Chapter 2 - Linear Functions

Algebra II

Table of Contents

• 2.1- Solving Linear Equations and Inequalities• 2.2- Proportional Reasoning• 2.3- Graphing Linear Functions• 2.4- Writing Linear Functions

2.1- Solving Linear Equations and Inequalities

Algebra II

2-1 Algebra 2 (bell work)

An equation is a mathematical statement that two expressions are equivalent

The solution set of an equation is the value or values of the variable that make the equation true.

A liner equation is one variable can be written in the form ax=b, where a and b are constants and a ≠ 0

The local phone company charges $12.95 a month for the first 200 of air time, plus $0.07 for each additional minute.

If Nina’s bill for the month was $14.56, how many additional minutes did she use?

2-1

monthly charge plus

additional minute charge times

12.95 0.07

number of additional minutes total charge

+

=

Let m represent the number of additional minutes that Nina used.

m 14.56* =

Model

Example 1 Application

Solve

12.95 + 0.07m = 14.56

0.07m = 1.61

0.07 0.07

m = 23

Nina used 23 additional minutes.

–12.95 - 12.95

2-1The local phone company charges $12.95 a month for the first 200 of air time, plus $0.07 for each additional minute.

If Nina’s bill for the month was $14.56, how many additional minutes did she use?

Let m represent the number of additional minutes that Nina used.

Stacked cups are to be placed in a pantry.

One cup is 3.25 in. high and each additional cup raises the stack 0.25 in.

How many cups fit between two shelves 14 in. apart?

one cupplus

additional cup height times

3.250.25

number of additional cups

total height

+

=

c 14.00* =

2-1

3.25 + 0.25c = 14.00

0.25c = 10.75

0.25 0.25

c = 43

44 cups fit between the 14 in. shelves.

Solve

–3.25 –3.25

2-1Stacked cups are to be placed in a pantry.

One cup is 3.25 in. high and each additional cup raises the stack 0.25 in.

How many cups fit between two shelves 14 in. apart?

Solve 4(m + 12) = –36

Method 1

The quantity (m + 12) is multiplied by 4, so divide by 4 first.

4(m + 12) = –36

4 4

m + 12 = –9

m = –21

–12 –12

2-1 Example 2 Solving Equations with the Distributive Property

Distribute before solving

4m + 48 = –36

4m = –84

–48 –48

=4m –84

4 4

m = –21

Solve 4(m + 12) = –36

Method 2

Method 1

3(2 – 3p) = 42

3 3

Solve 3(2 –3p) = 42.

–3p = 12

2 – 3p = 14 –2 –2

–3 –3

p = –4

2-1

Method 2

6 – 9p = 42

–9p = 36

–6 –6

=–9p 36

–9 –9

p = –4

Solve 3(2 – 3p) = 42

–15 + 12r = –9

12r = 6

+15 +15

=12r 6

12 12

r =

Solve –3(5 – 4r) = –92-1

If there are variables on both sides of the equation

1. Simplify each side.

2. Collect all variable terms on one side and all constants terms on the other side.

3. Isolate the variables

Just Read

Solve 3(w + 7) – 5w = w + 12

–2w + 21 = w + 12

+2w +2w

21 = 3w + 12

9 = 3w

3 33 = w

–12 –12

2-1

–11k + 25 = –6k – 10+11k +11k

25 = 5k – 10

35 = 5k

5 5

7 = k

+10 + 10

Solve 3k– 14k + 25 = 2 – 6k – 12

Example 3 Solving Equations with Variables on Both Sides

Define the following words

1. Identity – An equation that is true for all values of the variable, such as x = x 1. Answer is all real numbers, or R, or ( - ∞ , ∞ )

2. Contradiction – An equation that has no solution, such as 3 = 5. Can also use the empty set symbol

2-1 Algebra 2 (bell work)

Solve 3v – 9 – 4v = –(5 + v).

