1. Graphing Linear Equations and Inequalities
CHAPTER 4
Graphing Linear Equations and Inequalities
4.1 The Rectangular Coordinate System
4.2 Graphing Linear Equations
4.3 Graphing Using Intercepts
4.4 Slope-Intercept Form
4.5 Point-Slope Form
4.6 Graphing Linear Inequalities
4.7 Introduction to Functions and Function Notation
Copyright © 2011 Pearson Education, Inc.

2. The Rectangular Coordinate System
4.1
The Rectangular Coordinate System
1. Determine the coordinates of a given point.
2. Plot points in the coordinate plane.
3. Determine the quadrant for a given ordered pair.
4. Determine whether the graph of a set of data points is linear.
Copyright © 2011 Pearson Education, Inc.

3. Copyright © 2011 Pearson Education, Inc.
Axis: A number line used to locate a point in a plane.
the origin
Positive numbers are to the right and up from the origin.
Negative numbers are to the left and down from the origin.
The notation for writing ordered pairs is: (horizontal coordinate, vertical coordinate).
Copyright © 2011 Pearson Education, Inc.
Slide 4- 3

4. Copyright © 2011 Pearson Education, Inc.
Identifying the Coordinates of a Point
To determine the coordinates of a given point in the rectangular system,
1. Follow a vertical line from the point to the x-axis (horizontal axis). The number at this position on the
x-axis is the first coordinate.
2. Follow a horizontal line from the point to the y-axis (vertical axis). The number at this position on the y axis is the second coordinate.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 4

5. Copyright © 2011 Pearson Education, Inc.
Example 1
Write the coordinates for each point shown. A B D F E C
We can get to point A by moving to the right 1 and then up 4. A: (1, 4)
B: (3, 2)
Left 3, up 2
C: (2, 2)
Left 2, down 2
D: (4, 0)
Right 4
E: (0, 2)
Down 2
F: (4, 3)
Right 4, down 3
Copyright © 2011 Pearson Education, Inc.
Slide 4- 5

6. Copyright © 2011 Pearson Education, Inc.
Plotting a Point
To graph or plot a point given its coordinates,
1. Beginning at the origin, (0, 0), move to the right or left along the x-axis the amount indicated by the first coordinate.
2. From that position on the x-axis, move up or down the amount indicated by the second coordinate.
3. Draw a dot to represent the point described by the coordinates.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 6

7. Copyright © 2011 Pearson Education, Inc.
Example 2
Plot the point described by the coordinates.
a. (4, 3)
b. (2, 4)
c. (0, 3)
(2, 4)
(0, 3)
(4, –3 ): Begin at the origin and move to the right 4, then down 3.
(–2, 4): Begin at the origin and move to the left 2, then up 4.
(0, 3): Begin at the origin and move up 3.
(4, 3)
Copyright © 2011 Pearson Education, Inc.
Slide 4- 7

8. Copyright © 2011 Pearson Education, Inc.
The quadrants are numbered using Roman numerals.
Note that the signs of the coordinates determine the quadrant in which a point lies.
Quadrant II
(, +)
Quadrant I
(+, +)
Quadrant III
(, )
Quadrant IV
(+, )
Copyright © 2011 Pearson Education, Inc.
Slide 4- 8

9. Copyright © 2011 Pearson Education, Inc.
Identifying Quadrants
To determine the quadrant for a given ordered pair, consider the signs of the coordinates.
(+, +) means the point is in quadrant I
(–, +) means the point is in quadrant II
(–, –) means the point is in quadrant III
(+, –) means the point is in quadrant IV
Note: If either coordinate is 0, the point is on an axis and not in the quadrant.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 9

10. Copyright © 2011 Pearson Education, Inc.
Example 3
State the quadrant in which each point is located.
a. (–14, 45)
Answer Quadrant II because the first coordinate is negative and the second coordinate is positive.
b. (6, –2 ½ )
Answer Quadrant IV because the first coordinate is positive and the second coordinate is negative.
c. (0, –15 )
Answer Since the x-coordinate is 0, this point is on the y-axis and is not in a quadrant.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 10

11. Copyright © 2011 Pearson Education, Inc.
In many problems, data are listed as ordered pairs.
If we plot each pair of data as an ordered pair we can see whether the points form a straight line. If they form a straight line they are linear.
Points that do not form a straight line are nonlinear.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 11

12. Copyright © 2011 Pearson Education, Inc.
Example 4
The following points show the path of an object over time. Plot the points with the time along the horizontal axis and the distance along the vertical axis. Then state whether the path is linear or nonlinear.
Time (hours)
Distance
(miles) 1 62 2
124 3
186 4
248
The data points form a straight line; the path of the object is linear.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 12

13. Identify the coordinates of the point shown below.
4.1

14. Identify the coordinates of the point shown below.
4.1

15. In which quadrant is the point (5, –2) located?
a) I
b) II
c) III
d) IV
4.1

16. In which quadrant is the point (5, –2) located?
a) I
b) II
c) III
d) IV
4.1

17. Graphing Linear Equations
4.2
Graphing Linear Equations
1. Determine whether a given pair of coordinates is a solution to a given equation with two unknowns.
2. Find solutions for an equation with two unknowns.
3. Graph linear equations.
Copyright © 2011 Pearson Education, Inc.

