Splash Screen.

Slides:



Advertisements
Similar presentations
B ELL W ORK : F RIDAY, S EPTEMBER 5 TH Name the property illustrated by –15b + 15b = 0. Solve 2(c – 5) – 2 = 8 + c. Solve |3x – 5| + 4 = 14. Solve 2b –
Advertisements

Splash Screen.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 2–1) CCSS Then/Now New Vocabulary Example 1:Identify Linear Functions Example 2:Real-World Example:
Splash Screen. Concept 1 Example 1 Domain and Range State the domain and range of the relation. Then determine whether the relation is a function. If.
Splash Screen. Lesson Menu Five-Minute Check (over Chapter 1) CCSS Then/Now New Vocabulary Key Concept: Functions Example 1:Domain and Range Key Concept:
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 6–1) CCSS Then/Now New Vocabulary Key Concept: Inverse Relations Example 1:Find an Inverse Relation.
Splash Screen. Lesson Menu Five-Minute Check (over Chapter 8 ) CCSS Then/Now New Vocabulary Key Concept: Quadratic Functions Example 1: Graph a Parabola.
Section 2.1 Notes: Relations and Functions
Splash Screen. Lesson Menu Five-Minute Check (over Chapter 1) Then/Now New Vocabulary Key Concept: Functions Example 1:Domain and Range Key Concept: Vertical.
CCSS Content Standards F.IF.4 For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 8–3) CCSS Then/Now New Vocabulary Key Concept: Vertical and Horizontal Asymptotes Example 1:Graph.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 4–1) CCSS Then/Now New Vocabulary Example 1:Two Real Solutions Key Concept: Solutions of a Quadratic.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Linear Relations and Functions
Five-Minute Check (over Lesson 1-5) Mathematical Practices Then/Now
Splash Screen.
Splash Screen.
Splash Screen.
Five-Minute Check (over Lesson 1–2) Mathematical Practices Then/Now
Five-Minute Check (over Lesson 2–2) Mathematical Practices Then/Now
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Five-Minute Check (over Lesson 2–1) Mathematical Practices Then/Now
Splash Screen.
Five-Minute Check (over Chapter 1) Mathematical Practices Then/Now
Presentation transcript:

Splash Screen

Five-Minute Check (over Chapter 1) CCSS Then/Now New Vocabulary Key Concept: Functions Example 1: Domain and Range Key Concept: Vertical Line Test Example 2: Real-World Example Example 3: Graph a Relation Example 4: Evaluate a Function Lesson Menu

Name the property illustrated by –15b + 15b = 0. A. Additive Identity B. Multiplicative Inverse C. Inverse Property of Addition D. Symmetry 5-Minute Check 2

Solve 2(c – 5) – 2 = 8 + c. A. –4 B. 4 C. 10 D. 20 5-Minute Check 3

Solve 2b – 5 ≤ –1. Graph the solution set on a number line. A. {b | b ≤ 2} B. {b | b < 2} C. {b | b ≥ 2} D. {b | b > 2} 5-Minute Check 5

Which algebraic equation shows the sentence four plus a number divided by six is equal to the product of twelve and the same number? A. eans B. eans C. eans D. eans 5-Minute Check 6

Mathematical Practices Content Standards F.IF.4 For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. F.IF.5 Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. Mathematical Practices 1 Make sense of problems and persevere in solving them. 7 Look for and make use of structure. CCSS

You identified domains and ranges for given situations. Analyze relations and functions. Use equations of relations and functions. Then/Now

one-to-one function onto function discrete relation continuous relation vertical line test independent variable dependent variable function notation Vocabulary

Concept 1

The relation is {(1, 2), (3, 3), (0, –2), (–4, –2), (–3, 1)}. Domain and Range State the domain and range of the relation. Then determine whether the relation is a function. If it is a function, determine if it is one-to-one, onto, both, or neither. The relation is {(1, 2), (3, 3), (0, –2), (–4, –2), (–3, 1)}. Answer: The domain is {–4, –3, 0, 1, 3}. The range is {–2, 1, 2, 3}. Each member of the domain is paired with one member of the range, so this relation is a function. It is onto, but not one-to-one. Example 1

State the domain and range of the relation shown in the graph State the domain and range of the relation shown in the graph. Is the relation a function? A. domain: {–2, –1, 0, 1} range: {–3, 0, 2, 3} Yes, it is a function. B. domain: {–3, 0, 2, 3} range: {–2, –1, 0, 1} Yes, it is a function. C. domain: {–2, –1, 0, 1} range: {–3, 0, 2, 3} No, it is not a function. D. domain: {–3, 0, 2, 3} range: {–2, –1, 0, 1} No, it is not a function. Example 1

Concept 2

TRANSPORTATION The table shows the average fuel efficiency in miles per gallon for SUVs for several years. Graph this information and determine whether it represents a function. Is this relation discrete or continuous? Example 2

Use the vertical line test Use the vertical line test. Notice that no vertical line can be drawn that contains more than one of the data points. Example 2

Answer: Yes, this relation is a function Answer: Yes, this relation is a function. Because the graph consists of distinct points, the relation is discrete. Example 2

HEALTH The table shows the average weight of a baby for several months during the first year. Graph this information and determine whether it represents a function. Example 2

A. Yes, this relation is a. B. No, this function. relation. is not a A. Yes, this relation is a B. No, this function. relation is not a function. C. Yes, this relation is a D. No, this function. relation is not a function. Example 2

Then graph the ordered pairs. Graph a Relation Graph y = 3x – 1 and determine the domain and range. Then determine whether the equation is a function, is one-to-one, onto, both, or neither. State whether it is discrete or continuous. Make a table of values to find ordered pairs that satisfy the equation. Choose values for x and find the corresponding values for y. Then graph the ordered pairs. Example 3

Find the domain and range. Graph a Relation Find the domain and range. Since x can be any real number, there is an infinite number of ordered pairs that can be graphed. All of them lie on the line shown. Notice that every real number is the x-coordinate of some point on the line. Also, every real number is the y-coordinate of some point on the line. Answer: The domain and range are both all real numbers. Example 3

Graph a Relation Determine whether the relation is a function and state whether it is discrete or continuous. This graph passes the vertical line test. Every x-value is paired with exactly one unique y-value, and every y-value corresponds to an x-value. Answer: Yes, the equation y = 3x – 1 represents a function. The function is both one-to-one and onto. Since the domain and range are both all real numbers, the relation is continuous. Example 3

Graph y = 2x + 5. A. B. C. D. Example 3

A. Given f(x) = x3 – 3, find f(–2). Evaluate a Function A. Given f(x) = x3 – 3, find f(–2). f(x) = x3 – 3 Original function f(–2) = (–2)3 – 3 Substitute. = –8 – 3 or –11 Simplify. Answer: f(–2) = –11 Example 4A

B. Given f(x) = x3 – 3, find f(2t). Evaluate a Function B. Given f(x) = x3 – 3, find f(2t). f(x) = x3 – 3 Original function f(2t) = (2t)3 – 3 Substitute. = 8t3 – 3 (2t)3 = 8t3 Answer: f(2t) = 8t3 – 3 Example 4B

A. Given f(x) = x2 + 5, find f(–1). B. –3 C. 3 D. 6 Example 4A

B. Given f(x) = x2 + 5, find f(3a). A. 3a2 + 5 B. a2 + 8 C. 6a2 + 5 D. 9a2 + 5 Example 4B

End of the Lesson

Pages 65 – 67 #11 – 31 odd, 35, 45, 50, 53, 55