1. Splash Screen

2. Key Concept: Real Numbers Example 1: Use Set-Builder Notation
Five-Minute Check
Then/Now
New Vocabulary
Key Concept: Real Numbers
Example 1: Use Set-Builder Notation
Example 2: Use Interval Notation
Key Concept: Function
Key Concept: Vertical Line Test
Example 3: Identify Relations that are Functions
Example 4: Find Function Values
Example 5: Find Domains Algebraically
Example 6: Real-World Example: Evaluate a Piecewise-Defined Function
Lesson Menu

3. Find the value of x 2 + 4x + 4 if x = –2.
B. 0
C. 4
D. 16
5–Minute Check 1

4. Find the value of x 2 + 4x + 4 if x = –2.
B. 0
C. 4
D. 16
5–Minute Check 1

5. Solve 5n + 6 = –3n – 10.
A. –8
B. –2 C.
D. 2
5–Minute Check 2

6. Solve 5n + 6 = –3n – 10.
A. –8
B. –2 C.
D. 2
5–Minute Check 2

7. Evaluate |x – 2y| – |2x – y| – xy if x = –2 and y = 7.
B. 9
C. 19
D. 41
5–Minute Check 3

8. Evaluate |x – 2y| – |2x – y| – xy if x = –2 and y = 7.
B. 9
C. 19
D. 41
5–Minute Check 3

9. Factor 8xy 2 – 4xy. A. 2x(4xy 2 – y) B. 4xy(2y – 1) C. 4xy(y 2 – 1)
D. 4y 2(2x – 1)
5–Minute Check 4

10. Factor 8xy 2 – 4xy. A. 2x(4xy 2 – y) B. 4xy(2y – 1) C. 4xy(y 2 – 1)
D. 4y 2(2x – 1)
5–Minute Check 4

11. A. B. C. D.
5–Minute Check 5

12. A. B. C. D.
5–Minute Check 5

13. Describe subsets of real numbers.
You used set notation to denote elements, subsets, and complements. (Lesson 0-1)
Describe subsets of real numbers.
Identify and evaluate functions and state their domains.
Then/Now

14. piecewise-defined function relevant domain
set-builder notation
interval notation
function
function notation
independent variable
dependent variable
implied domain
piecewise-defined function
relevant domain
Vocabulary

15. Key Concept 1

16. A. Describe {2, 3, 4, 5, 6, 7} using set-builder notation.
Use Set-Builder Notation
A. Describe {2, 3, 4, 5, 6, 7} using set-builder notation.
The set includes natural numbers greater than or equal to 2 and less than or equal to 7.
This is read as the set of all x such that 2 is less than or equal to x and x is less than or equal to 7 and x is an element of the set of natural numbers.
Answer:
Example 1

17. A. Describe {2, 3, 4, 5, 6, 7} using set-builder notation.
Use Set-Builder Notation
A. Describe {2, 3, 4, 5, 6, 7} using set-builder notation.
The set includes natural numbers greater than or equal to 2 and less than or equal to 7.
This is read as the set of all x such that 2 is less than or equal to x and x is less than or equal to 7 and x is an element of the set of natural numbers.
Answer:
Example 1

18. B. Describe x > –17 using set-builder notation.
Use Set-Builder Notation
B. Describe x > –17 using set-builder notation.
The set includes all real numbers greater than –17.
Answer:
Example 1

19. B. Describe x > –17 using set-builder notation.
Use Set-Builder Notation
B. Describe x > –17 using set-builder notation.
The set includes all real numbers greater than –17.
Answer:
Example 1

20. C. Describe all multiples of seven using set-builder notation.
Use Set-Builder Notation
C. Describe all multiples of seven using set-builder notation.
The set includes all integers that are multiples of 7.
Answer:
Example 1

21. C. Describe all multiples of seven using set-builder notation.
Use Set-Builder Notation
C. Describe all multiples of seven using set-builder notation.
The set includes all integers that are multiples of 7.
Answer:
Example 1

