Every Student Successful - Fontana Unified School District...Fontana Unified School District Every...

18
Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning Secondary Honors IM3 May 2020 School Name: _____________ Student ID#: ______________ Math Teacher Name: _____________ Period: ____ May 2020

Transcript of Every Student Successful - Fontana Unified School District...Fontana Unified School District Every...

Page 1: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities

Offline Distance Learning

Secondary

Honors IM3 May 2020

School Name: _____________ Student ID#: ______________

Math Teacher Name: _____________ Period: ____

May 2020

Page 2: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

May 2020

Page 3: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

Modeling with Functions

Function Family

Function Name Algebraic Shape of Graph

Linear 𝑦 = π‘šπ‘₯ + 𝑏

Exponential 𝑦 = π‘Žπ‘π‘₯

Quadratic 𝑦 = π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐

Polynomial 𝑦 = π‘Žπ‘›π‘₯𝑛 + π‘Žπ‘›βˆ’1π‘₯π‘›βˆ’1 + π‘Žπ‘›βˆ’2π‘₯π‘›βˆ’2 + β‹― π‘Ž0

May 2020

Page 4: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

Rational 𝑦 =

1

π‘₯

Absolute Value 𝑦 = |π‘₯|

Logarithmic 𝑦 = π‘™π‘œπ‘”π‘(π‘₯)

Trigonometric 𝑦 = π‘Ž sin(𝑏π‘₯)

or

𝑦 = π‘Ž cos (𝑏π‘₯)

Radical 𝑦 = √π‘₯𝑛

May 2020

Page 5: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

Function Transformation Rules

Examples: The parent function is 𝑦 = π‘₯

a. 𝑦 = π‘₯ + 2

the function shifted up by 2

b. 𝑦 = π‘₯ βˆ’ 2

the function shifted down by 2

c. 𝑦 = 2π‘₯

the function compressed horizontally by 2

d. π’š = βˆ’π’™

the function reflected over the y – axis

May 2020

Page 6: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

May 2020

Page 7: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

Try It:

Graph the following quadratic equations on the grid. The equation 𝑦 = π‘₯2 has been graphed for you. For each new

equation explain what the number 3 does to the graph 𝑦 = π‘₯2. Pay attention to the y-intercept, the x-intercept(s), and

the rate of change. Identify what changes in the graph and what stays the same.

a. 𝑦1 = π‘₯2 + 3

b. 𝑦2 = π‘₯2 βˆ’ 3

c. 𝑦3 = (π‘₯ βˆ’ 3)2

d. 𝑦4 = (π‘₯ + 3)2 + 3

e. 𝑦1 = 3π‘₯2

Practice Problems: Sketch the graph of the parent function and the graph of the transformed function on the same set of axes.

1.

2.

3. 4.

May 2020

Page 8: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

May 2020

Page 9: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

Composing and Decomposing

Composing Functions: Applying one function to the results of another. A composite function is created

when one function is substituted into another function.

It can be written as (𝑓 β—‹ 𝑔)(π‘₯) which means 𝑓(𝑔(π‘₯)). This can be read as β€œ 𝑓 of 𝑔 of π‘₯”

Examples:

1. Given: 𝑓(π‘₯) = 2π‘₯2 + 1 π‘Žπ‘›π‘‘ 𝑔(π‘₯) = π‘₯ βˆ’ 4. 𝐹𝑖𝑛𝑑 𝑓(𝑔(π‘₯)).

𝑓(𝑔(π‘₯)) = 𝑓(π‘₯ βˆ’ 4) [𝑠𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 π‘₯ βˆ’ 4 π‘–π‘›π‘‘π‘œ 𝑔(π‘₯)]

= 2(π‘₯ βˆ’ 4)2 + 1 [ πΉπ‘œπ‘Ÿ π‘’π‘£π‘’π‘Ÿπ‘¦ π‘₯ 𝑖𝑛 𝑓(π‘₯), 𝑠𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 π‘₯ βˆ’ 4]

= 2(π‘₯2 βˆ’ 8π‘₯ + 16) + 1 [ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› 𝑏𝑦 𝑒π‘₯π‘π‘Žπ‘›π‘‘π‘–π‘›π‘” (π‘₯ βˆ’ 4)2]

= 2π‘₯2 βˆ’ 16π‘₯ + 32 + 1 [ π·π‘–π‘ π‘‘π‘Ÿπ‘–π‘π‘’π‘‘π‘’ 2]

𝑓(𝑔(π‘₯)) = 2π‘₯2 βˆ’ 16π‘₯ + 33 [π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦]

2. Given: 𝑓(π‘₯) = 4π‘₯ + 9 π‘Žπ‘›π‘‘ 𝑔(π‘₯) =π‘₯βˆ’9

4. 𝐹𝑖𝑛𝑑 𝑔(𝑓(π‘₯)).

