopping

distance is 160 feet, or the width of a football field. At 65 miles per hour, the distance required is 410 feet, or approximately the length of one and one-tenth football fields. Figure (a) on the next slide shows

stopping distances for cars at various speeds on dry roads and on wet roads. Figure (b) on the next slide uses a line graph to represent stopping distances at various speeds on dry roads.

Example 3: An Application Involving

Example 3: An Application Involving

Function Notation

a. Use the line graph to estimate a car’s required stopping distance at 60 miles per hour on dry

Example 3: An Application Involving

Function Notation

Solution a. The required stopping distance at 60

miles per hour is estimated using the point shown below. The second coordinate of this point extends slightly more than midway between 300 and 400

on the vertical axis. Thus, 360 is a reasonable

estimate. We conclude that at 60 miles per hour on dry

Example 3: An Application Involving

Function Notation

models a car’s required stopping distance, f x( ), _{in feet, on}

dry pavement traveling at x miles per hour. Use this function to find the required stopping distance at 60 miles per hour. Round to the nearest foot.

2 ( ) 0.0875 – 0.4 66.6 f x  x x   2   (60) 0.0875 60 – 0.4 60 66.6 f       0.0875 3600 – 0.4 60 66.6   315 24 66.6    357.6 358  

The model indicates that the required stopping distance on dry pavement at 60 miles per hour is approximately 358 feet.

Example 4: Graphing Functions

Graph the functions f x( )  2 x and g x( )  2x  4 in the same

rectangular coordinate system. Select integers for x from −2 to 2, inclusive.

Solution: For each function we use tables to display the

Example 4: Graphing Functions

Next, plot the five points for each function and connect them.

Vertical Line Test for Functions

If any vertical line intersects a graph in more than one point, the graph does not define y as a function of x.

Example 5: Using the Vertical Line

Test

Use the vertical line test to identify graphs in which y is a function of x.

Example 5: Using the Vertical Line

Test

Example 6: Analyzing the Graph of a

Function

The given graph illustrates the body temperature from 8 a.m. through 3 p.m. Let x be the number of hours after 8 a.m. and y be the body temperature at time x.

a. What is the temperature at 8 a.m.? b. During which period of time is your temperature decreasing?

c. Estimate your minimum

temperature during the time period shown. How many hours 8 a.m. does this occur? At what time does this

Example 6: Analyzing the Graph of a

Function

d. During which period of time is your temperature increasing? e. Part of the graph is shown as a horizontal line segment. What does this mean about your

temperature and when does this occur?

Example 6: Analyzing the Graph of a

Function

Solution:

a. The temperature at 8 a.m. is when x is 0, since no time has passed when it is 8 a.m. Thus, the

temperature at 8 a.m. is 101°F. b. The temperature is

decreasing when the graph falls from left to right. This occurs

between x = 0 and x = 3. Thus, the temperature is decreasing between the times 8 a.m. and 11 a.m.

Example 6: Analyzing the Graph of a

Function

c. The minimum temperature can be found by locating the lowest point on the graph. This point lies above 3 on the horizontal axis. The y-coordinate falls midway between 98 and 99, at approximately 98.6. Thus, the minimum temperature is 98.6°F at 11 a.m.

d. The temperature is increasing when the graph rises from left to right. This occurs between x = 3 and x = 5. Thus, the temperature

Example 6: Analyzing the Graph of a

Function

e. The horizontal line segment shown indicates that the

temperature is neither increasing nor decreasing. The temperature remains the same at 100°F,

between x = 5 and x = 7. Thus, the temperature is at a constant 100°F between 1 p.m. and 3 p.m.

f. By the vertical line test we can see that no vertical line will intersect the graph more than once. So, the body temperature is a function of time. Each hour

Thinking Mathematically

Chapter 7

Algebra: Graphs,

Functions, and

Linear Systems

Section 7.2 Linear Functions and Their

Graphs

Objectives

1. Use intercepts to graph a linear equation.

2. Calculate slope.

3. Use the slope and y-intercept to graph a line.

4. Graph horizontal and vertical lines.

5. Interpret slope as a rate of change.

Graphing Using Intercepts

All equations of the form Ax  By  C are straight lines when graphed, as long as A and B are not both

zero, and are called linear equations in two

variables.

Locating Intercepts

To locate the x-intercept, set y = 0 and solve the equation for x.