18. Copyright © 2011 Pearson Education, Inc.
Checking a Potential Solution for an Equation with Two Variables
To determine whether a given ordered pair is a solution for an equation with two variables,
1. Replace the variables in the equation with the corresponding coordinates.
2. Verify that the equation is true.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 18

19. Copyright © 2011 Pearson Education, Inc.
Example 1a
Determine whether the ordered pair is a solution for the equation. (3, 5); y = 4x – 3
Solution
y = 4x – 3
5 = 4(–3) – 3 Replace x with –3 and y with 5.
5 ? –12 – 3
5  – Because the equation is not true, (–3, 5) is not a solution for y = 4x – 3.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 19

20. Copyright © 2011 Pearson Education, Inc.
Example 1b
Determine whether the ordered pair is a solution for the equation. (4, 5); 3x – y = 7
Solution
3x – y = 7
3(4) – 5 = 7 Replace x with 4 and y with 5.
12 – 5 ? 7
7 = Because the equation is true, (4, 5) is a solution for 3x – y = 7.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 20

21. Copyright © 2011 Pearson Education, Inc.
Finding Solutions to Equations with Two Variables
To find a solution to a linear equation in two variables,
1. Choose a value for one of the variables (any value).
2. Replace the corresponding variable with your chosen value.
3. Solve the equation for the value of the other variable.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 21

22. Example 2 Find three solutions for the equation. 4x + y = 5 Solution
To find a solution we replace one of the variables with a chosen value and then solve for the other variable.
Choose x = 0
4x + y = 5
4(0) + y = 5
y = 5
Solution (0, 5)
Choose x = 1 4x + y = 5 4(1) + y = y = y = 1
Solution (1, 1)
Choose x = 2 4x + y = 5 4(2) + y = y = y = –3
Solution (2, 3)
Copyright © 2011 Pearson Education, Inc.
Slide 4- 22

23. Copyright © 2011 Pearson Education, Inc.
continued
The solutions can be summarized in a table:
Keep in mind that there are an infinite number of correct solutions for a given equation in two variables. x y
Ordered Pair 5
(0, 5) 1
(1, 1) 2 3
(2, 3)
Copyright © 2011 Pearson Education, Inc.
Slide 4- 23

24. Copyright © 2011 Pearson Education, Inc.
Equations in two variables have an infinite number of solutions. Because of this fact, there is no way that all solutions to an equation can be found. However, all the solutions can be represented using a graph.
The graph of the solutions of every linear equation will be a straight line.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 24

25. Copyright © 2011 Pearson Education, Inc.
Graphing Linear Equations
To graph a linear equation:
1. Find at least two solutions to the equation.
2. Plot the solutions as points in the rectangular coordinate system.
3. Connect the points to form a straight line.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 25

26. Copyright © 2011 Pearson Education, Inc.
Example 3
Graph 4x + y = 5.
Solution We found three solutions to this equation in the previous example.
4x + y = 5 x y
Ordered Pair 5
(0, 5) 1
(1, 1) 2 3
(2, 3)
Plot each point and then connect the points to form a straight line.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 26

27. Copyright © 2011 Pearson Education, Inc.
Example 4
Graph y = 2.
Solution y is equal to a constant
To establish ordered pairs, we can rewrite the equation as 0x + y = 2
y is always 2 no matter
what we choose for x.
y = 2 x y
Ordered Pair 2
(0, 2) 1
(1, 2)
(3, 2)
Copyright © 2011 Pearson Education, Inc.
Slide 4- 27

28. Copyright © 2011 Pearson Education, Inc.
Horizontal Lines
The graph of y = c where, c is a real-number constant, is a horizontal line parallel to the x-axis that passes through the y-axis at a point with coordinates (0, c).
Copyright © 2011 Pearson Education, Inc.
Slide 4- 28

29. Copyright © 2011 Pearson Education, Inc.
Example 5
Graph x = 2.
Solution x is equal to a constant.
To establish ordered pairs, we can rewrite the equation as x + 0y = 2.
x is always 2 no matter
what we choose for y.
x = 2 x y
Ordered Pair 2
(2, 0) 1
(2, 1)
(2, 2)
Copyright © 2011 Pearson Education, Inc.
Slide 4- 29

