relations and functions algebra 1 lesson 5-2 (for help, go to lessons 1–9 and 1–2.) graph each...

Relations and FunctionsALGEBRA 1 LESSON 5-2ALGEBRA 1 LESSON 5-2

(For help, go to Lessons 1–9 and 1–2.)

Graph each point on a coordinate plane.

1. (2, –4) 2. (0, 3) 3. (–1, –2) 4. (–3, 0)

Evaluate each expression.

5. 3a – 2 for a = –5 6. for x = 3 7. 3x2 for x = 6x + 3–6

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Solutions

1. 2.

3. 4.

5. 3a – 2 for a = –5: 3(–5) – 2 = –15 – 2 = –17

6. for x = 3: = = –1

7. 3x2 for x = 6: 3 • 62 = 3 • 36 = 108

x + 3–6

3 + 3–6

6 –6

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ALGEBRA 1 LESSON 5-2ALGEBRA 1 LESSON 5-2

Find the domain and the range of the ordered pairs.

age weight 14 120 12 110 18 126 14 125 16 124

domain: {12, 14, 16, 18} List the values in order. Do not repeat values.

range: {110, 120, 124, 125, 126}

Relations and Functions

8-6,7


Use the vertical-line test to determine whether the relation

{(3, 2), (5, –1), (–5, 3), (–2, 2)} is a function.

A vertical line would not pass through more than one point, so the relation is a function.

Step 1: Graph the ordered pairs on a coordinate plane.

Step 2: Pass a pencil across the graph. Keep your pencil straight to represent a vertical line.


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Determine whether each relation is a function.

a. {(4, 3), (2, –1), (–3, –3), (2, 4)}

The domain value 2 corresponds to two range values, –1 and 4.

b. {(–4, 0), (2, 12), (–1, –3), (1, 5)}

There is no value in the domain that corresponds to more than one value of the range.



The relation is not a function.

The relation is a function.

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a. Evaluate ƒ(x) = –5x + 25 for x = –2.


ƒ(–2) = 10 + 25 Simplify.ƒ(–2) = 35

ƒ(x) = –5x + 25ƒ(–2) = –5(–2) + 25 Substitute –2 for x.

b. Evaluate y = 4x2 + 2 for x = 3.

y = 36 + 2 Simplify.y = 38

y = 4x2 + 2y = 4(3)2 + 2 Substitute 3 for x.

y = 4(9) + 2 Simplify the powers first.


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Evaluate the function rule ƒ(g) = –2g + 4 to find the range for

the domain {–1, 3, 5}.

The range is {–6, –2, 6}.

ƒ(g) = –2g + 4ƒ(5) = –2(5) + 4ƒ(5) = –10 + 4ƒ(5) = –6

ƒ(g) = –2g + 4ƒ(–1) = –2(–1) + 4ƒ(–1) = 2 + 4ƒ(–1) = 6

ƒ(g) = –2g + 4ƒ(3) = –2(3) + 4ƒ(3) = –6 + 4ƒ(3) = –2



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1. a. Find the domain and range of the ordered pairs (1, 3), (–4, 0), (3, 1), (0, 4), (2, 3).

b. Use mapping to determine whether the relation is a function.domain: {–4, 0, 1, 2, 3} range: {0, 1, 3, 4}

The relation is a function.

2. Use the vertical-line test to determine whether each relation is a function.

a.

no

b.

yes

3. Find the range of the function ƒ(g) = 3g – 5 for the domain {–1.5, 2, 4}.{–9.5, 1, 7}

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Function Rules, Tables, and Graphs

ALGEBRA 1 LESSON 8-6,7ALGEBRA 1 LESSON 8-6,7

(For help, go to Lesson 5-2.)

Graph the data in each table.

x y

–3 –7

–1 –1

0 2

2 8

1. x y

–3 4

–2 0

0 –2

2 4

2. x y

–4 –3

0 –2

2 –1.5

4 –1

3.

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1. Graph the points: 2. Graph the points:(–3, –7) (–3, 4)(–1, –1) (–2, 0)(0, 2) (0, –2)(2, 8) (2, 4)

3. Graph the points:(–4, –3)(0, –2)(2, –1.5)(4, –1)

Solutions

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Step 2: Plot the points for the ordered pairs.

Step 3: Join the points to form a line.


Model the function rule y = + 2 using a table of values

and a graph.

13 x

Step 1: Choose input value for x. Evaluate to find y

x (x, y)

–3 y = (–3) + 2 = 1 (–3, 1)

0 y = (0) + 2 = 2 (0, 2)

3 y = (3) + 2 = 3 (3, 3)

y = x + 21313

13

13


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At the local video store you can rent a video game for $3.

It costs you $5 a month to operate your video game player. The total

monthly cost C(v) depends on the number of video games v you rent.

Use the function rule C(v) = 5 + 3v to make a table of values and a

graph.

v C(v) = 5 + 3v (v, C(v))

0 C(0) = 5 + 3(0) = 5 (0, 5)

1 C(1) = 5 + 3(1) = 8 (1, 8)

2 C(2) = 5 + 3(2) = 11 (2, 11)

ALGEBRA 1 LESSON 8-6,7ALGEBRA 1 LESSON 8-6,7Function Rules, Tables, and Graphs

8-6,7

a. Graph the function y = |x| + 2.

