1. Chapter 3
Graphs and Functions
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8

2. Graphing Equations 3.1 Learning Objectives: Plot ordered pairs
Determine whether an ordered pair of numbers is a solution to an equation in two variables
Graph linear equations
Graph nonlinear equation

3. Graphing an Ordered Pair
y-axis
Quadrant: II
Quadrant: I
The origin is the point (0,0)
x-axis
Quadrant: III
Quadrant: IV
Where would the following points be located:
(2,4) (-3,-5) (-4,1) (2,0) (0,-5)
Quad I
Quad III
Quad II
X-axis
Y-axis

4. Example 1a Determine whether (3, –2) is a solution of 2x + 5y = −4.
2x + 5y = −4 Replace x with 3 and y with –2
2(3) + 5(–2) = −4
6 + (–10) = −4
−4 = −4
true
So (3, −2) is a solution of 2x + 5y = −4.

5. Example 1b Determine whether (–1, 6) is a solution of 3x – y = 5.
3x – y = 5 Replace x with –1 and y with 6
3(–1) – 6 = 5.
–3 – 6 = 5
–9 = 5
false
So (–1, 6) is not a solution of 3x – y = 5.

6. Linear Equations in Two Variables
A linear equation in two variables is an equation that can be written in the form
Ax + By = C
where A, B, and C are integers. A is positive. The greatest common factor for A, B, and C is one.
This form is called standard form.
Finding x- and y-Intercepts
To find an x-intercept, find x when y = 0.
To find a y-intercept, find y when x = 0.

7. Example 2 Graph the linear equation 2x – y = −4. X-intercept:
For this example, we will graph using the x- & y-intercepts
X-intercept:
Find x when y = 0
2x – 0 = -4
2x = -4
x = -2
X-int: (-2,0) x y
Y-intercept:
Find y when x = 0
2(0) – y = -4
– y = -4
y = 4
Y-int: (0,4)

8. Example 3 Graph the linear equation y = x + 3. y x
For this one, we will use slope and y-intercept to graph it.
Start with the y-intercept
b = 3,
written as (0,3)
The slope is
m = ¾

9. Example 4 x y 5 5 x y 2 4 1 1 2
This graph is not a line. Find ordered pairs and plot them on the graph. Connect the points with a smooth curve.
This curve is given a special name, a parabola. The equation is known as a quadratic equation.

10. Example 5 x y 5 5 x y 2 2 1 1
This graph is not a line. Find ordered pairs and plot them on the graph. Connect the points.
We see that this graph is V-shaped.

11. Example 6 Graph y = 4 Points on the line will include:5 5
Graph y = 4
Points on the line will include:
any point where the y-value is 4
(0,4), (1,4), (2,4),…
This is a horizontal line
The slope of a horizontal line is zero

12. Example 7 Graph x = 4 Points on the line will include:y 5 5
Graph x = 4
Points on the line will include:
any point where the x-value is 4
(4,0), (4,1), (4,2),…
This is a vertical line
The slope of a vertical line is undefined

13. Example 8 Are each of the following equations linear or not: Y = 2X
Not linear (quadratic)
Not linear (absolute value)

14. 3.1 Summary
Learning Objectives:
Plot ordered pairs
Determine whether an ordered pair of numbers is a solution to an equation in two variables
Graph linear equations
Graph nonlinear equation
Key Vocabulary:
Rectangular coordinate system
Cartesian
Axis
Origin
Quadrant
Ordered pair
Coordinate
Point
Solution
Standard form
Intercept

15. Introduction to Functions
3.2
Introduction to Functions
Learning Objectives:
Define relation, domain, and range
Identify functions
Use the vertical line test for functions
Find the domain and range of a function
Use function notation 15

16. I prefer to ask: Are the x-coordinates all different?
Relation, Domain, and Range A relation is a set of ordered pairs. The domain of the relation is the set of all first coordinates of the ordered pairs. The range of the relation is the set of all second coordinates of the ordered pairs.
A function is a relation in which each first coordinate in the ordered pair corresponds to exactly one second coordinate.
I prefer to ask: Are the x-coordinates all different?
Vertical Line Test
If no vertical line can be drawn so that it intersects a graph more than once, the graph is the graph of a function.

