1. The Absolute Value Function
Section 3.4
The Absolute Value Function

2. Objectives:
Define and graph translations, reflections, and scalar transformations of the absolute value function.
Solve equations involving absolute value symbols by using graphing and algebraic methods.

3. Absolute Value The absolute value is the distance from zero.
Bars are used to show absolute value.
l -2 l = 2 and l 2 l = 2 -2 2
2 from 0
2 from 0

4. Example 1: Simplify
|9| + |-9|

5. Example 1: Simplify
|9| + |-9|
|9| + |-9| = 9 + 9

6. Example 1: Simplify
|9| + |-9|
|9| + |-9| = 9 + 9
= 18

7. Example 1: Simplify
|9| + |-9|
|9| + |-9| = 9 + 9
= 18
|13| - |-2|

8. |9| + |-9| |9| + |-9| = 9 + 9 = 18 |13| - |-2| |13| - |-2| = 13 – 2
Example 1: Simplify
|9| + |-9|
|9| + |-9| = 9 + 9
= 18
|13| - |-2|
|13| - |-2| = 13 – 2

9. |9| + |-9| |9| + |-9| = 9 + 9 = 18 |13| - |-2| |13| - |-2| = 13 – 2
Example 1: Simplify
|9| + |-9|
|9| + |-9| = 9 + 9
= 18
|13| - |-2|
|13| - |-2| = 13 – 2
= 11

10. Properties of Absolute Value
|-a| = |a|
|ab| = |a| |b|
a = |a|
b |b| b ≠ 0

11. Absolute Value is defined by:

12. Solving absolute value equations
First, isolate the absolute value expression.
Set up two equations to solve.
For the first equation, drop the absolute value bars and solve the equation.
For the second equation, drop the bars, negate the opposite side, and solve the equation.
Always check the solutions!!!!!!!!!!!!!!!!!

13. Example 2: Solve 6|5x + 2| = 312 6|5x + 2| = 312 |5x + 2| = 52
Isolate the absolute value expression by dividing by 6.
6|5x + 2| = 312
|5x + 2| = 52
Set up two equations to solve.
5x + 2 = 52 5x + 2 = -52
5x = x = -54
x = 10 or x = -10.8
Check: 6|5x + 2| = |5x + 2| = 312
6|5(10)+2| = |5(-10.8)+ 2| = 312
6|52| = |-52| = 312
312 = = 312

14. Example 3: Solve 3|x + 2| -7 = 14 3|x + 2| -7 = 14 3|x + 2| = 21
Isolate the absolute value expression by adding 7 and dividing by 3.
3|x + 2| -7 = 14
3|x + 2| = 21
|x + 2| = 7
Set up two equations to solve.
x + 2 = x + 2 = -7
x = or x = -9
Check: 3|x + 2| - 7 = |x + 2| -7 = 14
3|5 + 2| - 7 = |-9+ 2| -7 = |7| - 7 = |-7| -7 = 14
= = 14
14 = = 14

15. The graph of the absolute value function is V-shaped and opens up.
To the right of
x = 0 the line is y = x
To the left of
x=0 the line is
y = -x
The graph is symmetric with the y-axis
Domain: (-∞,∞)
Range: [0, ∞).

16. Calculator:
To graph the absolute value on a calculator, go to the MATH menu. Then go to NUM and down to abs(

17. Example 4: Using the Graphing Calculator
Graph f(x) = -│3 – x│. Find the domain and range.

18. What do you think the inverse of the absolute value function would look like?

19. Homework:
Pg # 140 Exercises 14 – 21 and 24 – 29

20. This will help us quickly sketch graphs of certain functions.
We will now see how certain transformations (operations) of a function change its graph.
This will help us quickly sketch graphs of certain functions.
The transformations are: (1) shifts, (2) reflections, and (3) stretching.

21. Translations: When you slide a function on the graph up, down, left, or right.

22. Vertical shifts Moves the graph up or down
Impacts only the “y” values of the function
No changes are made to the “x” values

23. Vertical Translation For c > 0,
The graph of y = f(x) + c is the graph of y = f(x) shifted up c units
The graph of y = f(x)  c is the graph of y = f(x) shifted down c units.

24. The shape of the graph is not changed, only its position.

25. Example 5:
Graph the function g(x) = │x│+ 1. a) Explain the relationship between the graph of g and the graph of f(x) =│x│. b) Solve │x│+ 1 = 7 graphically. c) Solve │x│+ 1 = 7 algebraically.

26. Horizontal shifts Moves the graph left or right
Impacts only the “x” values of the function
No changes are made to the “y” values

27. Horizontal Translation
For c > 0,
The graph of y = f(x  c) is the graph of y = f(x) shifted right c units.
The graph of y = f(x + c) is the graph of y = f(x) shifted left c units.
y = (x - 2)²

28. The shape is not changed, but the graph is shifted left or right.
Example:
Shift 3 units left.

29. Another Example:

30. Note the shift to the right 2 units.
Example:
Note the shift to the right 2 units.

31. Example 6:
Graph the function j(x) = │x + 1│. a) Explain the relationship between the graph of j and the graph of f(x) =│x│. b) Solve │x + 1│= 3 graphically. c) Solve │x + 1│= 3 algebraically.

