Algebra 1 Chapter 5 Notes Pack

Page 1

Algebra 1 • Lesson 5.1* (Domain and Range) Day 51 Notes Before we begin exploring domain and range, we need a working definition for relation. A relation is a ! ! !

!

!

!

!

of

Q2. The domain of a relation is the set of all !! values (i.e., ! ! ! values).

.

Discovering Domain • Graphically (Q1-Q3): Q1. The domain of the first relation is {1, 3, 4, 9 } . Find the

Q3. The domain of the first relation is { !3,!2,!1 } . Find

domain of the second and third relations.

the domain of the second and third relations.

Page 1 of 4!

Updated: October 31, 2009


Algebra 1 • Lesson 5.1* (Domain and Range) Day 51 Notes Discovering Range • Graphically (Q4-Q6): Q4. The range of the first relation is

{!5, 0, 2, 4 } . Find the

range of the second and third relations.

Q5. The range of the first relation is

{ 2, 4, 6 } . Find the

range of the second and third relations.

Q6. The range of a relation is the set of all ! ! values (i.e., ! ! ! values). Page 2 of 4!

Updated: October 31, 2009


Algebra 1 • Lesson 5.1* (Domain and Range) Day 51 Notes Discovering Domain • Numerically (Q7-Q8): Q7. The domain of the first relation is

{ 2, 3, 4, 5 } . Find the

domain of the second, third, and fourth relations.

Q9. The range of the first relation is { !7, 0, 2, 3 } . Find the range of the second, third, and fourth relations.

x

y

x

y

2

–7

2

–7

3

3

3

3

4

0

4

0

5

2

5

2

x

y

x

y

6

4

6

4

11

3

11

3

16

2

16

2

21

1

21

1

x

y

x

y

1.3

–1.3

1.3

–1.3

1.7

–1.7

1.7

–1.7

2.1

–2.1

2.1

–2.1

2.5

–2.5

2.5

–2.5

x

y

x

y

1 2

2

1 2

2

3 4

4 3

3 4

4 3

7 8

8 7

7 8

8 7

15 16

16 15

15 16

16 15

Q8. The domain of a relation is the set of all !! values (i.e., ! ! ! values).

Page 3 of 4!

Discovering Range • Numerically (Q9-Q10):

Q10. The range of a relation is the set of all !! values (i.e., ! ! ! values).

Updated: October 31, 2009


Algebra 1 • Lesson 5.1* (Domain and Range) Day 51 Notes Domain and Range • Graphically (Q11-Q14): Q11. Find the domain and range of the relation below.

Q13. Find the domain and range of the relation below.

Domain: !

!

!

!

!

Domain: !

!

!

!

!

!

!

!

!

!

!

!

!

!

!

!

!

Range: ! !

!

!

!

!

Range: ! !

!

!

!

!

!

!

!

!

!

!

!

!

!

!

!

!

We can also say…

We can also say…

Domain: !

!

!

!

!

Domain: !

!

!

!

!

Range: ! !

!

!

!

!

Range: ! !

!

!

!

!

Q12. Find the domain and range of the relation below.

Q14. Find the domain and range of the relation below.

Domain: !

!

!

!

!

Domain: !

!

!

!

!

Range: ! !

!

!

!

!

Range: ! !

!

!

!

!

We can also say…

We can also say…

Domain: !

!

!

!

!

Domain: !

!

!

!

!

Range: ! !

!

!

!

!

Range: ! !

!

!

!

!

Page 4 of 4!

Updated: October 31, 2009


Algebra 1 • Lesson 5.2* (Relations and Functions • Introduction) Day 52 Notes A function is a rule or relationship where there is exactly one output value for each input value. In other words, a relation (graph, table, formula, etc.) is a function if each input gives exactly one output.

Q5. Does the relationship ( social security number, person ) represent a function? If not, provide a counterexample (that is, an example of one input that has two or more outputs).