30. Copyright © 2011 Pearson Education, Inc.
Vertical Lines
The graph of x = c where, c is a real-number constant, is a vertical line parallel to the y-axis that passes through the x-axis at a point with coordinates (c, 0).
Copyright © 2011 Pearson Education, Inc.
Slide 4- 30

31. Which ordered pair is a solution of the equation 3x – 2y = –6?
b) (1, 5)
c) (2, 0)
d) (0, 3)
4.2

32. Which ordered pair is a solution of the equation 3x – 2y = –6?
b) (1, 5)
c) (2, 0)
d) (0, 3)
4.2

33. Graph 4x – y = –4 a) b) c) d)
4.2

34. Graph 4x – y = –4 a) b) c) d)
4.2

35. Graphing Using Intercepts
4.3
Graphing Using Intercepts
1. Given an equation, find the coordinates of the x- and y-intercepts.
2. Graph linear equations using intercepts.
Copyright © 2011 Pearson Education, Inc.

36. Copyright © 2011 Pearson Education, Inc.
Notice that the line intersects the x-axis as (4, 0) and the y-axis at (0, –2). These points are called intercepts.
The point where a graph
intersects the x-axis is called
the x-intercept.
intersects the y-axis is called
the y-intercept.
x-intercept (4, 0)
y-intercept (0, 2)
Copyright © 2011 Pearson Education, Inc.
Slide 4- 36

37. Copyright © 2011 Pearson Education, Inc.
Finding the x- and y-intercepts
To find an x-intercept,
To find a y-intercept,
1. Replace y with 0 in the given equation.
1. Replace x with 0 in the given equation.
2. Solve for x.
2. Solve for y.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 37

38. Example 1 For the equation 7x – 2y = 14, find the x- and y-intercepts.
Solution
For the x-intercept, replace y with 0 and solve for x.
7x – 2y = 14
7x – 2(0) = 14
7x = 14
x = 2
x-intercept: (2, 0)
For the y-intercept, replace x with 0 and solve for y.
7x – 2y = 14
7(0) – 2y = 14
–2y = 14
y = –7
y-intercept: (0, –7)
Copyright © 2011 Pearson Education, Inc.
Slide 4- 38

39. Example 2 For the equation , find the x- and y-intercepts.
Solution
For the y-intercept, replace x with 0 and solve for y.
y-intercept: (0, 0)
For the x-intercept, replace y with 0 and solve for x.
x-intercept: (0, 0)
Copyright © 2011 Pearson Education, Inc.
Slide 4- 39

40. Copyright © 2011 Pearson Education, Inc.
Intercepts for y = mx
If an equation can be written in the form y = mx, where m is a real number other than 0, then the x- and y-intercepts are at the origin, (0, 0).
Copyright © 2011 Pearson Education, Inc.
Slide 4- 40

41. Example 3 For the equation y = 3x + 4, find the x- and y-intercepts.
Solution
x-intercept:
y = 3x + 4
0 = 3x + 4
–4 = 3x
y-intercept:
y = 3x + 4
y = 3(0) + 4
y = 4
y-intercept: (0, 4)
Copyright © 2011 Pearson Education, Inc.
Slide 4- 41

42. Example 4 For the equation y = 5, find the x- and y-intercepts.
Solution
The graph of y = 5 is an horizontal line parallel to the x-axis that passes through the y-axis at the point (0, 5). Notice that this point is the y-intercept. Because the line is parallel to the x-axis, it will never intersect the x-axis. Therefore, there is no x-intercept.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 42

43. Copyright © 2011 Pearson Education, Inc.
The y-intercept for y = mx + b
If an equation is in the form y = mx + b, where m and b are real numbers, then the y-intercept will be (0, b).
Copyright © 2011 Pearson Education, Inc.
Slide 4- 43

44. Copyright © 2011 Pearson Education, Inc.
Intercepts for y = c
The graph of an equation in the form y = c, where c is a nonzero real number, has no x-intercept and the y-intercept is (0, c). The graph of y = 0 is the x-axis.
Intercepts for x = c
The graph of an equation in the form x = c, where c is a nonzero real-number, has no y-intercept and the x-intercept is (c, 0). The graph of x = 0 is the y-axis.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 44

45. Example 5 Graph using the x- and y-intercepts: 7x – 2y = 14
Solution
The intercepts were found in a previous example to be (2, 0) and (0, –7).
We plot the points and connect them with a line.
It is helpful to find a third solution and verify that all three points can be connected to form a straight line.
7x – 2y = 14
Copyright © 2011 Pearson Education, Inc.
Slide 4- 45