3v – 9 – 4v = –(5 + v)

–9 – v = –5 – v

+ v + v

–9 ≠ –5 x

The equation has no solution. The solution set is the empty set, which is represented by the symbol .

2-1 Example 4 Identifying Identities and ContradictionsSolve 2(x – 6) = –5x – 12 + 7x

2x – 12 = 2x – 12

–2x –2x

–12 = –12 The solutions set is all real number, or .

2-1 Solve 3(2 –3x) = –7x – 2(x –3)

3(2 –3x) = –7x – 2(x –3)

6 – 9x = –9x + 6

+ 9x +9x

6 = 6

The solutions set is all real numbers, or .

Solve 5(x – 6) = 3x – 18 + 2x

The equation has no solution. The solution set is the empty set, which is represented by the symbol

5(x – 6) = 3x – 18 + 2x

5x – 30 = 5x – 18

–5x –5x

–30 ≠ –18 x

These properties also apply to inequalities expressed with >, ≥, and ≤.

Just ReadIf you multiply or divide by a negative number, flip the inequality sign

2-1

Solve and graph 8a –2 ≥ 13a + 8

8a – 2 ≥ 13a + 8

–13a –13a

–5a – 2 ≥ 8

+2 +2

–5a ≥ 10

–5 –5

–5a ≤ 10

a ≤ –2

2-1 Example 5 Solving Inequalities

Solve and graph x + 8 ≥ 4x + 17

x + 8 ≥ 4x + 17

–x –x

8 ≥ 3x +17

–17 –17

–9 ≥ 3x

3 3 –9 ≥ 3x

–3 ≥ x or x ≤ –3

HW pg.94

• 2.1-– Day 1: 2-11, 21-25 (Odd), 41, 42, 45, 47– Day 2: 12-20, 34-39, 49, 62-65– Ch: 52

– Follow All HW Guidelines or ½ off

2.2- Proportional Reasoning

Algebra II

2-2

Copy boxed parts below

Solve each proportion.

A.

206.4 = 24p

=

=16 24 p 12.9

16 24

p 12.9

206.4 24p

24 24

8.6 = p

14 c 88 132

=

=

=

B.

14 c 88 132

88c = 1848

=88c 184888 88

c = 21

2-2Example 1

Solve each proportion.

A.

924 = 84y

=

=

y 77

12 84

y 77 12 84

11 = y

15 2.5 x 7

=

924 84y 84 84

=

=

B.

15 2.5 x 7

2.5x =105

=2.5x 1052.5 2.5

x = 42

2-2 Optional

Percent is a ratio that means per hundred.

For example:

30% = 0.30 =

Remember!

30100

Because percents can be expressed as ratios, you can use the proportion to solve percent problems.

2-2

A poll taken one day before an election showed that 22.5% of voters planned to vote for a certain candidate.

If 1800 voters participated in the poll, how many indicated that they planned to vote for that candidate?

Method 1 Use a proportion.

So 405 voters are planning to vote for that candidate.

Method 2 Use a percent equation.

22.5% 0.225

Percent (as decimal) whole = part

0.225 1800 = x

405 = x

x = 405

2-2Example 2 Solving Percent Problems

At Clay High School, 434 students, or 35% of the students, play a sport.

How many students does Clay High School have?

Method 1 Use a proportion.

Cross multiply.

Solve for x.

Clay High School has 1240 students.

Method 2 Use a percent equation.

0.35x = 434

35% = 0.35

x = 1240

Percent (as decimal) whole = part

x = 1240

100(434) = 35x

2-2

Ryan ran 600 meters and counted 482 strides. How long is Ryan’s stride in inches? (Hint: 1 m ≈ 39.37 in.)

Use a proportion to find the length of his stride in meters.

600 m 482 strides

x m 1 stride

=

600 = 482x

x ≈ 1.24 m

Ryan’s stride length is approximately 49 inches.