To locate the y-intercept, set x = 0 and solve the equation for y.

Example 1: Using Intercepts to Graph

a Linear Equation

Graph: 3x  2y  6.

Solution:

Find the x-intercept by letting y = 0 and

solving for x.

3x  2y  6 3x   2 0 6

Find the y-intercept by letting x = 0 and

solving for y.

3x  2y  6 3 0 2  y  6

Example 1: Using Intercepts to Graph

a Linear Equation

The x-intercept is 2, so the line passes through the point (2, 0). The y-intercept is 3, so the line passes through the point (0, 3).

Now, we verify our work by checking for x = 1. Plug x = 1 into the given linear equation. We leave this to the student. For x = 1, the y-coordinate should be 1.5.

Slope

The slope of the line through the distinct points

1, 1) 2, 2) (x y and (x y is 2 1 2 1 Change in rise Change in run y x y y x x    

Example 2: Using the Definition of

Slope

Find the slope of the line passing through the pair of points: (−3, −1) and (−2, 4).

Solution: Let (x y₁, ₁)   ( 3, 1 ) and (x y₂, ₂)  ( 2, 4 .)

We obtain the slope such that

2 1 2 1 Change in 4 ( 1) 5 5. Change in 2 ( 3) 1             y y y m x x x

The Slope-Intercept Form of the

Equation of a Line

The slope-intercept form of the equation of a

nonvertical line with slope m and y-intercept b is

. y  mx b

The Slope-Intercept Form of the

Equation of a Line

Graphing

y



mx b



1. Plot the point containing the intercept on the y-axis. This is the point (0, b).

2. Obtain a second point using the slope m. Write m as a fraction, and use rise over run, starting at

the point containing the y-intercept, to plot this point.

3. Use a straightedge to draw a line through the two points. Draw arrowheads at the end of the line to show that the line continues indefinitely in both

Example 3: Graphing by Using the

Slope and y-Intercept

Graph the linear function 2 2 3

 

y x by using the slope

and y-intercept.

Solution: Since the graph is given in slope-intercept

Example 3: Graphing by Using the

Slope and y-Intercept

Step 1 Plot the point containing the y-intercept on the y-axis.

The y-intercept is (0, 2).

Step 2 Obtain a second point

using the slope, m. The slope as a fraction is already given:

. run rise 3 2 _  m

We plot the second point at (3, 4).

Step 3 Use a straightedge to draw a line through the two

Example 4: Graphing by Using the

Slope and the y-Intercept

Graph the linear function _2x  _5y  ₀ by using the slope and y-intercept.

Solution: We put the equation in slope-intercept

form by solving for y.

2 5 0 2 2 5 2 0 5 2 x y x x y x y x         

Example 4: Graphing by Using the

Slope and the y-Intercept

Next, we find the slope and

y-intercept:

Start at y-intercept (0, 0) and obtain a second point by using the slope. 2 2 rise 5 5 run      m

We obtain (5, −2) as the second point and use a straightedge to draw the line through these points.

Equations of Horizontal and Vertical

Lines

The graph of y = b or ( )

f x  b is a horizontal

line. The y-intercept is b.

The graph of x = a is a vertical line. The x-intercept is a.

Example 5: Graphing a Horizontal

Line

Graph y = −4 in the rectangular coordinate system.

Solution: All ordered pairs have y-coordinates

that are −4. Any value can be used for x. We graph the three ordered pairs in the table: (−2,−4), (0, −4), and (3,−4). Then use a straightedge to

Example 5: Graphing a Horizontal

Line

Example 6: Graphing a Vertical

Line

Graph x = 2 in the rectangular coordinate system.

Solution: All ordered

pairs have the

x-coordinate 2. Any value can be used for y. We graph the ordered pairs (2, −2), (2, 0), and (2, 3). Drawing a line that passes through the three points gives the vertical line.

Example 6: Graphing a Vertical

Line

Horizontal and Vertical Lines

The graph of y = b or f x( )  b

is a horizontal line. The y-intercept is b.

The graph of x = a is a vertical line. The

Slope as Rate of Change

Slope is defined as a ratio of a change in y to a corresponding change in x.

Slope can be interpreted as a rate of change in an applied situation.

Example 7: Slope as a Rate of

Change

The line graphs in figure show the percentage of

American men and women ages 20 to 24 who were married from 1970 through 2010. Find the slope of the line segment

representing women.