46. Copyright © 2011 Pearson Education, Inc.
continued
Choose x = 3 as a check.
7x – 2y = 14
7(3) – 2y = 14
–2y = –7
y = 3.5
Table of Solutions x y
x-intercept: 2
y-intercept: 2
check 3
3.5
7x – 2y = 14
Copyright © 2011 Pearson Education, Inc.
Slide 4- 46

47. Example 6a Graph using x- and y-intercepts.
(0, 0) is the x- and y-intercept.
We need to find one more point and a third point as a check.
Choose x = 4. Choose x = 4.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 47

48. Copyright © 2011 Pearson Education, Inc.
continued
Table of Solutions x y
x- and y-intercept:
solution 4 3
check 4 3
Plot the solutions and connect the points to form the line.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 48

49. Example 6b Graph using x- and y-intercepts.
(0, 2) is the y-intercept.
We need to find one more point and the x-intercept.
Choose x = 4. Let y = 0.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 49

50. Copyright © 2011 Pearson Education, Inc.
continued
Table of Solutions x y
x-intercept
2.6
y-intercept 2
check 4 5
Plot the solutions and connect the points to form the line.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 50

51. What is the x-intercept for 3x – y = 6?
b) (0, –6)
c) (–6, 0)
d) (0, 2)
4.3

52. What is the x-intercept for 3x – y = 6?
b) (0, –6)
c) (–6, 0)
d) (0, 2)
4.3

53. What is the y-intercept for y = 2x – 6?
b) (0, –6)
c) (–6, 0)
d) (0, 3)
4.3

54. What is the y-intercept for y = 2x – 6?
b) (0, –6)
c) (–6, 0)
d) (0, 3)
4.3

55. Copyright © 2011 Pearson Education, Inc.
4.4
Slope-Intercept Form
1. Compare lines with different slopes.
2. Graph equations in slope-intercept form.
3. Use slope-intercept form to write the equation of a line.
4. Find the slope of a line given two points on the line.
Copyright © 2011 Pearson Education, Inc.

56. Copyright © 2011 Pearson Education, Inc.
Example 1
Graph each of the following on the same grid.
y = x y = 3x y = 4x
Solution
Complete a table of values.
y = 3x
y = x
y = 4x
y = 3x
y = x
y = x
If x is
y = x
y = 3x
y = 4x 1 3 4 2 6 8
Copyright © 2011 Pearson Education, Inc.
Slide 4- 56

57. Copyright © 2011 Pearson Education, Inc.
Example 2
Graph each of the following on the same grid.
Solution
Complete a table of values.
If x is
y = x 1 2
Copyright © 2011 Pearson Education, Inc.
Slide 4- 57

58. Copyright © 2011 Pearson Education, Inc.
Example 3
Graph each of the following on the same grid.
Solution
Complete a table of values.
If x is
y = x 1 1 2 2
Copyright © 2011 Pearson Education, Inc.
Slide 4- 58

59. Copyright © 2011 Pearson Education, Inc.
If the coefficient of m increases, the graphs get steeper. Because the coefficient m affects how steep a line is, m is called the slope of the line.
If a slope of the line is a fraction between 0 and 1, then the smaller the fraction is, the less inclined or flatter the lines get.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 59

60. Copyright © 2011 Pearson Education, Inc.
Graphs of y = mx
Given an equation of the form y = mx, the graph of the equation is a line passing through the origin and having the following characteristics:
If m > 0, then the graph is a line that slants uphill from left to right.
If m < 0, then the graph is a line that slants downhill from left to right.
The greater the absolute value of m, the steeper the line.
m > 0
m < 0
Copyright © 2011 Pearson Education, Inc.
Slide 4- 60

61. Copyright © 2011 Pearson Education, Inc.
Slope: The ratio of the vertical change between any two points on a line to the horizontal change between these points.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 61

62. Copyright © 2011 Pearson Education, Inc.
Graphing Equations in Slope-Intercept Form
To graph an equation in slope-intercept form, y = mx + b,
1. Plot the y-intercept, (0, b).
2. Plot a second point by rising the number of units indicated by the numerator of the slope, m, then running the number of units indicated by the denominator of the slope, m.
3. Draw a straight line through the two points.
Note: You can check by locating additional points using the slope. Every point you locate using the slope should be on the line.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 62

63. Copyright © 2011 Pearson Education, Inc.
Example 4
For the equation determine the slope and the y-intercept. Then graph the equation.
Solution
m =
y-intercept: (0, 3)
Plot the y-intercept and then use the slope to find other points.
rise 2
(3, 1)
run 3
Copyright © 2011 Pearson Education, Inc.
Slide 4- 63

64. Copyright © 2011 Pearson Education, Inc.
Example 5
For the equation 2x + 5y = 20, determine the slope and the y-intercept. Then graph the equation.
Solution
Write the equation in slope-intercept form by isolating y.
2x + 5y = 20
5y = 2x  20
The slope is and the y-intercept is (0, –4).
Copyright © 2011 Pearson Education, Inc.
Slide 4- 64