If one input produces more than one output, the relation(ship) is not a function. These ideas take time to sink in, and are best explored using all four representations (verbal, graphical, numerical, and algebraic). Let!s get started! Functions • Verbally (Q1-Q9): Q1. Does the relationship (city, ZIP code ) represent a function? If not, provide a counterexample (that is, an example of one input that has two or more outputs).

Q2. Does the relationship

( person,

birth date ) represent a

Q6. Does the relationship ( person, passport number ) represent a function? If not, provide a counterexample (that is, an example of one input that has two or more outputs).

Q7. Does the relationship ( apartment building, tenant )

function? If not, provide a counterexample (that is, an example of one input that has two or more outputs).

represent a function? If not, provide a counterexample (that is, an example of one input that has two or more outputs).

Q3. Does the relationship ( last name, first name )

Q8. Does the relationship ( wedding ring, finger ) represent

represent a function? If not, provide a counterexample (that is, an example of one input that has two or more outputs).

a function? If not, provide a counterexample (that is, an example of one input that has two or more outputs).

Q4. Does the relationship ( state, capital ) represent a

Q9. Does the relationship ( game, final score ) represent a

function? If not, provide a counterexample (that is, an example of one input that has two or more outputs).

function? If not, provide a counterexample (that is, an example of one input that has two or more outputs).

Page 1 of 2"

Updated: November 1, 2009


Algebra 1 • Lesson 5.2* (Relations and Functions • Introduction) Day 52 Notes Functions • Graphically (Q10-Q11):

Functions • Numerically (Q12-Q14):

Q10. Does the relation shown below represent a function? Explain why or why not.

Q12. Does the relation shown below represent a function? Explain why or why not.

x

y

2

–7

3

3

4

0

5

2

Q13. Does the relation shown below represent a function? Explain why or why not.

Q11. Does the relation shown below represent a function? Explain why or why not.

Page 2 of 2!

x

y

6

4

6

3

11

2

11

1

Q14. Does the relation shown below represent a function? Explain why or why not.

x

y

1.1

–1.3

1.2

–1.3

1.3

–1.3

1.4

–1.3

Updated: November 1, 2009


Algebra 1 • Lesson 5.3* (Identifying Functions Quickly) Day 53 Notes Our goal today is to reinforce the concept of a function, and to develop ways to identify them quickly.

Q3. Does the relation shown below represent a function? Explain why or why not.

Functions • Graphically (Q1-Q5): Q1. Does the relation shown below represent a function? Explain why or why not.

Q4. Does the relation shown below represent a function? Explain why or why not.

Q11. Does the relation shown below represent a function? Explain why or why not.

Q5. How can you tell quickly which relations above are functions? Can you create a visual test?

Page 1 of 2!

Updated: November 1, 2009


Algebra 1 • Lesson 5.3* (Identifying Functions Quickly) Day 53 Notes Functions • Numerically (Q6-Q10): Q6. Does the relation shown below represent a function?

x

y

1

2

2

2

3

3

4

3

Q7. Does the relation shown below represent a function?

x

y

2

6

2

5

6

4

6

3

Q8. Does the relation shown below represent a function?

x

y

1

2

2

4

4

6

4

8

Q9. Does the relation shown below represent a function?

x

y

0

3

0

3

1

4

1

4

Q10. How can you tell quickly which relations above are functions? Can you create a quick “input-focused” test?

Page 2 of 2!

Updated: November 1, 2009


Algebra 1 • Lesson 5.4* (Squares and Square Roots) Day 54 Notes Squares (Q1-Q8):

2

Q1. The square of a number n is n ! n , which can also be

Q5. True or false: ( !a ) = a 2 .

written as n 2 . Find the square of each number in the table without using a calculator. number ( n )

( )

square n 2

Q6. True or false: !4 2 = !16 .

1 2 3 4

Q7. True or false: 62 = 36 .

5 6 7 8

Q8. True or false: !a 2 = a 2 .

9 10 Q2. Use your calculator to find the square of each number in the table. number ( n )

( )

square n 2

–3

Square Roots (Q9-Q16):

12

Q9. The positive square root of a number n is n . The square root (positive or negative) is the number you would have to square (multiply by itself) to obtain n . Find the positive square root of each number in the table without using a calculator. Then verify your results using a calculator.