17. Example 1
Determine the domain and range of the relation. Is the relation a function?
{(4,9), (–4,9), (2,3), (10, –5)}
Domain:
Range:
{4, –4, 2, 10}
{9, 3, –5}
Ask yourself: “Are the x-values all different?”
Answer: Yes, so yes it is a function

18. Example 2 Is the relation y = x2 – 2x a function?
What type of graph is this?
A parabola which opens up
Does it pass the vertical line test?
Answer: yes, so yes it is a function

19. Example 3 Is the relation x2 – y2 = 9 a function?
Each element of the domain (the x-values) would correspond with 2 different values of the range (both a positive and negative y-value), the relation is NOT a function.
For ex: let x = 4 and solve for x
42 – y2 = 9
– y2 = –7
y2 = 7
y =
This would yield two point (4, ) and (4, ) and the x-values are not all different, the relation is NOT a function

20. Example 4
Use the vertical line test to determine whether each graph is the graph of a function. x y x y
This is a function
This is NOT a function

21. Example 5
Use the vertical line test to determine whether each graph is the graph of a function. x y
This is NOT a function
Also, think about points on the line like: (-3,1), (-3,2), (-3,3)
Are the x-values all different? No
So it is NOT a function

22. Example 6 x y
Find the domain and range of the function graphed to the right.
Domain: 3 ≤ x ≤ 4
Domain
Range: 4 ≤ y ≤ 2
Range

23. Example 7 x y
Find the domain and range of the function graphed to the right.
Range: y ≥ – 2
Range
Domain: all real numbers
Domain

24. Example 8
If y = x2 – 2x, re-write the equation in function notation and then find f(–3).
f(x) = x2 – 2x
f(–3) = (–3)2 – 2(–3)
= 9 – (–6)
= 15

25. Example 9 Given the graph of the following function, find each. f(5) =y
f(5) = 7
f(x)
f(4) = 3
f(5) = 1

26. Example 9 Given the graph of the following function, find each.y
f(x)
For how many x-values does f(x) = 2?
three
x ≈-3.5, x=0, x≈3.2

27. 3.2 Summary
Learning Objectives:
Define relation, domain, and range
Identify functions
Use the vertical line test for functions
Find the domain and range of a function
Use function notation
Key Vocabulary:
Relation
Domain
Range
Function
Vertical line test
Function notation
Dependent/independent variable

28. Graphing Linear Functions
3.3
Graphing Linear Functions
Learning Objectives:
Graph linear functions given in slope-intercept form
Graph linear functions by finding intercepts
Graph vertical and horizontal lines 28

29. Finding x- and y-Intercepts To find an x-intercept find x when y = 0 To find a y-intercept find y when x = 0

30. Example 1
Graph g(x) = 3x – 1. Compare this graph with the graph of f(x) = 3x. x y
Notice the graphs are the parallel.
The graph of g(x) is f(x) shifted down one unit.
f(x) = 3x
g(x) = 3x – 1

31. Example 2
Graph the linear functions f(x) = –4x and g(x) = −4x – 5 on the same set of axis. x y
f(x) = −4x
g(x) = −4x – 5
Notice the graphs are the same except the graph of g(x) = −4x – 5 is shifted down five units.

32. Example 3
Find the y-intercept of
The y-intercept of is

33. Example 4 • Find the intercepts and graph: 4 = x – 3y.
The computer will typically want integer points
Solve for y and use the slope
Find the intercepts and graph: 4 = x – 3y.
x-intercept: find x, when y = 0
Then 4 = x – 3y becomes
4 = x – 3(0)
4 = x
x – 3y = 4
–3y = –x + 4
y = 1/3 x – 4/3 x y
So the x-intercept is (4,0).
y-intercept: find y, when x = 0.
Then 4 = x – 3y becomes
4 = 0 – 3y
4 = – 3y.
(7, 1) •
(4, 0)
(0, )
So the y-intercept is (0, ).

34. • • • Find x-intercept and y-intercept and graph – x + 3y = 5
Add on to notes:
Find x-intercept and y-intercept and graph – x + 3y = 5
x-intercept: find x, when y = 0
Then – x + 3y = 5 becomes
– x + 3(0) = 5
x = –5 x y
So the x-intercept is (-5,0).
y-intercept: find y, when x = 0.
Then – x + 3y = 5 becomes
(0) + 3y = 5
y = 5/3 • • •
So the y-intercept is (0,5/3).
Solve the equation for y to find the slope of the line
– x + 3y = 5
3y = x + 5
y = 1/3x + 5/3

35. Example 5 Find the intercepts and graph x = – 3
This line can have any points as long as x = -3
Points like: (-3,1), (-3,2), (-3,3) x y
(-3, 0)
X-intercept:
(-3,0)
Y-intercept:
none

