1. Chapter 2 - Linear Functions
Algebra II

2. Table of Contents 2.1- Solving Linear Equations and Inequalities
2.2- Proportional Reasoning
2.3- Graphing Linear Functions
2.4- Writing Linear Functions

3. 2.1- Solving Linear Equations and Inequalities
Algebra II

4. 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

5. Let m represent the number of additional minutes that Nina used.
2-1
Example 1
Application
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?
Let m represent the number of additional minutes that Nina used.
Model
additional minute charge
number of additional minutes
monthly charge
total charge
plus
times = =
14.56
12.95 +
0.07 * m

6. 2-1
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?
Let m represent the number of additional minutes that Nina used.
Solve
m = –
0.07m =
0.07
0.07
m = 23
Nina used 23 additional minutes.

7. number of additional cups
2-1
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?
additional cup height
number of additional cups
one cup
total height
plus
times =
3.25 =
14.00 +
0.25 * c

8. Stacked cups are to be placed in a pantry.
2-1
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?
Solve
c =
– –3.25
0.25c = 10.75
0.25
c = 43
44 cups fit between the 14 in. shelves.

9. 2-1
Example 2
Solving Equations with the Distributive Property
Solve 4(m + 12) = –36
Solve 4(m + 12) = –36
Method 1
Method 2
The quantity (m + 12) is multiplied by 4, so divide by 4 first.
Distribute before solving
4m + 48 = –36
–48 –48
4(m + 12) = –36
4m = –84
m + 12 = –9 =
4m –84
–12 –12
m = –21
m = –21

10. 2-1
Solve 3(2 –3p) = 42.
Solve 3(2 – 3p) = 42
Method 1
Method 2
6 – 9p = 42
3(2 – 3p) = 42
– –6
–9p = 36
2 – 3p = 14
– –2 =
–9p
– –9
–3p = 12
p = –4
– –3
p = –4

11. 2-1
Solve –3(5 – 4r) = –9
– r = –9
12r = 6 =
12r
r =

12. Just Read
If there are variables on both sides of the equation
Simplify each side.
Collect all variable terms on one side and all constants terms on the other side.
Isolate the variables

13. 2-1
Example 3
Solving Equations with Variables on Both Sides
Solve 3k– 14k + 25 = 2 – 6k – 12
Solve 3(w + 7) – 5w = w + 12
–2w + 21 = w + 12
–11k + 25 = –6k – 10
+11k k
+2w w
25 = 5k – 10
21 = 3w + 12
– –12
= k
= 3w
7 = k
3 = w

14. Define the following words
2-1
Algebra 2 (bell work)
Define the following words
Identity – An equation that is true for all values of the variable, such as x = x
Answer is all real numbers, or R, or ( - ∞ , ∞ )
Contradiction – An equation that has no solution, such as 3 = 5. Can also use the empty set symbol

15.  2-1 Example 4 Identifying Identities and Contradictions
Solve 2(x – 6) = –5x – x
Solve 3v – 9 – 4v = –(5 + v).
2x – 12 = 2x – 12
3v – 9 – 4v = –(5 + v)
–2x –2x
–9 – v = –5 – v
+ v v
–12 = –12 
The solutions set is all real number, or .
–9 ≠ –5 x
The equation has no solution. The solution set is the empty set, which is represented by the symbol .

16.  2-1 Solve 3(2 –3x) = –7x – 2(x –3) Solve 5(x – 6) = 3x – 18 + 2x
6 = 6 x
–30 ≠ –18
The solutions set is all real numbers, or .
The equation has no solution. The solution set is the empty set, which is represented by the symbol

17. 2-1
Just Read
If you multiply or divide by a negative number, flip the inequality sign
These properties also apply to inequalities expressed with >, ≥, and ≤.

18. 2-1
Example 5
Solving Inequalities
Solve and graph 8a –2 ≥ 13a + 8
Solve and graph x + 8 ≥ 4x + 17
8a – 2 ≥ 13a + 8
x + 8 ≥ 4x + 17
–13a –13a
–x –x
–5a – 2 ≥ 8
8 ≥ 3x +17
– –17
–5a ≥ 10
–9 ≥ 3x
–5a ≤ 10
–9 ≥ 3x
– –5
–3 ≥ x or x ≤ –3
a ≤ –2

19. HW pg.94
2.1-
Day 1: 2-11, (Odd), 41, 42, 45, 47
Day 2: 12-20, 34-39, 49, 62-65
Ch: 52
Follow All HW Guidelines or ½ off

20. 2.2- Proportional Reasoning
Algebra II

21. 2-2
Copy boxed parts below

22. 2-2
Example 1
Solve each proportion. c = p A. B. = p c = =
206.4 = 24p
88c = 1848 p =
88c =
8.6 = p
c = 21

23. 2-2
Optional
Solve each proportion. y A. = B. = x y x = =
924 = 84y
2.5x =105 y = =
2.5x
11 = y
x = 42

24. 2-2
Because percents can be expressed as ratios, you can use the proportion to solve percent problems.
Percent is a ratio that means per hundred.
For example:
30% = 0.30 =
Remember! 30
100

25. Percent (as decimal)  whole = part
2-2
Example 2
Solving Percent Problems
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.
Method 2 Use a percent equation.
Percent (as decimal)  whole = part
0.225  1800 = x
405 = x
x = 405
So 405 voters are planning to vote for that candidate.