Describe what the slope represents.

Example 7: Slope as a Rate of

Change

Solution: We let x represent a year and y the percentage of

married women ages 20–24 in that year. The two points shown on the line segment for women have the following coordinates (1970, 65) and (2010, 21). Change in 21 65 Change in 2010 1970 44 1.1 40 y m x        

Modeling Data with the Slope-Intercept

Form of the Equation of a Line

Linear functions are useful for modeling data that fall on or near a line.

Thinking Mathematically

Chapter 7

Algebra: Graphs,

Functions, and

Linear Systems

Section 7.3 Systems of Linear

Equations in Two Variables

Objectives

1. Decide whether an ordered pair is a solution of a linear system.

2. Solve linear systems by graphing.

3. Solve linear systems by substitution.

4. Solve linear systems by addition.

5. Identify systems that do not have exactly one ordered-pair solution.

Systems of Linear Equations & Their

Solutions

Two linear equations are called a system of linear

equations or a linear system.

A solution to a system of linear equations in two

variables is an ordered pair that satisfies both

Example 1: Determining Whether an

Ordered Pair is a Solution of a System

Determine whether (1, 2) is a solution of the system:

2 3 4 2 4 x y x y     

Solution: Because 1 is the x-coordinate and 2 is

the y-coordinate of (1, 2), we replace x with 1 and

Example 1: Determining Whether an

Ordered Pair is a Solution of a System

? ? 2 3 4 2(1) 3(2) 4 2 6 4 4 4, x  y          True ? ? 2 4 2(1) 2 4 2 2 4 4 4, x  y      True

The pair (1, 2) satisfies both equations; it makes

each equation true. Thus, the pair is a solution of the system.

Solving Linear Systems by Graphing

For a system with one solution, the coordinates of

the point of intersection of the lines is the system’s solution.

Example 2: Solving Linear Systems

by Graphing

We find the solution by graphing both x  2y  2 and

– 2 6

x y  in the same rectangular coordinate system. We will use intercepts to graph each equation.

Example 2: Solving Linear Systems

by Graphing

2

2 :

x



y



The line passes through (2,0). y-intercept: Set x = 0. 0 2 2 2 2 1 y y y    

The line passes through (0,1).

Example 2: Solving Linear Systems

by Graphing

The line passes through (6,0). y-intercept: Set x = 0. 0 2 6 2 6 3 y y y     

The line passes through (0,−3). We will graph x  2y  6 as a red line.

Example 2: Solving Linear Systems

by Graphing

We see the two graphs

intersect at (4,−1). Hence, this is the solution to the system.

We can check this by

substituting in (4,−1) into

each equation and verifying That the solution is true for both equations.

Solving Linear Systems by the

Substitution Method

Solving Linear Systems By Substitution

1. Solve either of the equations for one variable in terms of the other. (If one of the equations is already in this form, you can skip this step.)

2. Substitute the expression found in step 1 into the other equation. This will result in an equation in one variable.

3. Solve the equation containing one variable.

4. Back-substitute the value found in step 3 into the

equation from step 1. Simplify and find the value of the remaining variable.

Example 3: Solving a System by

Substitution

Solve by the substitution method:

1 4 3 24. y x x y      Solution:

Step 1 Solve either of the equations for one

Example 3: Solving a System by

Substitution

Step 2 Substitute the expression from step 1 into the other equation.

This gives us an equation in one variable, namely

4x    3( x 1) 24.

Example 3: Solving a System by

Substitution

Step 3 Solve the resulting equation containing one variable. 4 3( 1) 24 4 3 3 24 7 3 24 7 21 3 x x x x x x x           

Example 3: Solving a System by

Substitution

Step 4 (continued)

With x = 3 and y = −4, the proposed solution is (3, −4).

Step 5 Check. Use this ordered pair to verify that this

solution makes each equation true. We leave this to the student.

Solving Linear Systems by the

Addition Method

Solving Linear Systems By Addition

1. If necessary, rewrite both equations in the form Ax  By  C.

2. If necessary, multiply either equation or both equations by appropriate nonzero numbers so that the sum of the

x-coefficients or the sum of the y-x-coefficients is 0.

3. Add the equations in step 2. The sum is an equation in one variable.

4. Solve the equation in one variable.

Example 4: Solving a System by the

Addition Method

Solve by the addition method:

3 2 48 9 8 24. x y x y      Solution:

Step 1 Rewrite both equations in the form

.