36. Example 6 Find the intercepts and graph y = 3
This line can have any points as long as y = 3
Points like: (1,3), (2,3), (3,3) x y
(0, 3)
X-intercept:
none
Y-intercept:
(0,3)

37. Example 7 Graph x = 3y y Y-intercept: (0,0) Solve for y x = 3y
Homework: system says to graph with x- and y-intercept
Graph x = 3y x y
(0, 0)
Y-intercept:
(0,0)
Solve for y
x = 3y
1/3x = y
x = 3y

38. 3.3 Summary
Learning Objectives:
Graph linear functions given in slope-intercept form
Graph linear functions by finding intercepts
Graph vertical and horizontal lines
Key Vocabulary:
Linear function
Vertical line
Horizontal line

39. The Slope of a Line 3.4 Learning Objectives:
Find the slope given two points
Find the slope given an equation of a line
Find the slope of horizontal and vertical lines
Parallel and Perpendicular slopes 39

40. Slope of a line passing through the points (x1, y1) and (x2, y2) is:
Slope-Intercept Form: y = mx + b,
m is the slope and (0,b) is the y-intercept
Slope of a horizontal line is zero
Slope of a vertical line is undefined

41. Example 1 Find the slope of the line containing the points
(4, −3) and (2, 2). Graph the line.
Notice this is an example of a negative slope. The graph of the line moves downward, or decreases, as we go from left to right.

42. Example 2 Find the slope and y-intercept of the line –3x + y = −5.
Write the equation in slope-intercept form: y = mx + b
Solve for y.
–3x + y = −5
slope is 3
y-intercept is (0, – 5)
+ 3x x
y = 3x – 5

43. Example 3 Find the slope and y-intercept of the line 2x – 6y = 12.
Write the equation in slope-intercept form: y = mx + b
Solve for y.
2x – 6y = 12
– 2x –2x
– 6y = – 2x + 12
________
– – 6
slope is
y-intercept is (0, –2)
y = x – 2

44. Example 4 The slope of a vertical line is undefined.
Find the slope of the line x = −7.
To find the slope, let’s find two ordered pair solutions.
The solutions must have an x-value of −7.
Let’s use (−7, 0) and (−7, 4).
The slope of a vertical line is undefined.

45. Example 5 The slope of the horizontal line is zero.
Find the slope of the line y = −4.
To find the slope, let’s find two ordered pair solutions.
The solutions must have an y-value of −4.
Let’s use (0, −4) and (6, –4).
The slope of the horizontal line is zero.

46. What type of slope does each of the following graphs represent:
Positive slope
Up and to the right
Negative slope
Down and to the right
Undefined slope
Up and to the right zero
Zero slope
Up zero to the right

47. Parallel Lines: Perpendicular Lines:
Have the same slopes but different y-intercepts
Perpendicular Lines:
Have the opposite and reciprocal slopes
The product of slopes of perpendicular lines = 1

48. Example 8
Determine whether the following lines are parallel, perpendicular, or neither.
–5x + y = –6
x + 5y = 5
Solve both equations for y.
Equation 1
y = 5x – 6
Equation 2
x + 5y = 5
The first equation has a slope of 5 and the second equation has a slope of , the lines are perpendicular.

49. 3.4 Summary
Learning Objectives:
Find the slope given two points
Find the slope given an equation of a line
Find the slope of horizontal and vertical lines
Parallel and Perpendicular slopes
Key Vocabulary:
Slope
Rate of Change
Slope-intercept form
Find the slope given an equation of a line
Find the slope of horizontal and vertical lines
Parallel and Perpendicular slopes

50. Equations of Lines 3.5 Learning Objectives:
Use the slope-intercept form to write the equation of a line
Graph a line using the slope and y-intercept
Use the point-slope form to write the equation of a line
Write equations of vertical and horizontal lines
Write equations of parallel and perpendicular lines
Parallel lines have the same slope
Perpendicular lines have opposite and reciprocal slopes 50

51. Slope-Intercept Form of an equation of a line is:
y = mx + b
slope of m
y-intercept of (0, b).
Point-Slope Form of an equation of line is:
y – y1 = m(x – x1)
the line passes through (x1,y1)

52. Example 1
Write the equation of the line with y-intercept (0, 5) and slope of .
slope:
y-intercept:

53. Example 2 Graph Y-intercept: (0,-3) Slope: ¼ Plot: (0, −3)
Rise up 1 unit and run to the right 4 units.
(0,-3)

54. Example 3
Find an equation of a line with slope –2, through the point (–11, –12). Write the equation in slope-intercept form.
Substitute the slope and point into the point-slope form of an equation.
y – (–12) = –2(x – (–11))
y + 12 = –2(x + 11)
y + 12 = –2x – 22
– – 12
y = –2x – 34

55. Example 4
Find the equation of the line through (–4, 0) and (6, –1). Write the equation in standard form.
Find the slope.
Substitute the slope and one of the points into the point-slope form.