≈ 1.24 m

1 stride length39.37 in.

1 m49 in.

1 stride length

2-2Example 3 Fitness Application (Dimensional Analysis)

Use a proportion to find the length of his stride in meters.

Luis ran 400 meters in 297 strides. Find his stride length in inches.

x ≈ 1.35 m

400 = 297x

400 m 297 strides

x m 1 stride

=

Luis’s stride length is approximately 53 inches.

≈ 1.35 m

1 stride length39.37 in.

1 m53 in.

1 stride length

2-2 Optional

= height of ∆XAB width of ∆XAB

height of ∆XYZ width of ∆XYZ

=

3 x

9 6

9x = 18, so x = 2

BA

X

YZ

2-2Example 4 Scaling Geometric Figures in the Coordinate Plane

Math Joke

• Q: Why were the similar triangles weighing themselves

• A: they were finding their Scale

The tree in front of Luka’s house casts a 6-foot shadow at the same time as the house casts a 22-foot shadow.

If the tree is 9 feet tall, how tall is the house?

=6

9 h

22

=Shadow of tree Height of tree

Shadow of house Height of house

6h = 198

h = 33

The house is 33 feet high.

9 ft

6 ft

h ft

22 ft

2-2

A 6-foot-tall climber casts a 20-foot long shadow at the same time that a tree casts a 90-foot long shadow.

How tall is the tree?

= 20

6 h

90

=Shadow of climber Height of climber

Shadow of treeHeight of tree

20h = 540

h = 27

The tree is 27 feet high.

6 ft

20 ft

h ft

90 ft

2-2 Optional

HW pg. 100

• 2.2-– 7, 9- 11, 13, 18, 19, 21, 22, 25, 33, 36, 38, 68-71 – Ch: 35

– Follow All HW Guidelines or ½ off

1. Copy down the definition of a linear function

Functions with a constant rate of change are called linear functions.

A linear function can be written in the form f(x) = mx + b, where x is the independent variable and m and b are constants. (m = slope) (b = y-intercept)

The graph of a linear function is a straight line made up of all points that satisfy y = f(x).

2-3

Algebra II (Bell work)

2.3- Graphing Linear Functions

Algebra II

Determine whether the data set could represent a linear function.

x –2 0 2 4

f(x) 2 1 0 –1

+2

–1

+2

–1

+2

–1

The rate of change, = , is constant

So the data set is linear.

2-3Example 1 Recognizing Linear Functions

Determine whether the data set could represent a linear function.

x 2 3 4 5

f(x) 2 4 8 16

+1

+2

+1

+4

+1

+8

The rate of change, 2 ≠ 4 ≠ 8, is not constant.

So the data set is not linear.

2-3

Determine whether the data set could represent a linear function.

x 10 8 6 4

f(x) 7 5 1 –7

–2

–2

–2

–4

–2

–8

The rate of change, ,

is not constant. So the data set is not linear.

2-3

The constant rate of change for a linear function is its slope. The slope of a

linear function is the ratio or m= =

2-3

Graph the line with slope m= that passes through (–1, –3).

2-3Graphing Lines using Slope and a PointExample 2

Graph the line with slope m = that passes through (0, 2).2-3

Copy the definitions below

The y-intercept is the y-coordinate of a point where the line crosses the y-axis.

The x-intercept is the x-coordinate of a point where the line crosses the x-axis.

2-3

Find the x-intercept: 4x – 2y = 16

Find the intercepts of 4x – 2y = 16, and graph the line.

4x – 2(0) = 16

Find the y-intercept: 4x – 2y = 16

4x = 16x = 4

4(0) – 2y = 16–2y = 16

y = –8

x-intercept

y-intercept

2-3Example 3 Graphing Lines Using the Intercepts

Find the x-intercept: 6x – 2y = –24

Find the intercepts of 6x – 2y = –24, and graph the line.