ALGEBRA 1 LESSON 8-6,7ALGEBRA 1 LESSON 8-6,7Function Rules, Tables, and Graphs

8-6,7

Make a table of values.

x y = |x| + 2 (x, y)

–3 y = |–3| + 2 = 5 (–3, 5)

–1 y = |–1| + 2 = 3 (–1, 3)

0 y = |0| + 2 = 2 (0, 2)

1 y = |1| + 2 = 3 (1, 3)

3 y = |3| + 2 = 5 (3, 5)

Then graph the data.

(continued)


b. Graph the function ƒ(x) = x2 + 2.


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x ƒ(x) = x2 + 2 (x, y)

–2 ƒ(–2) = 4 + 2 = 6 (–2, 6)

–1 ƒ(–1) = 1 + 2 = 3 (–1, 3)

0 ƒ(0) = 0 + 2 = 2 ( 0, 2)

1 ƒ(1) = 1 + 2 = 3 ( 1, 3)

2 ƒ(2) = 4 + 2 = 6 ( 2, 6)

Make a table of values. Then graph the data.



3. Graph ƒ(x) = 2x2 – 2.

1. Model y = –2x + 4 with a table of values and a graph.

2. Graph y = |x| – 2.

x y = –2x + 4 (x, y)

–1 y = –2(–1) + 4 = 6 (–1, 6)

0 y = –2(0) + 4 = 4 (0, 4)

1 y = –2(1) + 4 = 2 (1, 2)

2 y = –2(2) + 4 = 0 (2, 0)

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Writing a Function RuleALGEBRA 1 LESSON 8-6,7ALGEBRA 1 LESSON 8-6,7

(For help, go to Lesson 8-6,7.)

Model each rule with a table of values.

1. f(x) = 5x – 1 2. y = –3x + 4 3. g(t) = 0.2t – 7

4. y = 4x + 1 5. f(x) = 6 – x 6. c(d) = d + 0.9

Evaluate each function rule for n = 2.

7. A(n) = 2n – 1 8. f(n) = –3 + n – 1 9. g(n) = 6 – n

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Solutions

3.

4.

2.

5.

1.

6.

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Solutions (continued)

7. A(n) = 2n – 1 for n = 2:

A(2) = 2(2) – 1 = 4 – 1 = 3

8. f(n) = –3 + n – 1 for n = 2:

f(2) = –3 + 2 – 1 = –1 – 1 = –2

9. g(n) = 6 – n for n = 2:

g(2) = 6 – 2 = 4

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Write a function rule for each table.

Ask yourself, “What can I do to 2 to get 8, 4 to get 10, ...?”

A rule for the function is ƒ(x) = x + 6.

Relate: equals plus 6

Write: = + 6

ƒ(x)

ƒ(x)

x

x

x ƒ(x)

2 8

4 10

6 12

8 14

a.

Writing a Function Rule

You add 6 to each x-value to get the ƒ(x) value.

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(continued)

x y

1 2

2 5

3 10

4 17

b. Ask yourself, “What can I do to 1 to get 2, 2 to get 5, . . . ?”

A rule for the function is y = x2 + 1.

Relate: equals plus 1

Write: = + 1x2

y

y

x times itself


You multiply each x-value times itself and add 1 to get the y value.

8-6,7


The journalism class makes $25 per page of advertising in

the yearbook. If the class sells n pages, how much money will it earn?

a. Write a function rule to describe this relationship.

The function rule P(n) = 25n describes the relationship between the number of pages sold and the money earned.

Relate: is 25 times money earned number of pages sold

Define: Let = number of pages sold.n

Let = money earned.P(n)

Write: = 25 • nP(n)


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(continued)

b. The class sold 6 pages of advertising. How much money did the class make?

P(6) = 25 • 6 Substitute 6 for n.

P(6) = 150 Simplify.

The class made $150.


P(n) = 25 • n

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The choir spent $100 producing audio tapes of its last

performance and will sell the tapes for $5.50 each. Write a rule to

describe the choir’s profit as a function of the number of tapes sold.


The function rule P(t) = 5.5t – 100 describes the profit as a function of the number of tapes sold.

Relate: is $5.50 times minus cost of tape production

total profit tapes sold

Define: Let = number of tapes sold.t

Let = total profit. P(t)

Write: = 5.5 • – 100tP(t)


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1. Write a function rule for each table.

2. Write a function rule to describe each relationship.a. the total cost T(c) of c pounds of apples at $.82 a pound

b. a scale model s of a moth m that is 6 times the actual size of the moth

3. You borrow $60 to buy a bread-making machine. You charge customers $1.50 a loaf for your special bread. Write a rule to describe your profit as a function of the number n of loaves sold.

x ƒ(x)

–1 –4

0 –3

1 –2

2 –1

b.x y

–1 –5

0 0

1 5

2 10

a.

y = 5x ƒ(x) = x – 3

T(c) = 0.82c

s(m) = 6m

P(n) = 1.5n - 60

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relations and functions algebra 1 lesson 5-2 (for help, go to lessons 1–9 and 1–2.) graph each...

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