22. Describe {6, 7, 8, 9, 10, …} using set-builder notation.
Example 1

23. Describe {6, 7, 8, 9, 10, …} using set-builder notation.
Example 1

24. A. Write –2 ≤ x ≤ 12 using interval notation.
Use Interval Notation
A. Write –2 ≤ x ≤ 12 using interval notation.
The set includes all real numbers greater than or equal to –2 and less than or equal to 12.
Answer:
Example 2

25. A. Write –2 ≤ x ≤ 12 using interval notation.
Use Interval Notation
A. Write –2 ≤ x ≤ 12 using interval notation.
The set includes all real numbers greater than or equal to –2 and less than or equal to 12.
Answer: [–2, 12]
Example 2

26. B. Write x > –4 using interval notation.
Use Interval Notation
B. Write x > –4 using interval notation.
The set includes all real numbers greater than –4.
Answer:
Example 2

27. B. Write x > –4 using interval notation.
Use Interval Notation
B. Write x > –4 using interval notation.
The set includes all real numbers greater than –4.
Answer: (–4, )
Example 2

28. C. Write x < 3 or x ≥ 54 using interval notation.
Use Interval Notation
C. Write x < 3 or x ≥ 54 using interval notation.
The set includes all real numbers less than 3 and all real numbers greater than or equal to 54.
Answer:
Example 2

29. C. Write x < 3 or x ≥ 54 using interval notation.
Use Interval Notation
C. Write x < 3 or x ≥ 54 using interval notation.
The set includes all real numbers less than 3 and all real numbers greater than or equal to 54.
Answer:
Example 2

30. Write x > 5 or x < –1 using interval notation.B.
C. (–1, 5) D.
Example 2

31. Write x > 5 or x < –1 using interval notation.B.
C. (–1, 5) D.
Example 2

32. Key Concept 3

33. Key Concept 3a

34. Identify Relations that are Functions
A. Determine whether the relation represents y as a function of x. The input value x is the height of a student in inches, and the output value y is the number of books that the student owns.
Answer:
Example 3

35. Answer: No; there is more than one y-value for an x-value.
Identify Relations that are Functions
A. Determine whether the relation represents y as a function of x. The input value x is the height of a student in inches, and the output value y is the number of books that the student owns.
Answer: No; there is more than one y-value for an x-value.
Example 3

36. B. Determine whether the table represents y as a function of x.
Identify Relations that are Functions
B. Determine whether the table represents y as a function of x.
Answer:
Example 3

37. B. Determine whether the table represents y as a function of x.
Identify Relations that are Functions
B. Determine whether the table represents y as a function of x.
Answer: No; there is more than one y-value for an x-value.
Example 3

38. C. Determine whether the graph represents y as a function of x.
Identify Relations that are Functions
C. Determine whether the graph represents y as a function of x.
Answer:
Example 3

39. C. Determine whether the graph represents y as a function of x.
Identify Relations that are Functions
C. Determine whether the graph represents y as a function of x.
Answer: Yes; there is exactly one y-value for each x-value. Any vertical line will intersect the graph at only one point. Therefore, the graph represents y as a function of x.
Example 3

40. D. Determine whether x = 3y 2 represents y as a function of x.
Identify Relations that are Functions
D. Determine whether x = 3y 2 represents y as a function of x.
To determine whether this equation represents y as a function of x, solve the equation for y.
x = 3y 2 Original equation
Divide each side by 3.
Take the square root of each side.
Example 3

41. Identify Relations that are Functions
This equation does not represent y as a function of x because there will be two corresponding y-values, one positive and one negative, for any x-value greater than 0.
Let x = 12.
Answer:
Example 3

42. Answer: No; there is more than one y-value for an x-value.
Identify Relations that are Functions
This equation does not represent y as a function of x because there will be two corresponding y-values, one positive and one negative, for any x-value greater than 0.
Let x = 12.
Answer: No; there is more than one y-value for an x-value.
Example 3