Ax  By  C Both equations are already in this form.

Variable terms appear on the left and constants appear on the right.

Step 2 If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x-coefficients or the sum of the y-coefficients is 0.

Example 4: Solving a System by the

Addition Method

Step 3 Add the equations.

Step 4 Solve the equation in one variable. We

solve 14y  168 by dividing both sides by −14.

14 168

Example 4: Solving a System by the

Addition Method

Step 5 Back-substitute and find the value for the other variable. 3 2 48 3 2(12) 48 3 24 48 3 24 8 x y x x x x        

Step 6 Check. The solution to the system is (8, 12).

We can check this by verifying that the solution is true for both equations. We leave this to the student.

Linear Systems Having No Solution

or Infinitely Many Solutions

The number of solutions to a system of two linear equations in two variables is given by one of the following:

Number of Solutions What This Means Graphically

Exactly one ordered-pair solution The two lines intersect at one point. No Solution The two lines are parallel.

Example 5: A System with No Solution

Solve the system:

4 6 12 6 9 12. x y x y    

Solution: Because no variable is isolated, we will use the

addition method.

The false statement 0 = 12 indicates that the system has no solution. The solution is the empty set, .

Example 6: A System with Infinitely

Many Solutions

Solve the system:

3

2

15

5

10.

y

x

x

y









Solution: Because the variable y is isolated in

3 2,

y  x  the first equation, we will use the

Example 6: A System with Infinitely

Many Solutions

3 2

y  x  _15x  _5y ₁₀ Substitute 3x 2 for y.

15x 5(3x  2) 10 The substitution results in an equation in one variable.

15x 15x 10 10

Apply the distributive property. Simplify.

The statement 10 = 10 is true. Hence, this indicates that the system has infinitely many solutions.

Modeling with Systems of Equations:

Making Money (and Losing It)

Revenue and Cost Functions

A company produces and sells x units of a product.

Revenue Function R x( )  (price per unit sold)x

Cost Function C x( )  fixed cost + (cost per unit produced)x The point of intersection of the graphs of the revenue and cost functions is called the break-even point.

Example 7: Finding a Break-Even

Point

A company is planning to manufacture radically

different wheelchairs. Fixed cost will be $500,000 and it will cost $400 to produce each wheelchair. Each

wheelchair will be sold for $600.

a. Write the cost function, C, of producing x wheelchairs.

b. Write the revenue function, R, from the sale of x wheelchairs.

Example 7: Finding a Break-Even

Point

Solution:

a. The cost function is the sum of the fixed cost and the variable cost.

b. The revenue function is the money generated from the sale of x wheelchairs.

Example 7: Finding a Break-Even

Point

c. The break even point occurs where the graphs of

C and R intersect. Thus, we find this point by

solving the system

( ) 500, 000 400 ( ) 600 , C x x R x x    or 500, 000 400 600 . y x y x   

Example 7: Finding a Break-Even

Point

Using substitution, we substitute 600x in for y in the first equation:

600 500, 000 400 200 500, 000 2500 x x x x    

Back-substituting 2500 for x in either of the system’s equations (or functions), we obtain

Example 7: Finding a Break-Even

Point

The break-even point is (2500, 1,500,000). This

means that the company will break even if it produces and sells 2500 wheelchairs for $1,500,000.

The Profit Function

The profit, _{( ),P x} generated after producing and selling x units of a product is given by the profit function

( ) ( ) ( ),

P x  R x C x

where R and C are the revenue and cost, respectively. The profit function, _{P x( ),}

for the previous example is ( ) ( ) ( )

Thinking Mathematically

Chapter 7

Algebra: Graphs,

Functions, and

Linear Systems

Section 7.4 Linear Inequalities in

Two Variables

Objectives

1. Graph a linear inequality in two variables.

2. Use mathematical models involving linear inequalities.

Linear Inequalities in Two Variables

and Their Solutions

If we change the symbol = in the equation Ax  By  C to

, , ,

,

  

or



variables.

For example, x  y 2 and 3x 5y 15 are linear

inequalities in two variables.

A solution of an inequality in two variables, x and y, is an ordered pair of real numbers such that when the

The Graph of a Linear Inequality in

Two Variables

The graph of an inequality in two variables is the set of all points whose coordinates satisfy the inequality.