6x – 2(0) = –24

Find the y-intercept: 6x – 2y = –24

6x = –24 x = –4

6(0) – 2y = –24 –2y = –24

y = 12

x-intercept

y-intercept

2-3

Linear functions can also be expressed as linear equations of the form y = mx + b.

When a linear function is written in the form y = mx + b

The function is said to be in slope-intercept form because m is the slope of the graph and b is the y-intercept.

Notice that slope-intercept form is the equation solved for y.

So slope-intercept form is: y = mx + b or f(x) = mx + b, where m = slope, b = y-int

2-3Algebra II (Bell work)

Math Joke

• Q: Why was the student afraid of the y-intercept

• A: She thought she’d be stung by the b

Solve for y first.

Write the function –4x + y = –1 in slope-intercept form.

–4x + y = –1

y = 4x – 1 +4x +4x

The line has y-intercept –1 and slope 4, which is . Plot the point (0, –1). Then move up 4 and right 1 to find other points.

2-3Example 4 Graphing Functions in Slope-Intercept Form

Solve for y first.

Write the function in slope-intercept form. Then

graph the function.

The line has y-intercept 8 and slope . Plot the point (0, 8). Then move down 4 and right 3 to find other points.

2-3

Solve for y first.

Write the function 5x = 15y + 30 in slope-intercept form. Then graph the function.

5x = 15y + 30

5x – 30 = 15y

–30 –30

The line has y-intercept –2 and slope . Plot the point (0, –2). Then move up 1 and right 3 to find other points.

2-3

Vertical and Horizontal Lines

Vertical Lines Horizontal Lines

The line x = a is a vertical line at a.

The line y = b is a horizontal line at b.

The slope of a vertical line is undefined.

The slope of a horizontal line is zero.

2-3

Determine if each line is vertical or horizontal.

A. x = 2

B. y = –4

This is a vertical line located at the x-value 2. (Note that it is not a function.)

This is a horizontal line located at the y-value –4.

x = 2

y = –4

2-3Example 5 Graphing Vertical and Horizontal Lines

Determine if each line is vertical or horizontal.

A. y = –5

B. x = 0.5

This is a horizontal line located at the y-value –5.

This is a vertical line located at the x-value 0.5.

x = 0.5

y = –5

2-3

HW pg. 109

• 2.3-– Day 1: 3 – 12, 52, 57, 67-79– Day 2: 13- 21 (21 No Graph), 33, 35, 47, 48, 53,

55ab

– Follow All HW Guidelines or ½ off – Don’t forget to write assignments on hw sheet (or

planner)

2.4

Algebra II (Bell work)

Copy the boxed parts below

2.4- Writing Linear Functions

Algebra II

Write the equation of the graphed line in slope-intercept form.

Step 1 Identify the y-intercept.The y-intercept b is 1.

Step 2 Find the slope.

Slope is = = – . Riserun

–34

34

3

–44

–3

2.4Example 1 Writing the Slope-Intercept Form of the Equation of the Line

Step 3 Write the equation in slope-intercept form.

y = mx + b

3

4

y = – x + 1

Write the equation of the graphed line in slope-intercept form.

Step 1 Identify the y-intercept.The y-intercept b is 3.

Slope is = . riserun

34

Write the equation in slope-intercept form.

y = mx + b

3

4

y = x + 3 m = and b = 3. 3

4

The equation of the line is 3

4

y = x + 3.

2.4

Find the slope of the line through (–1, 1) and (2, –5).

Let (x1, y1) be (–1, 1) and (x2, y2) be (2, –5).

The slope of the line is –2.

2.4

Example 2 Finding the Slope of a Line Given Two or More Points

Find the slope of the line.

x 4 8 12 16

y 2 5 8 11Let (x1, y1) be (4, 2) and (x2, y2) be (8, 5).

The slope of the line is .34

2.4

Find the slope of the line shown.