43. Determine whether 12x 2 + 4y = 8 represents y as a function of x.
A. Yes; there is exactly one y-value for each x-value.
B. No; there is more than one y-value for an x-value.
Example 3

44. Determine whether 12x 2 + 4y = 8 represents y as a function of x.
A. Yes; there is exactly one y-value for each x-value.
B. No; there is more than one y-value for an x-value.
Example 3

45. To find f (3), replace x with 3 in f (x) = x 2 – 2x – 8.
Find Function Values
A. If f (x) = x 2 – 2x – 8, find f (3).
To find f (3), replace x with 3 in f (x) = x 2 – 2x – 8.
f (x) = x 2 – 2x – 8 Original function
f (3) = 3 2 – 2(3) – 8 Substitute 3 for x.
= 9 – 6 – 8 Simplify.
= –5 Subtract.
Answer:
Example 4

46. To find f (3), replace x with 3 in f (x) = x 2 – 2x – 8.
Find Function Values
A. If f (x) = x 2 – 2x – 8, find f (3).
To find f (3), replace x with 3 in f (x) = x 2 – 2x – 8.
f (x) = x 2 – 2x – 8 Original function
f (3) = 3 2 – 2(3) – 8 Substitute 3 for x.
= 9 – 6 – 8 Simplify.
= –5 Subtract.
Answer: –5
Example 4

47. B. If f (x) = x 2 – 2x – 8, find f (–3d).
Find Function Values
B. If f (x) = x 2 – 2x – 8, find f (–3d).
To find f (–3d), replace x with –3d in f (x) = x 2 – 2x – 8.
f (x) = x 2 – 2x – 8 Original function
f (–3d) = (–3d)2 – 2(–3d) – 8 Substitute –3d for x.
= 9d 2 + 6d – 8 Simplify.
Answer:
Example 4

48. B. If f (x) = x 2 – 2x – 8, find f (–3d).
Find Function Values
B. If f (x) = x 2 – 2x – 8, find f (–3d).
To find f (–3d), replace x with –3d in f (x) = x 2 – 2x – 8.
f (x) = x 2 – 2x – 8 Original function
f (–3d) = (–3d)2 – 2(–3d) – 8 Substitute –3d for x.
= 9d 2 + 6d – 8 Simplify.
Answer: 9d 2 + 6d – 8
Example 4

49. C. If f (x) = x2 – 2x – 8, find f (2a – 1).
Find Function Values
C. If f (x) = x2 – 2x – 8, find f (2a – 1).
To find f (2a – 1), replace x with 2a – 1 in f (x) = x 2 – 2x – 8.
f (x) = x 2 – 2x – 8 Original function
f (2a – 1) = (2a – 1)2 – 2(2a – 1) – 8 Substitute 2a – 1 for x.
= 4a 2 – 4a + 1 – 4a + 2 – 8 Expand (2a – 1)2 and 2(2a – 1).
= 4a 2 – 8a – 5 Simplify.
Answer:
Example 4

50. C. If f (x) = x2 – 2x – 8, find f (2a – 1).
Find Function Values
C. If f (x) = x2 – 2x – 8, find f (2a – 1).
To find f (2a – 1), replace x with 2a – 1 in f (x) = x 2 – 2x – 8.
f (x) = x 2 – 2x – 8 Original function
f (2a – 1) = (2a – 1)2 – 2(2a – 1) – 8 Substitute 2a – 1 for x.
= 4a 2 – 4a + 1 – 4a + 2 – 8 Expand (2a – 1)2 and 2(2a – 1).
= 4a 2 – 8a – 5 Simplify.
Answer: 4a 2 – 8a – 5
Example 4

51. If , find f (6). A. B. C. D.
Example 4

52. If , find f (6). A. B. C. D.
Example 4

53. A. State the domain of the function .
Find Domains Algebraically
A. State the domain of the function
Because the square root of a negative number cannot be real, 4x – 1 ≥ 0. Therefore, the domain of g(x) is all real numbers x such that x ≥ , or
Answer:
Example 5