𝑔(𝑓(π‘₯)) = 𝑔(4π‘₯ + 9) [𝑠𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 4π‘₯ + 9 π‘–π‘›π‘‘π‘œ 𝑓(π‘₯)]

=(4π‘₯+9)βˆ’9

4 [ πΉπ‘œπ‘Ÿ π‘’π‘£π‘’π‘Ÿπ‘¦ π‘₯ 𝑖𝑛 𝑔(π‘₯), 𝑠𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 4π‘₯ + 9]

= 4π‘₯

4 [ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ ]

𝑔(𝑓(π‘₯)) = π‘₯ [ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ ]

3. Given: 𝑓(π‘₯) = 4π‘₯ + 9 π‘Žπ‘›π‘‘ 𝑔(π‘₯) =π‘₯βˆ’9

4. 𝐹𝑖𝑛𝑑 𝑓(𝑓(π‘₯)).

𝑓(𝑓(π‘₯)) = 𝑓(4π‘₯ + 9) [𝑠𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 4π‘₯ + 9 π‘–π‘›π‘‘π‘œ 𝑓(π‘₯)]

= 4(4π‘₯ + 9) + 9 [ πΉπ‘œπ‘Ÿ π‘’π‘£π‘’π‘Ÿπ‘¦ π‘₯ 𝑖𝑛 𝑓(π‘₯), 𝑠𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 4π‘₯ + 9]

= 16π‘₯ + 36 + 9 [ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ ]

𝑓(𝑓(π‘₯)) = 16π‘₯ + 45 [ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ ]

4. Given: 𝑓(π‘₯) = 4π‘₯ + 9 π‘Žπ‘›π‘‘ 𝑔(π‘₯) =π‘₯βˆ’9

4. 𝐹𝑖𝑛𝑑 𝑓(𝑓(3)).

𝑓(𝑓(π‘₯)) = 16π‘₯ + 45 [𝑀𝑒 π‘˜π‘›π‘œπ‘€ π‘‘β„Žπ‘–π‘  π‘“π‘Ÿπ‘œπ‘š 𝑒π‘₯π‘Žπ‘šπ‘™π‘’ 3]

𝑓(𝑓(3)) = 16(3) + 45 [𝑠𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 3 π‘–π‘›π‘‘π‘œ π‘₯]

= 48 + 45 [ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ ]

𝑓(𝑓(3)) = 93 [ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ ]

May 2020

Page 10: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

May 2020

Page 11: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

Practice Problems:

1. Let 𝑓(π‘₯) = 2π‘₯2 βˆ’ 4 π‘Žπ‘›π‘‘ 𝑔(π‘₯) = 5π‘₯. 𝐹𝑖𝑛𝑑 π‘’π‘Žπ‘β„Ž π‘Žπ‘›π‘‘ π‘ π‘–π‘šπ‘π‘™π‘–π‘“π‘¦.

a) (𝑓 β—‹ 𝑔)(1) b) (𝑔 β—‹ 𝑓)(1) c) (𝑓 β—‹ 𝑓)(βˆ’2) d) (𝑔 β—‹ 𝑔)(βˆ’1)

2. Let 𝑓(π‘₯) =8

π‘₯βˆ’3π‘Žπ‘›π‘‘ 𝑔(π‘₯) =

15

π‘₯+1. 𝐹𝑖𝑛𝑑 π‘’π‘Žπ‘β„Ž π‘Žπ‘›π‘‘ π‘ π‘–π‘šπ‘π‘™π‘–π‘“π‘¦.

a) (𝑓(𝑔(π‘₯)) b) (𝑔 β—‹ 𝑓)(π‘₯) c) (𝑓(𝑓(π‘₯)) d) (𝑔(𝑔(π‘₯))

3. Use your answers for a) and B) in problem 2 to calculate the two problems below.

a) (𝑓(𝑔(βˆ’1)) b) (𝑔 β—‹ 𝑓)(3)

May 2020

Page 12: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

May 2020

Page 13: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

Translating My Composition

Decomposing Functions: is a process by which you can break down one complex function into multiple

smaller functions. By doing this, you can solve for functions in shorter, easier-to-understand pieces. There

may be more than one way to decompose a composite function.

