1. Splash Screen

3. 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

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

7. 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

8. A. B. C. D.
5–Minute Check 5

9. 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

10. 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

11. natural numbers:
whole numbers:
Integers:
rational numbers:
irrational numbers:
real numbers:

12. Key Concept 1

13. 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 but 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

14. 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

15. 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

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

17. This concept also describes the behavior of “x” in a function.
Inequalities can be expressed geometrically or by using interval notation.
This concept also describes the behavior of “x” in a function.

18. Interval Notation < > ≤ ≥
( )
NOT included
[ ]
Included

19. Negative and Positive Infinity are never INCLUDED VALUES
The symbols and Indicate that the corresponding line segments extend infinitely far to the left or right.
Negative and Positive Infinity are never INCLUDED VALUES
Real Numbers (7)

20. IMPORTANT WORD MEANINGS
“AND”
Intersection- what two things have in common.

21. Bounded intervals

22. Use Interval Notation
The set includes all real numbers greater than or equal to –2 AND less than but equal to 12.
Answer: [–2, 12]
Example 2

23. 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

24. IMPORTANT WORD MEANINGS
Union- two things combined to make one set

25. Unbounded intervals

26. 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

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 OR all real numbers greater than or equal to 54.
Answer:
Example 2

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

30. Concept of Function and Relation
Relation: Any set of ordered pairs where Each value of the first variable is paired with one or more values of the second variable.

31. Concept of Function and Relation
A FUNCTION is a relation in which each element of the domain is paired with exactly one element of the range.

32. Another way of saying it is that there is one and only one output (y) with each input (x).
f(x) x y

33. Is the relation a function?
{(2, 3), (3, 0), (5, 2), (4, 3)}
YES, every domain is different!
f(x) 2 3
f(x) 3
f(x) 5 2
f(x) 4 3

34. Is the relation a function.
{(4, 1), (5, 2), (5, 3), (6, 6), (1, 9)}
f(x) 4 1
f(x) 5 2
NO, 5 is paired with 2 numbers!
f(x) 5 3
f(x) 6
f(x) 1 9

35. Identify Functions Verbally
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.
Key Concept 3

36. Identify Functions Numerically
Key Concept 3

37. Is this relation a function? {(1,3), (2,3), (3,3)}
Yes No

38. 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

39. Identify Functions Graphically
Key Concept 3a

40. Vertical-Line Test
If every vertical line intersects a graph only once, then the graph represents a function.
function
not a function

42. 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

43. Identify Functions Algebraically
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 Original equation
Divide each side by 3.
Take the square root of each side.

44. 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

45. 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

46. Evaluate functions. f(x) = –2.5x + 11 where x = –1 Find f(-1):
Key Skills

47. 3x-2 3 7 3(3)-2 -2 -8 3x-2 3(-2)-2 f(3)= f(-2)=
Given f(x) = 3x - 2, find:
f(3)=
f(-2)= 3
3x-2
3(3)-2 7 -2
3x-2
3(-2)-2 -8

48. Given h(z) = z2 - 4z + 9, find h(-3)
(-3)2 – 4(-3) + 9
9+12+9 -3 30
h(-3) = 30

49. 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

50. 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

51. 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

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

53. Find the Domain of a Function 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

54. 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

55. 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

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

57. Evaluate a Piecewise Function
Piecewise Function –a function defined by two or more functions over a specified domain.
What do they look like?
f(x) =
2x + 1 , x  0
x – 1 , x  0

58. Evaluating Piecewise Functions:
Evaluating piecewise functions is just like evaluating functions that you are already familiar with.
f(x) =
2x + 1 , x  0
x – 1 , x  0
Let’s calculate f(2).
You are being asked to find y when
x = 2. Since 2 is  0, you will only substitute into the second part of the function.
f(2) = 2 – 1 = 1

59. Evaluate the following:
f(x) =
3x - 2, x  -2
-x , -2  x  1
2x – 7x, x  1
f(-2) = 2
f(3) =
-12
f(-4) =
-14
f(1) = -6

60. 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

61. 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

62. 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

63. 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

64. 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

65. 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

66. 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

67. HOMEWORK
Pgs. 9-12
See Notes

68. End of the Lesson