32. Example 7: Write the new function. Shifting the function f(x) =
Vertical shift of 3 units up
Horizontal shift of 3 units left
HINT: x’s go the opposite direction than you might believe.

33. *Some graphs can be obtained from a combination of vertical and horizontal shifts

34. Combining a vertical & horizontal shift
Example of function that is shifted down 4 units and right 6 units from the original function.

35. Homework:
Vertical and Horizontal Shifts Worksheet

36. The second common type of transformation is a REFLECTION!

37. REFLECTIONS about AXES

38. Reflection about x-axis. Change the sign of each y-coordinate.
Reflections of Graphs
Reflection about x-axis. Change the sign of each y-coordinate.

39. Reflections about y-axis. Change the sign of each x coordinate.
Reflections of Graphs
Reflections about y-axis. Change the sign of each x coordinate.

40. The graph is a parabola with x-intercepts 2 and 1.
Example 8:
For the function f(x) = x2 + x  2 graph its reflection across the x-axis and across the y-axis.
The graph is a parabola with x-intercepts 2 and 1.

41. f(x) = x2 + x  2
To obtain its reflection across the x-axis, graph y = f(x), or y = (x2 + x  2).
The x-intercepts have not changed, but the y-coordinates change sign.

42. f(x) = x2 + x  2
To obtain the reflection across the y-axis let y = f(x), or y = (x)2  x  2.
The x-intercepts have changed to 1 and 2, but the y-coordinates are unchanged.

43. Example 9:
Describe the graph of s(x) = -│x│. It is a reflection of f(x) = │x│ over the x-axis.

44. Homework:
Reflections Worksheet P.7 Exercises 1, 2, 3, 4, 5(b), 5(f), 6(e), 9, 11, 12, 14

45. Scalar Transformations
Scalar transformations cause a distortion in the graph.
The basic shape of the graph is changed.
Graphs can shrink or stretch.

46. Vertical Stretching and Vertical Shrinking
The graph of y = cf(x) is a vertical stretching of y = f(x) if c > 1.
The graph of y = cf(x) is a vertical shrinking of y = f(x) if 0 < c < 1.

47. Horizontal Stretching and Shrinking
The graph of y = f(cx) is a horizontal shrinking of y = f(x) if c > 1.
The graph of y = f(cx) is a horizontal stretching of y = f(x) if 0 < c < 1.

48. Use the graph of y = f(x) to sketch the graph of each equation.
Example 10:
Use the graph of y = f(x) to sketch the graph of each equation.
a) y = 2f(x) b)
(2, 1)
(0, 2)
(2, 1)
y = f(x)

49. Multiply each y-coordinate by 2.
Solution:
(2, 1)
(0, 2)
(2, 1)
y = f(x)
a) y = 2f(x)
Vertical stretching
Multiply each y-coordinate by 2.
(2, 1  2) = (2, 2)
(0, 2  2) = (0, 4)
(2, 1  2) = (2, 2)

50. (2, 1)
(0, 2)
(2, 1)
y = f(x) b)
Horizontal stretching
Multiply each x coordinate by 2.

51. Example 11:
Graph the function j(x) = 2│x│ and h(x) = ½ │x│. a) Compare the graphs of j and h with the graph of f(x) =│x│. b) Describe the axis of symmetry, domain, and range of each.

52. Classwork/Homework Horizontal and Vertical Stretches/Shrinks Worksheet
P.7 Exercises: 5(a, c, d, e), 6(a, b, c, d, f), 7, 8, 10, 13

53. Combining Transformations
Transformations of graphs can be combined to create new graphs.
For example the graph of y = 2(x + 3)2 + 1 can be obtained by performing four transformations on the graph of y = x2.

54. Combining Transformations
y = 2(x + 3)2 + 1 can be obtained by performing four transformations on the graph of y = x2:
1. Shift the graph 3 units left: y = (x + 3)2
2. Vertically stretch the graph by a factor of 2: y = 2(x + 3)2
3. Reflect the graph across the x-axis:
y = 2(x + 3)2
4. Shift the graph upward 1 unit: y = 2(x + 3)2 + 1

55. Original Function y = x²

56. 1. Shift of the graph 3 units left: y = (x + 3)²

57. 2. Vertically stretch the graph by a factor of 2: y = 2(x + 3)²

58. 3. Reflect the graph across the x-axis: y = 2(x + 3)²

59. 4. Shift the graph upward 1 unit: y = 2(x + 3)² + 1

60. Vertical shift of one unit. Shift left 3 units.
y = 2(x + 3)2 + 1
Reflects about the x-axis.
Stretch vertically by a factor of 2.

61. Example 12: Describe how the graph of the equation g(x) = -|x + 3| - 2 can be obtained by transforming the graph of f(x) = |x|. Then graph the equation.
Reflect the graph across the x-axis.  -|x|
Shift the graph left 3 units.  -| x + 3|
Shift the graph down 2 units.  -|x + 3| -2

62. Example 13: Write the equation for
y = ½|x| + 3

63. Classwork:
Combining Transformations Worksheet

64. Homework:
Practice and Apply worksheet 3.4