The Graph of a Linear Inequality in

Two Variables

Graphing a Linear Inequality in Two Variables

1. Replace the inequality symbol with an equal sign and

graph the corresponding linear equation. Draw a solid line if the original inequality contains a  or symbol.

Draw a dashed line if the original inequality contains a < or > symbol.

2. Choose a test point from one of the half-planes. (Do not

choose a point on the line.) Substitute the coordinates of the test point into the inequality.

Example 1: Graphing a Linear

Inequality in Two Variables

Graph: 3x  5y 15.

Solution:

Step 1 We need to graph 3x  5y 15.

We can use intercepts to graph this line. We set y = 0 to find the x-intercept. 3x  5y  15 3x   5 0 15 3x  15 5 x  We set x = 0 to find the y-intercept. 3x  5y  15 3 0 5  y  15 5y 15   3 y  

Example 1: Graphing a Linear

Inequality in Two Variables

The x-intercept is 5, so the line passes through (5,0). The

y-intercept is −3, so the line passes through (0,−3).

Step 2 We choose (0,0) as a test

point.

3x 5y 15 3 0 5 0 15   

Example 1: Graphing a Linear

Inequality in Two Variables

Step 3 Since the statement is

false, we shade the

half-plane that does not include the test point (0,0).

Thus, the graph with the

shading is the solution to the given inequality.

Example 2: The Graph of a Linear

Inequality in Two Variables

Graph:

2

.

3

y

 

x

Step 1 We need to graph

2

.

3

y

 

x

Example 2: The Graph of a Linear

Inequality in Two Variables

Step 2 We choose a test point not on the line, (1, 1),

which lies in the half-plane above the line.

2 3 2 1 3 2 1 3 y x y        _True!

Step 3 Since the statement is true, then we shade

Graphing Linear Inequalities Without

Using Test Points

For the vertical line x = a:

If x > a, shade the half-plane to the right of x = a.

If x < a, shade the half-plane to the left of x = a.

For the horizontal line y = b:

• If y > b, shade the half-plane above y = b.

• If y < b, shade the half-plane below y = b.

Example 3: Graphing Inequalities

Without Using Test Points

Graph each inequality in a rectangular coordinate system:

a.

y

 

3

x



2

Solution:

Modeling with Systems of Linear

Inequalities

Just as two or more linear equations make up a system of linear equations, two or more linear inequalities make up a system of linear

inequalities. A solution of a system of linear

inequalities in two variables is an ordered pair that

Graphing Systems of Linear

Inequalities

The solution set of a system of linear inequalities

in two variables is the set of all ordered pairs that

Example 4: Graphing a System of

Linear Inequalities

Graph the solution set of the system: 1

x  y

2x  3y 12.

Solution: Replacing each inequality symbol with an

equal sign indicates that we need to graph x − y = 1 2x  3y 12.

and We can use intercepts to graph

Example 4: Graphing a System of

Example 4: Graphing a System of

Linear Inequalities

Now we are ready to graph the solution set of the system of linear inequalities.

Thinking Mathematically

Chapter 7

Algebra: Graphs,

Functions, and

Linear Systems

Objectives

1. Write an objective function describing a quantity that must be maximized or minimized.

2. Use inequalities to describe limitations in a situation.

Objective Functions in Linear

Programming

A method for solving problems in which a particular quantity that must be maximized or minimized is

limited by other factors is called linear

programming.

An objective function is an algebraic expression in two or more variables describing a quantity that must be maximized or minimized.

Example 1: Writing an Objective

Function

Bottled water and medical supplies are to be

shipped to victims of an earthquake by plane. Each container of bottled water will serve 10 people and each medical kit will aid 6 people. Let x represent the number of bottles of water to be shipped and y the number of medical kits. Write the objective

function that describes the number of people that can be helped.

Example 1: Writing an Objective

Function

Solution: Because each water serves 10 people

and each medical kit aids 6 people, we have

Using z to represent the number of people helped, the objective function is z 10x  6 .y

Constraints on Linear Programming

A constraint is expressed as a linear inequality. The list of constraints forms a system of linear inequalities.

Example 2: Writing a Constraint

Each plane can carry no more than 80,000 pounds. The bottled water weighs 20 pounds per container and each medical kit weighs 10 pounds. Let x

represent the number of bottles of water to be

shipped and y the number of medical kits. Write an inequality that describes this constraint.