Examples:

Let 𝑓(π‘₯) = π‘₯ + 5, 𝑔(π‘₯) = π‘₯2, β„Ž(π‘₯) = 3π‘₯ π‘Žπ‘›π‘‘ 𝑗(π‘₯) = 2π‘₯. Express each function as a composite of 𝑓, 𝑔

β„Ž, and/or j. 1. 𝐴(π‘₯) = π‘₯4 [Think: What multiple functions composed this?]

The two functions are

g(π‘₯) = π‘₯2 and g(π‘₯) = π‘₯2

Check:

𝑔(𝑔(π‘₯)) = 𝑔(π‘₯2)

= (π‘₯2)2

= π‘₯4

2. 𝐢(π‘₯) = 3π‘₯ + 15 [Think: What multiple functions composed this?]

The two functions are

β„Ž(π‘₯) = 3π‘₯ and 𝑓(π‘₯) = π‘₯ + 5

Check:

β„Ž(𝑓(π‘₯)) = β„Ž(π‘₯ + 5 )

= 3(π‘₯ + 5)

= 3π‘₯ + 15

3. 𝐢(π‘₯) = 3(π‘₯ + 5)2 [Think: What multiple functions composed this?]

The three functions are

β„Ž(π‘₯) = 3π‘₯ and 𝑓(π‘₯) = π‘₯ + 5 and 𝑔(π‘₯) = π‘₯2,

Check:

β„Ž(𝑔(𝑓(π‘₯)))) = β„Ž(𝑔(π‘₯ + 5 ) [ work inside first- substitute π‘₯ + 5 into 𝑓(π‘₯)]

= β„Ž((π‘₯ + 5)2) [ For every π‘₯ in 𝑔(π‘₯), substitute π‘₯ + 5]

= 3(π‘₯ + 5)2 [For every π‘₯ in β„Ž(π‘₯), substitute 𝑔(𝑓(π‘₯)) which is (π‘₯ + 5)2]

May 2020

Page 14: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

May 2020

Page 15: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

Practice Problems:

Let 𝑓(π‘₯) = π‘₯2, 𝑔(π‘₯) = 5π‘₯, π‘Žπ‘›π‘‘ β„Ž(π‘₯) = √π‘₯ + 2. Express each function as a composite of 𝑓, 𝑔 and/or β„Ž. 1. F(π‘₯) = π‘₯4

2. 𝐢(π‘₯) = 5π‘₯2

3. 𝑃(π‘₯) = π‘₯ + 2

4. 𝑅(π‘₯) = 5√π‘₯ + 10

5. 𝑄(π‘₯) = 25π‘₯

7. 𝐷(π‘₯) = √√π‘₯ + 2 + 2

8. 𝐡(π‘₯) = π‘₯ + 4√π‘₯ + 4

May 2020

Page 16: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

May 2020

Page 17: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

Different Combinations

SET Problems: Identifying the 2 functions that make up a composite function.

Find functions 𝑓 and 𝑔 so that 𝑓 β—‹ 𝑔 = 𝐻

1. 𝐻(π‘₯) = √π‘₯2 + 5π‘₯ βˆ’ 4

2. 𝐻(π‘₯) = (3 βˆ’1

π‘₯)2

3. 𝐻(π‘₯) = (3π‘₯ βˆ’ 7)4 4. 𝐻(π‘₯) = |5π‘₯2 βˆ’ 78|

5. 𝐻(π‘₯) = 2

3βˆ’π‘₯5 6. 𝐻(πœƒ) = (tan πœƒ)2

7. 𝐻(π‘₯) = 9(4π‘₯ βˆ’ 8) + 1 8. 𝐻(π‘₯) = √

1

6π‘₯

May 2020

Page 18: Every Student Successful - Fontana Unified School District...Fontana Unified School District Every Student Successful | Engaging Schools | Empowered Communities Offline Distance Learning

GO Problems: Finding function values given the graph. Use the graph to find all of the missing values.

1. 𝑓(β– ) = 8 2. 𝑓(β– ) = 5

3. 𝑓(β– ) = -1 4. 𝑓(β– ) = 0

5. 𝑓(βˆ’1) = 6. 𝑔(0) =

7. 𝑓(π‘₯) = 𝑔(π‘₯) 8. 𝑓(π‘₯) βˆ’ 𝑔(π‘₯) = 0

9. 𝑓(π‘₯) βˆ— 𝑔(π‘₯) = 0 10. 𝑓(2) + 𝑔(2) =

May 2020