Let (x1, y1) be (0,–2) and (x2, y2) be (1, –2).

The slope of the line is 0.

2.4

Because the slope of line is constant, it is possible to use any point on a line and the slope of the line to write an equation of the line in point-slope form.

2.4

Math Joke

Student: I’ll just draw a quick line by hand and guess the slope

Teacher: No, No. It’s point-slope form, not point-sloppy form

2.4

In slope-intercept form, write the equation of the line that contains the points in the table.

x –8 –4 4 8

y –5 –3.5 –0.5 1

First, find the slope. Let (x1, y1) be (–8, –5) and (x2, y2) be (8, 1).

Next, choose a point, and use either form of the equation of a line.

2.4Example 3 Writing Equations of Lines

Method A Point-Slope Form

Rewrite in slope-intercept form.

Using (8, 1):

y – y1 = m(x – x1)

2.4Method B Slope-intercept Form

Using (8, 1), solve for b.

b = –2

y = mx + b

1 = 3 + b

The equation of the line is .

Method B Slope-Intercept Form

Write the equation of the line in slope-intercept form through (–2, –3) and (2, 5).

First, find the slope. Let (x1, y1) be (–2,–3) and (x2, y2) be (2, 5).

y = mx + b

5 = (2)2 + b

5 = 4 + b

1 = b

Rewrite the equation using m and b.

y = mx + b y = 2x + 1

The equation of the line is y = 2x + 1.

2.4

Method A Point-Slope Form

Rewrite in slope-intercept form.

y – y1 = m(x – x1)

y – (3) = –5(x – 1)

y – 3 = –5(x – 1)

y – 3 = –5(x – 1)

y – 3 = –5x + 5

y = –5x + 8

2.4Algebra II (Bell Work)

Summarize the boxed parts below

m = 1.8

y – 2 = 1.8(x – 5)

y – 2 = 1.8x – 9

y = 1.8x – 7

Write the equation of the line in slope-intercept form.

parallel to y = 1.8x + 3 and through (5, 2)

2.4

perpendicular to and through (9, –2)

The slope of the given line is , so the slope of

the perpendicular line is the opposite reciprocal,

Write the equation of the line in slope-intercept form.

parallel to y = 5x – 3 and through (1, 4)

m = 5

y – 4 = 5(x – 1)

y – 4 = 5x – 5

y = 5x – 1

2.4

perpendicular to and through (9, –2)

The slope of the given line is , so the slope of

the perpendicular line is the opposite reciprocal, .

The table shows the rents and selling prices of properties from a game.

Selling Price ($)

Rent ($)

75 9

90 12

160 26

250 44

Express the rent as a function of the selling price.

Let x = selling price and y = rent.

Find the slope by choosing two points.

Let (x1, y1) be (75, 9) and (x2, y2) be (90, 12).

2.4Example 4 Entertainment Application

To find the equation for the rent function, use point-slope form.

y – y1 = m(x – x1)

2.4

Graph the relationship between the selling price and the rent.

How much is the rent for a property with a selling price of $230?

To find the rent for a property, use the graph or substitute its selling price of $230 into the function.

The rent for the property is $40.

y = 46 – 6

y = 40

2.4

Items Cost ($)

4 14.00

7 21.50

18

Express the cost as a linear function of the number of items.

Let x = items and y = cost.

Find the slope by choosing two points. Let (x1, y1) be (4, 14) and (x2, y2) be (7, 21.50).

2.4Optional

To find the equation for the number of items, use point-slope form.

y – y1 = m(x – x1)

y – 14 = 2.5(x – 4)

y = 2.5x + 4

2.4

HW pg. 120

• 2.4– Day 1: 1-8, 17, 27, 43, 52, 54– Day 2: 9-11, 19, 23-25, 38, 44, 49, – Ch: 28

– Follow All HW Guidelines or ½ off – Don’t forget to write assignments on hw sheet (or

planner)