65. Copyright © 2011 Pearson Education, Inc.
m = y-intercept: (0, –4)
continued
We begin at (0, –4) and then rise 2 and run 5.
rise 2
run 5
Copyright © 2011 Pearson Education, Inc.
Slide 4- 65

66. Copyright © 2011 Pearson Education, Inc.
Example 6
A line with the slope of 7 crosses the y-axis at the point (0, 4). Write the equation of the line.
Solution
We use y = mx + b, the slope-intercept form of the equation, replacing m with the slope 7 and the y-coordinate of the y-intercept 4.
y = 7x + 4
Copyright © 2011 Pearson Education, Inc.
Slide 4- 66

67. Copyright © 2011 Pearson Education, Inc.
The Slope Formula
Given two points (x1, y1) and (x2, y2), where x2  x1, the slope of the line connecting the two points is given by the formula
Rise:
y2 – y1
Run:
x2 – x1
Copyright © 2011 Pearson Education, Inc.
Slide 4- 67

68. Example 7 Find the slope of the line connecting the given points.
a. (4, 6) and (–2, 8)
Solution
b. (3, 8) and (–2, 8)
Solution
Copyright © 2011 Pearson Education, Inc.
Slide 4- 68

69. Copyright © 2011 Pearson Education, Inc.
Zero Slope
Two points with different x-coordinates and the same y-coordinates, (x1, c) and (x2, c), will form a line with a slope 0 (a horizontal line) and equation y = c.
Undefined Slope
Two points with the same x-coordinates and different y-coordinates, (c, y1) and (c, y2), will form a line with a slope that is undefined (a vertical line) and equation x = c.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 69

70. Copyright © 2011 Pearson Education, Inc.
Example 8
Find the slope of the line connecting the points (3, 8) and (3, −4).
Solution
Note that (3, 8) and (3, −4) connect to form a vertical line whose equation is x = 3. The slope of this line is found using the slope formula.
Because we never divide by 0, the slope is undefined.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 70

71. What is the slope of the line 3x + y = 6?
a) m = 3
b) m = 3 c)
d) m = 2
4.4

72. What is the slope of the line 3x + y = 6?
a) m = 3
b) m = 3 c)
d) m = 2
4.4

73. How would you graph the line y = 3x – 2?
a) Plot (0, –2), down 2, right 3
b) Plot (0, –2), down 3, right 1
c) Plot (0, 2), up 3, left 2
d) Plot (0, –2), up 3, right 1
4.4

74. How would you graph the line y = 3x – 2?
a) Plot (0, –2), down 2, right 3
b) Plot (0, –2), down 3, right 1
c) Plot (0, 2), up 3, left 2
d) Plot (0, –2), up 3, right 1
4.4

75. What is the slope of the line through the given points (5, 6) and (6, 3)?
a) m = 9
b) m = 9 c) d)
4.4

76. What is the slope of the line through the given points (5, 6) and (6, 3)?
a) m = 9
b) m = 9 c) d)
4.4

77. Copyright © 2011 Pearson Education, Inc.
4.5
Point-Slope Form
1. Use point-slope form to write the equation of a line.
2. Write the equation of a line parallel to a given line.
3. Write the equation of a line perpendicular to a given line.
Copyright © 2011 Pearson Education, Inc.

78. Copyright © 2011 Pearson Education, Inc.
You can use the point-slope form to write the equation of a line given any two points on the line.
Using the Point-Slope Form of the Equation of a Line
To write the equation of a line given its slope and any point, (x1, y1), on the line, use the point-slope form of the equation of a line, y – y1 = m(x – x1). If given a second point (x2, y2), and not the slope, we first calculate the slope using then use y – y1= m(x – x1).
Copyright © 2011 Pearson Education, Inc.
Slide 4- 78

79. Copyright © 2011 Pearson Education, Inc.
Example 1
A line with a slope of 4 crosses the y-axis at the point (0, 5). Write the equation in slope–intercept form.
Solution
y – y1= m(x – x1)
y – 5= 4(x – 0)
y – 5= 4x
y = 4x + 5
Copyright © 2011 Pearson Education, Inc.
Slide 4- 79

80. Copyright © 2011 Pearson Education, Inc.
Example 2
Write the equation of a line passing through the points (0, 6) and (3, 6). Write the equation in slope-intercept form.
Solution
Find the slope:
y – y1= m(x – x1)
y – (−6)= 4(x – 0)
y + 6 = 4x
y = 4x − 6
Copyright © 2011 Pearson Education, Inc.
Slide 4- 80

81. Copyright © 2011 Pearson Education, Inc.
Equations can also be written in standard form,
Ax + By = C, where A, B, and C are real numbers.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 81

