relations and functions grade 10 chapter 2.1

RELATIONS AND FUNCTIONS GRADE 10 CHAPTER 2.1

31/10/2013

CLASSWORK

• (Pages 64-67) Exercises 1 to 3, 4, 6, 8, 15, 16, 17, 21, 22, 23, 24 to 32, 33, 35, 37, 40, 41 to 44.

HOMEWORK

• Worksheet 2.1 – (See Weebly)

Over Chapter 1

A. AB. BC. CD. D

A B C D

0% 0%0%0%

A. –4

B. 4

C. 10

D. 20

Solve 2(c – 5) – 2 = 8 + c.

Over Chapter 1

A. AB. BC. CD. D

A B C D

0% 0%0%0%

Solve |3x – 5| + 4 = 14.

A.

B.

C.

D.

Over Chapter 1

A. AB. BC. CD. D

A B C D

0% 0%0%0%

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}

Over Chapter 1

A. AB. BC. CD. D

A B C D

0% 0%0%0%

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

You identified domains and ranges for given situations. (Lesson 0–1)

• Analyze relations and functions.

• Use equations of relations and functions.

• one-to-one function

• onto function• discrete relation• continuous relation• vertical line test• independent variable• dependent variable• function notation

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, 0), (–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.

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

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.

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?

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

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

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.

A. A

B. B

C. C

D. D

A B C D

0% 0%0%0%

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.

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.

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 allreal numbers.

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.

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

Graph y = 2x + 5.

A. B.

C. D.

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

Evaluate a Function

B. Given f(x) = x3 – 3, find f(2t).

Answer: f(2t) = 8t3 – 3

f(x) = x3 – 3 Original function

f(2t) = (2t)3 – 3 Substitute.

= 8t3 – 3 (2t)3 = 8t3

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. –4

B. –3

C. 3

D. 6

A. Given f(x) = x2 + 5, find f(–1).

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

B. Given f(x) = x2 + 5, find f(3a).

A. 3a2 + 5

B. a2 + 8

C. 6a2 + 5

D. 9a2 + 5