26. Percent (as decimal)  whole = part
2-2
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.
Method 2 Use a percent equation.
35% = 0.35
Percent (as decimal)  whole = part
0.35x = 434
Cross multiply.
100(434) = 35x
x = 1240
Solve for x.
x = 1240
Clay High School has 1240 students.

27. 600 m 482 strides x m 1 stride  ≈ 2-2 Example 3
Fitness Application (Dimensional Analysis)
Ryan ran 600 meters and counted 482 strides. How long is Ryan’s stride in inches? (Hint: 1 m ≈ 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
 ≈
1.24 m
1 stride length
39.37 in.
1 m
49 in.
Ryan’s stride length is approximately 49 inches.

28. 400 m 297 strides x m 1 stride 2-2 Optional
Luis ran 400 meters in 297 strides. Find his stride length in inches.
Use a proportion to find the length of his stride in meters.
400 m
297 strides
x m
1 stride =
400 = 297x
x ≈ 1.35 m
 ≈
1.35 m
1 stride length
39.37 in.
1 m
53 in.
Luis’s stride length is approximately 53 inches.

29. Scaling Geometric Figures in the Coordinate Plane
2-2
Example 4
Scaling Geometric Figures in the Coordinate Plane =
height of ∆XAB width of ∆XAB
height of ∆XYZ width of ∆XYZ Y Z = x B A X
9x = 18, so x = 2

30. Math Joke Q: Why were the similar triangles weighing themselves
A: they were finding their Scale

31. = Shadow of tree Height of tree Shadow of house Height of house 2-2
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?
9 ft
6 ft =
Shadow of tree
Height of tree
Shadow of house
Height of house = 6 9 h 22
h ft
22 ft
6h = 198
h = 33
The house is 33 feet high.

32. = Shadow of climber Height of climber Shadow of tree Height of tree
2-2
Optional
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?
6 ft
20 ft =
Shadow of climber
Height of climber
Shadow of tree
Height of tree = 20 6 h 90
h ft
90 ft
20h = 540
h = 27
The tree is 27 feet high.

33. 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

34. Algebra II (Bell work) Copy down the definition of a linear function
2-3
Algebra II (Bell work)
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).

35. 2.3- Graphing Linear Functions
Algebra II

36. Determine whether the data set could represent a linear function.
2-3
Example 1
Recognizing Linear Functions
Determine whether the data set could represent a linear function. +2 –1 +2 –1 +2 –1 x –2 2 4
f(x) 1 –1
The rate of change, = , is constant
So the data set is linear.

37. Determine whether the data set could represent a linear function.
2-3
Determine whether the data set could represent a linear function. +1 +2 +1 +4 +1 +8 x 2 3 4 5
f(x) 8 16
The rate of change, ≠ 4 ≠ 8, is not constant.
So the data set is not linear.

38. Determine whether the data set could represent a linear function.
2-3
Determine whether the data set could represent a linear function. –2 –2 –4 –2 –8 x 10 8 6 4
f(x) 7 5 1 –7
The rate of change, ,
is not constant. So the data set is not linear.

39. 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= =

40. Graph the line with slope m= that passes through (–1, –3).
2-3
Example 2
Graphing Lines using Slope and a Point
Graph the line with slope m= that passes through (–1, –3).

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

42. Copy the definitions below The y-intercept is the
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.

43. Find the intercepts of 4x – 2y = 16, and graph the line.
2-3
Example 3
Graphing Lines Using the Intercepts
Find the intercepts of 4x – 2y = 16, and graph the line.
Find the x-intercept: 4x – 2y = 16
4x – 2(0) = 16
x-intercept
y-intercept
4x = 16
x = 4
Find the y-intercept: 4x – 2y = 16
4(0) – 2y = 16
–2y = 16
y = –8

44. Find the intercepts of 6x – 2y = –24, and graph the line.
2-3
Find the intercepts of 6x – 2y = –24, and graph the line.
Find the x-intercept: 6x – 2y = –24
6x – 2(0) = –24
x-intercept
y-intercept
6x = –24
x = –4
Find the y-intercept: 6x – 2y = –24
6(0) – 2y = –24
–2y = –24
y = 12

45. 2-3
Algebra II (Bell work)
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

46. Math Joke Q: Why was the student afraid of the y-intercept
A: She thought she’d be stung by the b

47. Solve for y first. –4x + y = –1 +4x +4x y = 4x – 1
2-3
Example 4
Graphing Functions in Slope-Intercept Form
Write the function –4x + y = –1 in slope-intercept form.
Solve for y first.
–4x + y = –1
+4x x
y = 4x – 1
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.