Example 2: Writing a Constraint

Solution: Because each plane carries no more

than 80,000 pounds, we have

The plane’s weight constraint is described by the inequality

Solving Problems with Linear

Programming

Let z  ax by be an objective function that depends on x and y. Furthermore, z is subject to a number of constraints on x and y. If a maximum or minimum value exists, it can be determined as follows:

1. Graph the system of inequalities representing the constraints.

2. Find the value of the objective function at each corner, or vertex, of the graphed region. The

Example 3: Solving a Linear

Programming Problem

Determine how many bottles of water and how

many medical kits should be sent on each plane to maximize the number of earthquake survivors who can be helped.

We must maximize z 10x 6y subject to the following constraints:

20 10 80, 000 6000. x y x y     _{ } 

Example 3: Solving a Linear

Programming Problem

Step 1 Graph the system of inequalities representing the constraints.

20x 10y  80, 000 20x 10y  80, 000 x-intercept 4000 y-intercept 8000 20(0) 10(0)  80, 000 0  80, 000 6000 x  y 6000 x  y x-intercept 6000 y-intercept 6000 0 0  6000 0  6000

Example 3: Solving a Linear

Programming Problem

Step 1 Graph the system of inequalities representing the constraints.

20 10 80,000 6000 x y x y     _{ }   10   20 10 80,000 10 10 60,000 x y x y         Add: 10x  _20,000 2000 x  y  4000 (2000, 4000) Intersection :

Example 3: Solving a Linear

Programming Problem

Step 2 Find the value of the objective function at each corner of the graphed

region. The maximum and minimum of the objective function occur at one or more of the corner points.

(0,0) z 10(0) 6(0)  0

Example 3: Solving a Linear

Programming Problem

Thus, the maximum value of z is 44,000 and this occurs when x = 2000 and y = 4000.

In practical terms, this means that the maximum number of earthquake survivors who can be helped with each plane shipment is 44,000. This can be accomplished by sending 2000 water bottles and 4000 medical kits per plane.

Thinking Mathematically

Chapter 7

Algebra: Graphs,

Functions, and

Linear Systems

Section 7.6 Modeling Data:

Exponential, Logarithmic, and

Quadratic Functions

Objectives

1. Graph exponential functions.

2. Use exponential models.

3. Graph logarithmic functions.

4. Use logarithmic models.

5. Graph quadratic functions.

Scatter Plots & Regression Lines

Data presented in a visual form as a set of points is called a scatter plot.

A line that best fits the data points in a scatter plot is called a regression line.

For example, the graph displays the relationship between literacy and

child mortality. Each point represents a country.

Modeling with Exponential Functions

Definition of the Exponential Function

The exponential function f with base b is defined by

 

x x

y  b or f x  b

where b is a positive constant other than 1 (b > 0 and

1

Example 1: Graphing an Exponential

Function

Graph: f x

 

Solution: We start by

selecting numbers for x and finding ordered pairs. We make a table:

Example 1: Graphing an Exponential

Function

Next, plot the ordered pairs and connect them with a smooth curve.

Example 2: Comparing Linear and

Exponential Models

The graphs below show the world population for seven selected years from 1950 through 2010. One is a bar graph and the other is a scatter plot

Example 2: Comparing Linear and

Exponential Models

After entering the data in a calculator, the

graphing calculator displays the linear model, y  ax b , and the exponential model, y  abx, that best fit the data. a. Express each model in

function notation, with

numbers rounded to three decimal places

Example 2: Comparing Linear and

Exponential Models

c. By one projection, world population is expected to reach 8 billion in the year 2026. Which function

serves as a better model for this prediction?

Solution:

a. Using the figure from the graphing calculator, the functions

 

 





f x  x  and g x 

model world population, in billions, x years after 1949. The linear function is f and the exponential function g.

Example 2: Comparing Linear and

Exponential Models

b. The graph shows that the world population in 2000 was 6.1 billion. The year 2000 is 51 years after 1949. Hence, we substitute 51 for x in each function and then compare with the actual population in 2000.

 

 

 

 





Example 2: Comparing Linear and

Exponential Models

Since the world population was 6.1 billion, it seems both functions model world population well for 2000. c. Now, we compare the models to a world

population of 8 billion in the year 2026. We use 77 for x since 2026 is 77 years after 1949.

 

 

 

 

Example 2: Comparing Linear and

Exponential Models

 





 





 

It seems the linear function f x serves as a better

 

model for a projected world population of 8 billion by 2026.