–8 0 1.5 –2

number (n )

11 –7

positive square root

( n)

1

100

4

–15

9 2

Q3. True or false: ( !4 ) = !16 .

16 25 36 49 64

2

Q4. True or false: ( !6 ) = 36 .

81 100

Page 1 of 2!

Updated: November 1, 2009


Algebra 1 • Lesson 5.4* (Squares and Square Roots) Day 54 Notes Q10. The negative square root of a number n is ! n . The square root (positive or negative) is the number you would have to square (multiply by itself) to obtain n . Find the negative square root of each number in the table without using a calculator. Then verify your results using a calculator. number (n )

Graphs of Squares and Square Roots (Q17-Q19): Q17. Enter the table from Q1 into a list or spreadsheet on your calculator. Make a scatter plot. Describe and sketch what you see.

negative square root

(! n )

1 4 9 16 25 36 49 64

Q18. Enter the table from Q9 into a list or spreadsheet on your calculator. Make a scatter plot. Describe and sketch what you see.

81 100 Q11. Evaluate

49 .

Q12. Evaluate

100 .

Q13. Evaluate ! 1 .

Q14. Evaluate

Q19. Enter the table from Q10 into a list or spreadsheet on your calculator. Make a scatter plot. Describe and sketch what you see.

9.

Q15. Evaluate ! 1, 000, 000 .

Q16. Evaluate

Page 2 of 2!

!25 .

Updated: November 1, 2009


Algebra 1 • Lesson 5.5* (Cubes and Cube Roots) Day 55 Notes Review • Squares and Square Roots (Q1-Q7):

9=

Q1. We say that

because

Cube Roots (Q9-Q14):

2

(

)

= 9.

49 =

because

(

)

Q3. We say that

16 =

because

(

)

Q4. We say that ! 81 =

because

Q5. We say that ! 1 =

because

Q6. We say that ! 36 =

Q7. We say that !

!

!25 has !

!

(

2

because

)

= 49 .

2

= 16 . 2

(

) 2

(

because

Now we will find cube roots

2

Q2. We say that

)

= 81 .

= 1. 2

(

In the previous lesson, we found square roots by focusing on their connection to square numbers.

)

= 36 .

!

!

!

!

3

connection to cube numbers (perfect cubes). Q9. We say that

3

Q10. We say that

Q11. We say that

Q12. We say that

Q13. We say that

1=

because

3

(

)

=1. 3

3

64 =

because

(

)

3

!8 =

because

(

)

3

0=

3

!125 =

3

27 =

because

b/c

3

3

(

) 3

(

)

= !25 has ! Q14. We say that

!

( n ) by focusing on their

because

(

= !125 . 3

)

.

!64 has !

!

(

2

!

!

because

!

!

!

!

)

!

Q15.

4+ 9 =

Q16.

81 ! 49 =

= !64 has !

.

Cube Hunt (Q9): Q9. Type factor(9) on your calculator, and you will see that 9 is a square number (we sometimes say perfect square), since 9 = 32 .

Q17.

3

27 ! 1 =

Find the first five perfect cubes using a similar technique. Summarize your results in the space below. (Hint: The first perfect cube is 1, since 1 = 13 .)

Q18. True or false:

3

64 > 16

3

125 < 100

First perfect cube: 1 Second perfect cube:

Q19. True or false:

Third perfect cube: Fourth perfect cube:

Q20. True or false: 12 + 23 ! 4 ! 3 125 =

1+ 31

Fifth perfect cube:

Page 1 of 1!

= !8 .

=0.

Evaluating Roots/Radicals (Q15-Q20): Q8. We say that

= 64 .

Updated: November 1, 2009

= 27 .


Algebra 1 • Lesson 5.6* (Exponents 1) Day 56 Notes Introduction • Exponents (Q1-Q10):

Multiplying Powers (Q11-Q20):

Use your calculator to rewrite each expression as a power (that is, using a base and an exponent).