48. Write the function in slope-intercept form. Then graph the function.
2-3
Write the function in slope-intercept form. Then graph the function.
Solve for y first.
The line has y-intercept 8 and slope
. Plot the point (0, 8). Then move down 4 and right 3 to find other points.

49. Solve for y first. 5x = 15y + 30 –30 –30 5x – 30 = 15y
2-3
Write the function 5x = 15y + 30 in slope-intercept form. Then graph the function.
Solve for y first.
5x = 15y + 30
– –30
5x – 30 = 15y
The line has y-intercept –2 and slope Plot the point (0, –2). Then move up 1 and right 3 to find other points.

50. Vertical and Horizontal Lines
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.

51. Determine if each line is vertical or horizontal.
2-3
Example 5
Graphing Vertical and Horizontal Lines
Determine if each line is vertical or horizontal.
A. x = 2
This is a vertical line located at the x-value 2. (Note that it is not a function.)
x = 2
y = –4
B. y = –4
This is a horizontal line located at the y-value –4.

52. Determine if each line is vertical or horizontal.
2-3
Determine if each line is vertical or horizontal.
A. y = –5
This is a horizontal line located at the y-value –5.
x = 0.5
y = –5
B. x = 0.5
This is a vertical line located at the x-value 0.5.

53. HW pg. 109
2.3-
Day 1: 3 – 12, 52, 57, 67-79
Day 2: (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)

54. 2.4
Algebra II (Bell work)
Copy the boxed parts below

55. 2.4- Writing Linear Functions
Algebra II

56. 2.4
Example 1
Writing the Slope-Intercept Form of the Equation of the Line
Write the equation of the graphed line in slope-intercept form.
Step 1
Identify the y-intercept.
The y-intercept b is 1.
Step Find the slope.
Slope is = = –
Rise
run –3 4 3
Step 3
Write the equation in slope-intercept form. 3 –4 4 –3
y = mx + b 3 4
y = – x + 1

57. 2.4
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 =
rise
run 3 4
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.

58. 2.4
Example 2
Finding the Slope of a Line Given Two or More Points
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.

59. x 4 8 12 16 y 2 5 11 2.4 Find the slope of the line.
Let (x1, y1) be (4, 2) and (x2, y2) be (8, 5).
The slope of the line is . 3 4

60. 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.

61. 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.

62. 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

63. x –8 –4 4 8 y –5 –3.5 –0.5 1 2.4 Example 3 Writing Equations of Lines
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.

64. 2.4
Method A Point-Slope Form
Method B Slope-intercept Form
Using (8, 1):
Using (8, 1), solve for b.
y – y1 = m(x – x1)
y = mx + b
1 = 3 + b
b = –2
Rewrite in slope-intercept form.
The equation of the line is

65. 2.4
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).
Method A Point-Slope Form
Method B Slope-Intercept Form
y – y1 = m(x – x1)
y = mx + b
5 = (2)2 + b
y – (3) = –5(x – 1)
5 = 4 + b
y – 3 = –5(x – 1)
1 = b
Rewrite in slope-intercept form.
Rewrite the equation using m and b.
y – 3 = –5(x – 1)
y – 3 = –5x + 5
y = 2x + 1
y = mx + b
y = –5x + 8
The equation of the line is y = 2x + 1.

66. 2.4
Algebra II (Bell Work)
Summarize the boxed parts below

67. 2.4
Write the equation of the line in slope-intercept form.
perpendicular to
and through (9, –2)
parallel to y = 1.8x + 3 and through (5, 2)
The slope of the given line is , so the slope of
the perpendicular line is the opposite reciprocal,
m = 1.8
y – 2 = 1.8(x – 5)
y – 2 = 1.8x – 9
y = 1.8x – 7

68. 2.4
Write the equation of the line in slope-intercept form.
parallel to y = 5x – 3 and
through (1, 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,
m = 5
y – 4 = 5(x – 1)
y – 4 = 5x – 5
y = 5x – 1

69. Let x = selling price and y = rent.
2.4
Example 4
Entertainment Application
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).

70. 2.4
To find the equation for the rent function, use point-slope form.
y – y1 = m(x – x1)

71. 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.
y = 46 – 6
y = 40
The rent for the property is $40.

72. Items Cost ($) 4 14.00 7 21.50 18 2.4 Optional
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).

73. 2.4
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

74. 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)