82. Copyright © 2011 Pearson Education, Inc.
Example 3
The following data points relate velocity of an object as time passes. The graph shows that the points are in a line. Write the equation of the line in standard form.
Time (x)
(in seconds)
Velocity (y)
(in ft./sec.)
4.5 1 6 2
7.5 3
10.5
Copyright © 2011 Pearson Education, Inc.
Slide 4- 82

83. Copyright © 2011 Pearson Education, Inc.
continued
Solution
First, we find the slope of the line using any two ordered pairs in the slope formula. We will use (0, 4.5) and (1, 6).
Because (0,4.5) is the y-intercept, we can write the equation in slope-intercept form.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 83

84. Copyright © 2011 Pearson Education, Inc.
continued
When we write the equation in standard form, it is customary to write the equation so that the x-term is first with a positive coefficient.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 84

85. Copyright © 2011 Pearson Education, Inc.
Example 4
A line connects the points (2, 6) and (–4, 3). Write the equation of the line in the form Ax + By = C, where A, B, and C are integers and A > 0.
Solution
Find the slope:
Use point-slope form: y – y1 = m(x – x1)
Distribute to clear ( ).
Copyright © 2011 Pearson Education, Inc.
Slide 4- 85

86. Copyright © 2011 Pearson Education, Inc.
continued
Multiply both sides by the LCD, 2.
Subtract x from both sides to get x and y together.
Add 12 to both sides to get the constant terms together.
Multiply by –1 so that the coefficient of x is positive.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 86

87. Copyright © 2011 Pearson Education, Inc.
Parallel Lines
Nonvertical parallel lines have
equal slopes and different
y-intercepts. Vertical lines
are parallel.
y = 2x + 1
y = 2x – 3
Copyright © 2011 Pearson Education, Inc.
Slide 4- 87

88. Copyright © 2011 Pearson Education, Inc.
Example 5
Write the equation of a line that passes through (1, –5) and parallel to y = –3x Write the equation in slope-intercept form.
Solution
In y = –3x + 4, the slope is –3, so the slope of the line parallel will also be –3.
Use point-slope form. y – y1 = m(x – x1)
y – (5) = –3(x – 1)
y + 5 = –3x + 3
y = –3x – 2
y1 = 5, x1 = 1 and m = –3
Simplify.
Subtract 5 from both sides to isolate y.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 88

89. Copyright © 2011 Pearson Education, Inc.
Perpendicular Lines
The slope of a line perpendicular to a line with a slope of will be
Horizontal and vertical lines are perpendicular.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 89

90. Copyright © 2011 Pearson Education, Inc.
Example 6
Write the equation of a line that passes through (7, 1) and is perpendicular to 7x – 2y = –2. Write the equation in slope-intercept form.
Solution
Determine the slope of the line 7x – 2y = –2.
Slope of perpendicular line:
Copyright © 2011 Pearson Education, Inc.
Slide 4- 90

91. Copyright © 2011 Pearson Education, Inc.
continued
slope = ; point (7, 1)
y – y1 = m(x – x1)
Simplify.
Add 1 to both sides to isolate y.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 91

92. Copyright © 2011 Pearson Education, Inc.
Example 7a
Determine whether the given lines are parallel, perpendicular, or neither.
Solution
Because the slopes are opposites
(negatives) and reciprocals, the lines are perpendicular.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 92

93. Copyright © 2011 Pearson Education, Inc.
Example 7b
Determine whether the given lines are parallel, perpendicular, or neither.
Solution
Because the slopes of both lines are and the
y-intercepts differ, the lines are parallel.
Copyright © 2011 Pearson Education, Inc.
Slide 4- 93

94. Write the equation of the line in point-slope form given m = 2 and the point (4, 5).
a) y + 5 = −2(x + 4)
b) y − 5 = −2(x + 4)
c) y + 5 = −2(x – 4)
d) y − 5 = −2(x – 4)
4.5

95. Write the equation of the line in point-slope form given m = 2 and the point (4, 5).
a) y + 5 = −2(x + 4)
b) y − 5 = −2(x + 4)
c) y + 5 = −2(x – 4)
d) y − 5 = −2(x – 4)
4.5

96. What is the equation of the line connecting the points (4, 3) and (1, 7)?
a) y = 2x + 5
b) y = 2x + 5
c) y = 2x – 5 d)
4.5

97. What is the equation of the line connecting the points (4, 3) and (1, 7)?
a) y = 2x + 5
b) y = 2x + 5
c) y = 2x – 5 d)
4.5

98. What is the relationship between the two lines. 5x – 3y = 11
What is the relationship between the two lines? x – 3y = x + 5y = 8
a) parallel
b) perpendicular
c) neither
4.5

99. What is the relationship between the two lines. 5x – 3y = 11
What is the relationship between the two lines? x – 3y = x + 5y = 8
a) parallel
b) perpendicular
c) neither
4.5

100. Graphing Linear Inequalities
4.6
Graphing Linear Inequalities
1. Determine whether an ordered pair is a solution for a linear inequality with two variables.
2. Graph linear inequalities.
Copyright © 2011 Pearson Education, Inc.