56. Example 5
Find the equation of the line passing through points (–4, 3) and (2, 5). Write the equation using function notation.
Find the slope.
3(y – 3) = 1(x + 4)
3y – 9 = x + 4
3y = x + 13
3y = x + 13

57. Example 6
Find an equation of the line containing the point (4, 5) with undefined slope.
Find an equation of the horizontal line containing the point (4, 5).
The equation is y = 5.
The equation is x = 4.

58. Example 7
Find an equation of a line that contains the point (– 2, 4) and is parallel to the line x + 3y = 6.
Write the equation in standard form.
First, we need to find the slope of the given line.
x+ 3y = 6
3y =  x + 6
y = -1/3x + 2
3(y – 4) = -1(x + 2)
The slope of the given line is -1/3.
Parallel lines have the same slope. So we will use the slope of -1/3 for our new equation, together with the given point of (-2,4).
3y – 12 = – x – 2
+x x
x + 3y – 12 = – 2
y – y1 = m(x – x1)
y – 4 = -1/3(x – (-2))
x + 3y = 10

59. Example 8
Write a function that describes the line containing the point (3, 5) and is perpendicular to the line 3x + 2y = 7.
First, we need to find the slope of the given line.
3x + 2y = 7
2y =  3x + 7
y = x +
3y + 15 = 2x – 6
– – 15
3y = 2x – 21
Since perpendicular lines have slopes that are negative reciprocals of each other, we use the slope of 2/3 for our new equation, together with the given point of (3,-5).
3y = 2x – 21
3(y + 5) = 2(x – 3)

60. Example 9
Find the equation of the perpendicular bisector to the segment with endpoints at (-3,4) and (5,-6). Write the equation in standard form.
Use point-slope formula, but remember to take the opposite and reciprocal slope and use the midpoint.
Find the slope of the line segment
5(y + 1) = 4(x – 1)
5y + 5 = 4x – 4
-4x x
-4x + 5y + 5 = -4
-4x + 5y = -9
4x – 5y = 9
Find the midpoint of the line segment

61. 3.5 Summary Learning Objectives:
Use the slope-intercept form to write the equation of a line
Graph a line using the slope and y-intercept
Use the point-slope form to write the equation of a line
Write equations of vertical and horizontal lines
Write equations of parallel and perpendicular lines
Parallel lines have the same slope
Perpendicular lines have opposite and reciprocal slopes
Formulas:
Slope-intercept: y = mx + b
Point-slope: y – y1 = m(x – x1)
Standard form: Ax + By = C,
where A,B,&C are integers
A is positive
where A,B,&C’s GCD is 1
UIL Math:
Show shortcuts for the last two examples

62. 3.6
Graphing Piecewise-Defined Functions and Shifting and Reflecting Graphs of Functions
Learning Objectives:
Graph piecewise-defined functions
Vertical and horizontal shifts
Reflect graphs 62

63. Example 1
Evaluate f(2), f(−4), and f(0) for the function
Then write your results in ordered-pair form.
f(−4)
f(0)
f(2)
f(−4) = 3x +1
f(2) = x – 4
f(0) = 3x + 1
f(2) = 2 – 4
f(−4) = 3(−4) +1
f(0) = 3(0) + 1
f(2) = –2
f(−4) = −11
f(0) = 1
(−4, –11)
(0, 1)
(2, –2)

64. Example 2 Graph y x Graph each “piece” separately. x f (x) = 3x – 1 –1
What is the domain and range for the graph?
Domain: all real numbers
Range: y < -1 or y > 3
Graph
Graph each “piece” separately. x y x
f (x) = 3x – 1
(closed circle) –1 –2
– 1
– 4
– 7
Values  0 x
f (x) = x + 3
(open circle) 1 2 3 x
f (x) = x + 3 3 4 5 6
Values > 0

65. Example 3 X 2X X X–4 X 3X Graph y x -2 -4 -2 -6 -3 -6 -4 -4 -8 1 -3 1
closed
open -3 -6 -4 -4 -8 1 -3
open X 3X 1 3
closed 2 6 3 9