Use your calculator to simplify each product of powers. Type in what is written below, and write down the result.

Q1. When I enter x ! x ! x the calculator displays...

Q11. x 2 ! x 3

Q2. When I enter a ! a ! a ! a ! a the calculator displays...

Q12. a 5 ! a 3

Q13. b10 !b 5 Q3. When I enter b !b !b !b the calculator displays...

Q14. y 7 ! y 8 Q4. When I enter n ! n the calculator displays... Q15. c 270 !c 3 Q5. When I enter c !c !c !c !c the calculator displays... Q16. t 5 ! t 5 Q6. To display x 5 enter… Q17. n 30 ! n 20

Q7. To display a 3 enter…

Q18. w 2 ! w 3 ! w 4

Q19. x 3 ! x 7 ! x 11

Q8. To display n 7 enter…

Q20. x a ! x b Q9. To display b1 enter…

We summarize these results by saying that when

Q10. To display c 8 enter...

!

!

!

powers with the same

!

!

, we !

!

the exponents.

We summarize this by saying that x n is equivalent to an expression with ! !

factors of !

. Algebraically, ab ! ac =

Page 1 of 1!

Updated: November 1, 2009


Algebra 1 • Lesson 5.7* (Exponents 2) Day 57 Notes Simplifying Exponents (Q1-Q6): Use your calculator to rewrite each quotient as a power. Q1.

Q10.

x !x !x !x !x = x !x Q11.

Q2.

a !a !a !a !a !a !a = a !a !a !a !a Q12.

Q3.

Q4.

Q5.

y8 = y

b !b !b !b = b !b !b

y !y !y !y !y = y !y !y !y !y

c !c !c = c

t !t !t !t !t !t Q6. = t !t

Dividing Powers (Q7-Q12):

c10

=

c4

t6 t6

=

Dividing Powers Quickly (Q13-Q16): Based on your observations made while working on Q1 through Q12, simplify these quotients of powers quickly without a calculator. Verify your results at the end using a calculator. Q13.

Q14.

Q15.

Q16.

a 11

=

a7

b19

=

b6

c 77

=

c 67

d 100 d 99

=

Rewrite each numerator and denominator in expanded form (so that the problems look like Q1 through Q6). Then simplify without using a calculator. Q7.

Q8.

x7 x

2

a5 a3

=

We summarize these results by saying that when !

!

!

powers with the same

!

!

, we !

!

= Algebraically,

Q9.

b9 b4

the exponents.

ab ac

=

=

Bonus. What does x 0 equal? (Hint: Study Q4.) Page 1 of 1!

Updated: November 1, 2009


Algebra 1 • Lesson 5.8* (Exponents 3) Day 58 Notes Powers of Powers (Q1-Q8):

Review • Exponent Rules (Q9-Q18):

Use your calculator to rewrite each expression. Be sure to write a multiplication symbol in between each variable.

Simplify each expression using the exponent rules learned in the last three lessons.

2

Q9. a m ! a n =

3

Q10.

4

Q11. a m

Q1. ( a !b ) =

Q2. (c ! d ) =

(

2

)

4

)

(

)

(

)

Q6. g 10 ! h 11

Q7. a !b 2 !c 3

(

5

2

Q8. x ! y ! z

=

n

Q13.

=

3

4

x 31 x 28

=

2

( )

Q14. x 3

=

7

=

Q12. x 17 ! x 3 =

=

(

Q5. m 2 ! n 3

an

( )

Q3. (e ! f ) =

Q4. x 2 ! y

am

Q15. y 7 ! y 5 ! y 6 =

=

10

)

=

Q16.

=

We summarize these results by saying that when raising a power to a power, !

!

!

y8 y5

=

4

( )

Q17. y 7

=

each inside

exponent with the outside exponent.

(

Q18. a 2b 4c 3 c

( )

Algebraically, ab

Page 1 of 1!

5

)

=

=

Updated: November 1, 2009


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