101. Copyright © 2011 Pearson Education, Inc.
Linear inequalities have the same form as linear equations, except that they contain an inequality symbol instead of an equal sign. Remember that a solution for a linear equation in two variables is an ordered pair that makes the equation true.
Checking an Ordered Pair
To determine whether an ordered pair is a solution for an inequality, replace the variables with the corresponding coordinates and see if the resulting inequality is true. If it is, the ordered pair is a solution.
Copyright © 2011 Pearson Education, Inc.
Slide

102. Copyright © 2011 Pearson Education, Inc.
Example 1
Determine whether (3, –5) is a solution for y  3x + 4.
Solution
y  3x + 4
– 5  3(3) + 4
–5  9 + 4
–5  13
This statement is true, so (3, –5) is a solution.
Replace x with 3 and y with 5.
Copyright © 2011 Pearson Education, Inc.
Slide

103. Copyright © 2011 Pearson Education, Inc.
Graphing linear inequalities is very much like graphing linear equations.
Graphing Linear Inequalities
To graph a linear inequality in two variables,
1. Graph the related equation (boundary line). The related equation has an equal sign in place of the inequality symbol. If the inequality symbol is  or , draw a solid line. If the inequality symbol is < or >, draw a dashed line.
2. Choose an ordered pair on one side of the boundary line and test this ordered pair in the inequality. If the ordered pair satisfies the inequality, shade the region that contains it. If the ordered pair does not satisfy the inequality, shade the region on the other side of the boundary line.
Copyright © 2011 Pearson Education, Inc.
Slide

104. Copyright © 2011 Pearson Education, Inc.
Example 2a
Graph: y > 4x + 1
Solution
Graph the related equation
y = 4x + 1.
(0, 1) and (1, 3)
Select a test point: (0, 0)
y > 4x + 1
0 > 4(0) + 1
0 > 1 False
Shade other side of the boundary line.
Copyright © 2011 Pearson Education, Inc.
Slide

105. Copyright © 2011 Pearson Education, Inc.
Example 2b
Graph: x + 4y  8
Solution
Graph the related equation.
(0, 2) and (−8, 0)
Select a test point: (0, 0)
x – 4y  8
0 – 4(0)  8
0  8 True
Shade the region that contains the point.
Copyright © 2011 Pearson Education, Inc.
Slide

106. Copyright © 2011 Pearson Education, Inc.
Example 2c
Graph: x > 4
Solution
First graph the related line x = 4,
using a dashed line.
Select a test point: (0, 0)
x > 0
0 > 0 False
Shade other side of the
boundary line.
Copyright © 2011 Pearson Education, Inc.
Slide

107. Copyright © 2011 Pearson Education, Inc.
Example 2d
Graph: y  −1
Solution
First graph the related line y = −1,
using a solid line.
Select a test point: (0, 0)
0  −1
False
Shade other side of the
boundary line.
Copyright © 2011 Pearson Education, Inc.
Slide

108. Which point is a solution to the inequality 3x + 2y  12?
b) (0, 0)
c) (2, 5)
d) (4, 5)
4.6

109. Which point is a solution to the inequality 3x + 2y  12?
b) (0, 0)
c) (2, 5)
d) (4, 5)
4.6

110. Graph 2x + 3y  12. a) b) c) d)
4.6

111. Graph 2x + 3y  12. a) b) c) d)
4.6

112. Introduction to Functions and Function Notation
4.7
Introduction to Functions and Function Notation
1. Identify the domain and range of a relation.
2. Identify functions and their domains and ranges.
3. Find the value of a function.
4. Graph linear functions.
Copyright © 2011 Pearson Education, Inc.