The Role of e in Applied Exponential

Functions

An irrational number, symbolized by the letter

e

,

2.71828...

e



The number

e

 

Example 3: Alcohol and Risk of a Car

Accident

Medical research indicates that the risk of having a car accident increases exponentially as the

concentration of alcohol in the blood increases. The risk is modeled by

12.77

6 x,

R  e

where x is the blood alcohol concentration and R, given as a percent, is the risk of having a car

Example 3: Alcohol and Risk of a Car

Accident

Solution: We substitute 0.08 for x in the function.

12.77 6 x R  e   12.77 0.08 6 R  e

Putting this in the calculator, we get an

approximation of 16.665813. Rounding to one

decimal place, the risk of getting in a car accident is approximately 16.7% with a blood alcohol

Modeling with Logarithmic Functions

Definition of the Logarithmic Function

For x > 0 and b > 0, b ¹ 1, log_b

y  x _{is equivalent to} y _.

b  x

The function f x

 

with base b.

Location of Base and Exponent in Exponential and Logarithmic Forms

Example 4: Graphing a Logarithmic

Function

Graph: y  log₂ x.

Solution: Because y  log₂ x means 2y  x,

we will use the exponential form of the equation to obtain the function’s graph.

Example 5: Dangerous Heat: Temperature

in an Enclosed Vehicle

When the outside air temperature is anywhere from 72° to 96°F, the temperature in an enclosed vehicle climbs by 43°in the first hour. The bar graph and

Example 5: Dangerous Heat: Temperature

in an Enclosed Vehicle

After entering data in a graphing calculator, the calculator displays the logarithmic mode y  a bln , x where ln x

is called the natural logarithm. a. Express the model in function notation, with numbers rounded to one decimal place.

b. Use the function to find the temperature increase, to the

nearest degree, after 50 minutes. How well does the model resemble the actual increase?

Example 5: Dangerous Heat: Temperature

in an Enclosed Vehicle

Solution:

a. Using the calculator figure and rounding to one decimal place, the function

 

f x    x

Example 5: Dangerous Heat: Temperature

in an Enclosed Vehicle

Solution:

a. We substitute 50 for x and evaluate the function

 

 

 

Since the actual temperature increases 41° after 50 minutes, the function models the actual increase well.

Modeling with Quadratic Functions

A quadratic function is any function of the form

 

2 2

, y  ax  bx c or f x  ax  bx c

where a, b, and c are real numbers, with

a ¹

0.

The graph of any quadratic function is called a

parabola.

Modeling with Quadratic Functions

Vertex of a Parabola

The Vertex of a Parabola

The vertex of a parabola whose equation is

2 y  ax  bx  c _{occurs when}

2

b

x

a





Graphing Quadratic Equations

The graph of y  ax2 bx c or f x  ax2 bx c ,

called a parabola, can be graphed using the following steps:

1. Determine whether the parabola opens upward or downward. If a > 0, it opens upward. If a < 0, it opens downward.

2. Determine the vertex of the parabola. The x-coordinate is . 2

b a



The y-coordinate is found by substituting the x-coordinate into the parabola’s equation and evaluating y.

3. Find any x-intercepts by replacing y or f x( ) _{with 0. Solve the resulting} quadratic equation for x.

4. Find the y-intercept by replacing x with 0.

5. Plot the intercepts and the vertex.

Example 6: Graphing a Parabola

Graph the quadratic function: y  x2  2x 3.

Solution: We follow the steps:

1. Determine how the parabola opens. Since a is

the coefficient of x2 and a 1 in this case, then the parabola opens upward.

Example 6: Graphing a Parabola

We use the formula to find the x-coordinate:

( 2) 1 2 2(1) b x a      

We plug x = 1 into the original function to find the

y-coordinate: 2 2 3 y  x  x 

 

 

Example 6: Graphing a Parabola

3. Find the x-intercepts. Replace y with 0 in

2 2 3. y  x  x  2 2 3 y  x  x  2 0  x  2x  3







Example 6: Graphing a Parabola

4. Find the y-intercept. Replace x with 0 in

2 2 3. y  x  x  2 2 3 y  x  x 

 

Example 6: Graphing a Parabola

Steps 5. and 6. Plot the intercepts and vertex. Connect these points with a smooth curve.

Determine an Appropriate Function

for Modeling Data