Post on 03-Oct-2020
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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
May 2020
Modeling with Functions
Function Family
Function Name Algebraic Shape of Graph
Linear π¦ = ππ₯ + π
Exponential π¦ = πππ₯
Quadratic π¦ = ππ₯2 + ππ₯ + π
Polynomial π¦ = πππ₯π + ππβ1π₯πβ1 + ππβ2π₯πβ2 + β― π0
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Rational π¦ =
1
π₯
Absolute Value π¦ = |π₯|
Logarithmic π¦ = ππππ(π₯)
Trigonometric π¦ = π sin(ππ₯)
or
π¦ = π cos (ππ₯)
Radical π¦ = βπ₯π
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Function Transformation Rules
Examples: The parent function is π¦ = π₯
a. π¦ = π₯ + 2
the function shifted up by 2
b. π¦ = π₯ β 2
the function shifted down by 2
c. π¦ = 2π₯
the function compressed horizontally by 2
d. π = βπ
the function reflected over the y β axis
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May 2020
Try It:
Graph the following quadratic equations on the grid. The equation π¦ = π₯2 has been graphed for you. For each new
equation explain what the number 3 does to the graph π¦ = 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. π¦1 = π₯2 + 3
b. π¦2 = π₯2 β 3
c. π¦3 = (π₯ β 3)2
d. π¦4 = (π₯ + 3)2 + 3
e. π¦1 = 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.
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May 2020
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 (π β π)(π₯) which means π(π(π₯)). This can be read as β π of π of π₯β
Examples:
1. Given: π(π₯) = 2π₯2 + 1 πππ π(π₯) = π₯ β 4. πΉπππ π(π(π₯)).
π(π(π₯)) = π(π₯ β 4) [π π’ππ π‘ππ‘π’π‘π π₯ β 4 πππ‘π π(π₯)]
= 2(π₯ β 4)2 + 1 [ πΉππ ππ£πππ¦ π₯ ππ π(π₯), π π’ππ π‘ππ‘π’π‘π π₯ β 4]
= 2(π₯2 β 8π₯ + 16) + 1 [ ππππππππ¦ ππ’πππ‘πππ ππ¦ ππ₯πππππππ (π₯ β 4)2]
= 2π₯2 β 16π₯ + 32 + 1 [ π·ππ π‘ππππ’π‘π 2]
π(π(π₯)) = 2π₯2 β 16π₯ + 33 [ππππππππ¦]
2. Given: π(π₯) = 4π₯ + 9 πππ π(π₯) =π₯β9
4. πΉπππ π(π(π₯)).
π(π(π₯)) = π(4π₯ + 9) [π π’ππ π‘ππ‘π’π‘π 4π₯ + 9 πππ‘π π(π₯)]
=(4π₯+9)β9
4 [ πΉππ ππ£πππ¦ π₯ ππ π(π₯), π π’ππ π‘ππ‘π’π‘π 4π₯ + 9]
= 4π₯
4 [ ππππππππ¦ ]
π(π(π₯)) = π₯ [ ππππππππ¦ ]
3. Given: π(π₯) = 4π₯ + 9 πππ π(π₯) =π₯β9
4. πΉπππ π(π(π₯)).
π(π(π₯)) = π(4π₯ + 9) [π π’ππ π‘ππ‘π’π‘π 4π₯ + 9 πππ‘π π(π₯)]
= 4(4π₯ + 9) + 9 [ πΉππ ππ£πππ¦ π₯ ππ π(π₯), π π’ππ π‘ππ‘π’π‘π 4π₯ + 9]
= 16π₯ + 36 + 9 [ ππππππππ¦ ]
π(π(π₯)) = 16π₯ + 45 [ ππππππππ¦ ]
4. Given: π(π₯) = 4π₯ + 9 πππ π(π₯) =π₯β9
4. πΉπππ π(π(3)).