67. Vertical Shifts (Upward or Downward)
Let k be a Positive Number
Graph of
Same As
Moved
g(x) = f(x) + k
f(x)
k units upward
g(x) = f(x)  k
k units downward
Horizontal Shifts (To the Left or Right)
Let h be a Positive Number
Graph of
Same As
Moved
g(x) = f(x  h)
f(x)
h units to the right
g(x) = f(x + h)
h units to the left

68. Example 4 Begin with the graph of f(x) = x2.y 5 5
Begin with the graph of f(x) = x2.
Shift the original graph downward 3 units.
What is the domain and range for g(x) (the red graph)?
Domain: all real numbers
Range: y > –3

69. Example 5 Begin with the graph of f(x) = |x|.y 5 5
Begin with the graph of f(x) = |x|.
Shift the original graph to the left 2 units.
What is the domain and range for g(x) (the red graph)?
Domain: all real numbers
Range: y > 0

70. Reflections about the x-axis
The graph of g(x) = – f(x) is the graph of f(x) reflected about the x-axis.

71. Example 6
+ 4 x y 5 5
The graph is reflected about the x-axis, then moved two units left and four units up.

72. Example 7 Given the function What is the domain? What is the range?
Think about the parent function graph, and how it has been shifted
Right 20 and up 12 units
(20,12)
Domain: x > 20
Range: y > 12

73. 3.6 Summary Learning Objectives: Graph piecewise-defined functions
Vertical and horizontal shifts
Reflect graphs
Vocabulary:
Vertical shift
Horizontal shift
reflection

74. Graphing Linear Inequalities
3.7
Graphing Linear Inequalities
Learning Objectives:
Graph linear inequalities
Graph the intersection or union of two linear inequalities 74

75. Example 1 Graph 7x + y > –14. y
Graph y = –7x – 14 as a dashed line.
I used x-int of (0,-2) and the slope of -7
Pick a point not on the graph: (0,0)
(0, 0)
Test it in the original inequality.
7(0) + 0 > –14,
0 > –14
True, so shade the side containing (0,0).

76. Example 2 Graph 3x + 5y  –2. y (0, 0) x Graph as a solid line.
Graph as a solid line.
Points on the line (0,-2/5) (1, -1)
I used (1,-1) with a slope of -3/5
(0, 0)
Pick a point not on the graph: (0,0), but just barely
Test it in the original inequality.
3(0) + 5(0) > –2
0 > –2
False, so shade the side that does not contain (0,0).

77. Example 3 Graph 3x < 15. Graph 3x = 15 as a dashed line. (x = 5)y
Graph 3x = 15 as a dashed line. (x = 5)
Pick a point not on the graph: (0,0)
(0, 0)
Test it in the original inequality.
3(0) < 15
0 < 15
True, so shade the side containing (0,0).

78. Example 3 Graph the intersection of x  1 and y  2x – 1.
Graph each inequality. The intersection of the two graphs is all points common in both regions.

79. Example 4 Graph the union of
Graph each inequality. The union of both inequalities is all the shaded regions.

80. Warning!
3.7 Summary
Note that although all of our examples allowed us to select (0, 0) as our test point, that will not always be true.
If the boundary line contains (0,0), you must select another point that is not contained on the line as your test point.
Learning Objectives:
Graph linear inequalities
Graph the intersection or union of two linear inequalities
Vocabulary:
Boundary
Half-plane
Solution region

81. Stretching and Compression Graphs of Absolute Value
3.8
Stretching and Compression Graphs of Absolute Value
Learning Objectives:
Graph absolute value functions
Write an equation of an absolute value function graph 81

82. Example 1 Graph h(x) = 4|x| and x h(x) g(x) −2 8 1/2 −1 1 1/4 4 2 y
Find and plot ordered-pairs solutions for the functions.
h(x) = 4|x| x
h(x)
g(x) −2 8
1/2 −1 1
1/4 4 2

83. The Graph of the Absolute Value Function
f(x) = a│x – h│ + k
(h,k) and is V-shaped
Opens up if a is positive and down if a is negative
Shifts h units right/left and k units up/down
The larger the │a│ the narrower the graph
The smaller the │a│ the wider the graph

84. Example 2 Graph What type of graph is this? V-shape
Opens which direction?
downward
Vertex?
(−3, 1)
Steepness? –¼

85. Example 3 Write an equation of the absolute value function graphed.
The vertex is
(2,−1)
h = 2
k = −1
a = 2
The equation of the graph is

86. 3.8 Summary Learning Objectives: Graph absolute value functions
Write an equation of an absolute value function graph