113. Copyright © 2011 Pearson Education, Inc.
Relation: A set of ordered pairs.
Domain: The set of all input values (x-values) for a relation.
Range: The set of all output values (y-values) for a relation.
Copyright © 2011 Pearson Education, Inc.
Slide

114. Copyright © 2011 Pearson Education, Inc.
Example 1
Determine the domain and range of the relation {(3, 2), (4, 0), (5, 6), (6, 9)}.
Solution
Domain—set of all x-values
{3, 4, 5, 6}
Range—set of all y-values
{2, 0, 6, 9}
Copyright © 2011 Pearson Education, Inc.
Slide

115. Copyright © 2011 Pearson Education, Inc.
Determining Domain and Range
To determine the domain of a relation given its graph, answer the question: What are all of the x-values that have a corresponding y-value? To determine the range, answer the question: What are all of the y-values that have a corresponding x-value?
Copyright © 2011 Pearson Education, Inc.
Slide

116. Copyright © 2011 Pearson Education, Inc.
Example 2
Determine the domain and
range of the relation.
Solution
The graph begins at (2, 0).
Values along the x-axis begin at
2 and continue infinitely, so the domain is {x|x  2}.
The y-values begin at 0 and continue infinitely, so the range is {y|y  0}.
Copyright © 2011 Pearson Education, Inc.
Slide

117. Copyright © 2011 Pearson Education, Inc.
Function: A relation in which every value in the domain is paired with exactly one value in the range.
Domain Range
Each element in the domain has a single arrow pointing to an element in the range.
Copyright © 2011 Pearson Education, Inc.
Slide

118. Copyright © 2011 Pearson Education, Inc.
Every function is a relation, but not every relation is a function.
If any value in the domain is assigned to more than one value in the range, then the relation is not a function.
Domain Range 10 12
not a function
Copyright © 2011 Pearson Education, Inc.
Slide

119. Copyright © 2011 Pearson Education, Inc.
Example 3
Determine whether the relation is a function.
Domain Range
March 1 Donna
April 17 Dennis
Sept. 3 Catherine
October 9 Denise
Nancy
The relation is not a function because an element in the domain, Sept. 3, is assigned to two names in the range.
Copyright © 2011 Pearson Education, Inc.
Slide

120. Copyright © 2011 Pearson Education, Inc.
The Vertical Line Test
To determine whether a relation is a function from its graph, perform a vertical line test.
1. Draw or imagine vertical lines through each point in the domain.
2. If each vertical line intersects the graph in at most
one point, the graph is the graph of a function.
3. If any vertical line intersects the graph at two or more different points, then the graph is not the graph of a function.
Copyright © 2011 Pearson Education, Inc.
Slide

121. Copyright © 2011 Pearson Education, Inc.
Example
For each graph, determine the domain and range. Then state whether each relation is a function.
a. b.
Domain: {x|x 1} Range: all real numbers
Domain: all real numbers Range: {y  1}
Not a function
Function
Copyright © 2011 Pearson Education, Inc.
Slide

122. Copyright © 2011 Pearson Education, Inc.
When written as an equation, the notation for a function is a modification of an equation in two variables.
y = 3x + 4 could be written as f(x) = 3x + 4
f(x) is read as “a function in terms of x” or “f of x”
Copyright © 2011 Pearson Education, Inc.
Slide

123. Copyright © 2011 Pearson Education, Inc.
Finding the Value of a Function
Given a function f(x), to find f(a), where a is a real number in the domain of f, replace x in the function with a and calculate the value.
Copyright © 2011 Pearson Education, Inc.
Slide

124. Copyright © 2011 Pearson Education, Inc.
Example 5
For the function f(x) = 3x – 5, find the following.
a. f(2) b. f(4) c. f(a)
Solution
a. f(2) = 3x – 5
= 3(2) – 5
= 6 – 5
= 1
b. f(4) = 3x – 5
c. f(a) = 3x – 5
= 3(4) – 5
= 3(a) – 5
= 12 – 5
= 3a – 5
= 17
Copyright © 2011 Pearson Education, Inc.
Slide

125. Copyright © 2011 Pearson Education, Inc.
Example 6
Use the graph to find the value of the function.
a. f(4) b. f(6) c. f(8)
Solution
a. When x = 4, y = 0, so f(4) = 0.
b. When x = 6, y = 1.5,
so f(6) = 1.5.
c. When x = 8, y = 2, so f(8) = 2.
Copyright © 2011 Pearson Education, Inc.
Slide

126. Copyright © 2011 Pearson Education, Inc.
We create the graph of a function the same way that we create the graph of an equation in two variables.
Slope-intercept form: y = mx + b
Linear function: f(x) = mx + b
Copyright © 2011 Pearson Education, Inc.
Slide

127. Copyright © 2011 Pearson Education, Inc.
Example 7
Graph: f(x) = 2x + 1
Solution
We could make a table of values or use the fact that the slope is 2 and the y-intercept
is 1.
f(x) = 2x + 1 x
f(x) 1 3 2 5
Copyright © 2011 Pearson Education, Inc.
Slide

128. Is the relation a function?
a) yes
b) no
4.7

129. Is the relation a function?
a) yes
b) no
4.7

130. For the function f(x) = 2x2 + x – 4 find f(1).
a) 3
b) –4
c) 2
d) 1
4.7

131. For the function f(x) = 2x2 + x – 4 find f(1).
a) 3
b) –4
c) 2
d) 1
4.7