π(π(π₯)) = 16π₯ + 45 [π€π ππππ€ π‘βππ ππππ ππ₯ππππ 3]
π(π(3)) = 16(3) + 45 [π π’ππ π‘ππ‘π’π‘π 3 πππ‘π π₯]
= 48 + 45 [ ππππππππ¦ ]
π(π(3)) = 93 [ ππππππππ¦ ]
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May 2020
Practice Problems:
1. Let π(π₯) = 2π₯2 β 4 πππ π(π₯) = 5π₯. πΉπππ πππβ πππ π πππππππ¦.
a) (π β π)(1) b) (π β π)(1) c) (π β π)(β2) d) (π β π)(β1)
2. Let π(π₯) =8
π₯β3πππ π(π₯) =
15
π₯+1. πΉπππ πππβ πππ π πππππππ¦.
a) (π(π(π₯)) b) (π β π)(π₯) c) (π(π(π₯)) d) (π(π(π₯))
3. Use your answers for a) and B) in problem 2 to calculate the two problems below.
a) (π(π(β1)) b) (π β π)(3)
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May 2020
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 π(π₯) = π₯ + 5, π(π₯) = π₯2, β(π₯) = 3π₯ πππ π(π₯) = 2π₯. Express each function as a composite of π, π
β, and/or j. 1. π΄(π₯) = π₯4 [Think: What multiple functions composed this?]
The two functions are
g(π₯) = π₯2 and g(π₯) = π₯2
Check:
π(π(π₯)) = π(π₯2)
= (π₯2)2
= π₯4
2. πΆ(π₯) = 3π₯ + 15 [Think: What multiple functions composed this?]
The two functions are
β(π₯) = 3π₯ and π(π₯) = π₯ + 5
Check:
β(π(π₯)) = β(π₯ + 5 )
= 3(π₯ + 5)
= 3π₯ + 15
3. πΆ(π₯) = 3(π₯ + 5)2 [Think: What multiple functions composed this?]
The three functions are
β(π₯) = 3π₯ and π(π₯) = π₯ + 5 and π(π₯) = π₯2,
Check:
β(π(π(π₯)))) = β(π(π₯ + 5 ) [ work inside first- substitute π₯ + 5 into π(π₯)]
= β((π₯ + 5)2) [ For every π₯ in π(π₯), substitute π₯ + 5]
= 3(π₯ + 5)2 [For every π₯ in β(π₯), substitute π(π(π₯)) which is (π₯ + 5)2]
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May 2020
Practice Problems:
Let π(π₯) = π₯2, π(π₯) = 5π₯, πππ β(π₯) = βπ₯ + 2. Express each function as a composite of π, π and/or β. 1. F(π₯) = π₯4
2. πΆ(π₯) = 5π₯2
3. π(π₯) = π₯ + 2
4. π (π₯) = 5βπ₯ + 10
5. π(π₯) = 25π₯
7. π·(π₯) = ββπ₯ + 2 + 2
8. π΅(π₯) = π₯ + 4βπ₯ + 4
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May 2020
Different Combinations
SET Problems: Identifying the 2 functions that make up a composite function.
Find functions π and π so that π β π = π»
1. π»(π₯) = βπ₯2 + 5π₯ β 4
2. π»(π₯) = (3 β1
π₯)2
3. π»(π₯) = (3π₯ β 7)4 4. π»(π₯) = |5π₯2 β 78|
5. π»(π₯) = 2
3βπ₯5 6. π»(π) = (tan π)2
7. π»(π₯) = 9(4π₯ β 8) + 1 8. π»(π₯) = β
1
6π₯
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GO Problems: Finding function values given the graph. Use the graph to find all of the missing values.
1. π(β ) = 8 2. π(β ) = 5
3. π(β ) = -1 4. π(β ) = 0
5. π(β1) = 6. π(0) =
7. π(π₯) = π(π₯) 8. π(π₯) β π(π₯) = 0
9. π(π₯) β π(π₯) = 0 10. π(2) + π(2) =
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