54. A. State the domain of the function .
Find Domains Algebraically
A. State the domain of the function
Because the square root of a negative number cannot be real, 4x – 1 ≥ 0. Therefore, the domain of g(x) is all real numbers x such that x ≥ , or
Answer: all real numbers x such that x ≥ , or
Example 5

55. B. State the domain of the function .
Find Domains Algebraically
B. State the domain of the function
When the denominator of is zero, the expression is undefined. Solving t 2 – 1 = 0, the excluded values in the domain of this function are t = 1 and t = –1. The domain of this function is all real numbers except t = 1 and t = –1, or
Answer:
Example 5

56. B. State the domain of the function .
Find Domains Algebraically
B. State the domain of the function
When the denominator of is zero, the expression is undefined. Solving t 2 – 1 = 0, the excluded values in the domain of this function are t = 1 and t = –1. The domain of this function is all real numbers except t = 1 and t = –1, or
Answer:
Example 5

57. C. State the domain of the function .
Find Domains Algebraically
C. State the domain of the function
This function is defined only when 2x – 3 > 0. Therefore, the domain of f (x) is or
Answer:
Example 5

58. C. State the domain of the function .
Find Domains Algebraically
C. State the domain of the function
This function is defined only when 2x – 3 > 0. Therefore, the domain of f (x) is or
Answer: or
Example 5

59. State the domain of g (x) = .
A or [4, ∞)
B or [–4, 4]
C or (− , −4] D.
Example 5

60. State the domain of g (x) = .
A or [4, ∞)
B or [–4, 4]
C or (− , −4] D.
Example 5

61. Evaluate a Piecewise-Defined Function
A. FINANCE Realtors in a metropolitan area studied the average home price per square foot as a function of total square footage. Their evaluation yielded the following piecewise-defined function. Find the average price per square foot for a home with the square footage of square feet.
Example 6

62. Because 1400 is between 1000 and 2600, use to find p(1400).
Evaluate a Piecewise-Defined Function
Because 1400 is between 1000 and 2600, use to find p(1400).
Function for 1000 ≤ a < 2600
Substitute 1400 for a.
Subtract.
= 85 Simplify.
Example 6

63. Evaluate a Piecewise-Defined Function
According to this model, the average price per square foot for a home with a square footage of 1400 square feet is $85.
Answer:
Example 6

64. Answer: $85 per square foot
Evaluate a Piecewise-Defined Function
According to this model, the average price per square foot for a home with a square footage of 1400 square feet is $85.
Answer: $85 per square foot
Example 6

65. Evaluate a Piecewise-Defined Function
B. FINANCE Realtors in a metropolitan area studied the average home price per square foot as a function of total square footage. Their evaluation yielded the following piecewise-defined function. Find the average price per square foot for a home with the square footage of 3200 square feet.
Example 6

66. Because 3200 is between 2600 and 4000, use to find p(3200).
Evaluate a Piecewise-Defined Function
Because 3200 is between 2600 and 4000, use
to find p(3200).
Function for ≤ a <
Substitute 3200 for a.
Simplify.
Example 6

67. Evaluate a Piecewise-Defined Function
According to this model, the average price per square foot for a home with a square footage of 3200 square feet is $104.
Answer:
Example 6

68. Answer: $104 per square foot
Evaluate a Piecewise-Defined Function
According to this model, the average price per square foot for a home with a square footage of 3200 square feet is $104.
Answer: $104 per square foot
Example 6

69. ENERGY The cost of residential electricity use can be represented by the following piecewise function, where k is the number of kilowatts. Find the cost of electricity for 950 kilowatts.
A. $47.50
B. $48.00
C. $57.50
D. $76.50
Example 6

70. ENERGY The cost of residential electricity use can be represented by the following piecewise function, where k is the number of kilowatts. Find the cost of electricity for 950 kilowatts.
A. $47.50
B. $48.00
C. $57.50
D. $76.50
Example 6

71. End of the Lesson