Alg2 Lesson 2.2: Composition w/ Absolute Value

Lesson 2.2: Function

Composition with the Absolute

Value Function

Obj: I can compose 2 functions given a table of values or a graph.

EQ: How do I evaluate the compositions of absolute
value of inputs and outputs given a table and graph?

Roles:
Facilitator
Scribe
Resourcer
Includer

Lesson Goals:
● Creative Thinking
● Talk through controversies and conflict
● Recognize and reduce ambiguity
● Encourage thinking based on formulas and prior info
● Help explain ideas to each other
● Own your ideas and work
● Record ideas in your journal
● Answer Questions on Slides
● Follow your team roles

Facilitator

• Make sure that all peers are staying on task.

• Give advice or suggestions to resolve the problem.

• Be sure everyone is able to explain.

Scribe

• Make sure peers organize their results on their own papers.

• Remind peers to use color, arrows, and other math tools to
communicate your mathematics, reasons, and connections.

• Be ready to join the teacher for a huddle.

Resourcer

• Make sure peers are getting the materials needed.

• Make sure that all materials are put away neatly.

• Make sure that peers are logged in to the needed site.

• Help troubleshoot any technology difficulties that may arise.

Includer

• Make sure that all peers are talking about their work.

• Helps keep peers’ voice volume low.

• Encourages everyone to ask questions.

• Communicates conflicts or questions to the teacher.

Poll

How do you plan to contribute?

• Ask questions to teacher and peers.

• Join classmates that are on the same lesson topic.

• Write through notes on my journal.

• Answer the slides on here at least 2 times, to demonstrate mastery.

● Check off tasks & skills on calendar.

● Select skills to work on.

● Work on Deltamath.

Remember to work on the following too…

Handout 2.2: Function Composition w/ the Abs Value Function

Work with a partner to complete the handout. Be ready to share your responses!

Record your answers on the slides. Turn in the physical paper with work and reasoning shown! This includes graphing parts a, c, and g!

Problem 1

Graphing

1a. Graph the equation.

$$y=\frac{1}{2}x-3$$

Multiple Choice

1b. Given $$p\left(x\right)=\frac{1}{2}x-3$$ and $$a\left(x\right)=\left|x\right|$$ , define the function h, such that $$h\left(x\right)=a\left(p\left(x\right)\right)$$

• h(x) = 0.5|x| - 3

• h(x) = |0.5x - 3|

Multiple Choice

1d. For what values of x will h(x) = p(x)?

• x ≥ 6

• x ≤ 6

• x ≥ 0

• x ≤ 0

Drag and Drop

1e. The graph of y = h(x) is entirely ​

Placeholder

the ​

Placeholder

. This is because when x ​

Placeholder

6, the output values of p are negative and the function h is the absolute value of the output values of p. Therefore, the graph of h is a reflection over the ​

Placeholder

of the part of the graph of p that is ​

Placeholder

the y-axis.

Drag these tiles and drop them in the correct blank above

on or above

on or below

x-axis

y-axis

<

below

above

Multiple Choice

1f. Given $$p\left(x\right)=\frac{1}{2}x-3$$ and $$a\left(x\right)=\left|x\right|$$ , define the function k, such that $$k\left(x\right)=p\left(a\left(x\right)\right)$$

• k(x) = 0.5|x| - 3

• k(x) = |0.5x - 3|

Multiple Choice

1h. For what values of x will k(x) = p(x)?

• x ≥ 6

• x ≤ 6

• x ≥ 0

• x ≤ 0

Drag and Drop

1i. The graph of y = k(x) has reflection symmetry over the line ​

Placeholder

= 0. This is because the ​

Placeholder

values of p are negative for x < 0. The function ​

Placeholder

is the absolute value of the input values of ​

Placeholder

, so the graph of k is the reflection over the x-axis of the part of the graph of p that is to the ​

Placeholder

of the x-axis.

Drag these tiles and drop them in the correct blank above

x

y

input

output

k

p

right

h

Draw

Complete the symmetrical sunflower.

Problem 2

Graphing

2a. Graph y = q(x) = x² - 2x - 3 on the coordinate plane.

Multiple Choice

2b. Given $$q\left(x\right)=x^2-2x-3$$ and $$a\left(x\right)=\left|x\right|$$ , define the function m, such that $$m\left(x\right)=a\left(q\left(x\right)\right)$$

• m(x) = |x|² - 2|x| - 3

• m(x) = |x² - 2x - 3|

Multiple Choice

2d. For what values of x will m(x) = q(x)?

• x ≤ -1 or x ≥ 3

• -1 ≤ x ≤ 3

• x ≥ 0

• x ≤ 0

Drag and Drop

2e. The graph of y = m(x) is entirely ​

Placeholder

the ​

Placeholder

. This is because when ​

Placeholder

, the output values of ​q are negative and the function m is the absolute value of the output values of​ q. Therefore, the graph of m is a reflection over the ​

Placeholder

of the part of the graph of q that is below the ​

Placeholder

.

Drag these tiles and drop them in the correct blank above

on or above

on or below

−1 < x < 3

x < -1 or x > 3

x-axis

y-axis

Multiple Choice

2f. Given $$q\left(x\right)=x^2-2x-3$$ and $$a\left(x\right)=\left|x\right|$$ , define the function n, such that $$n\left(x\right)=q\left(a\left(x\right)\right)$$

• m(x) = |x|² - 2|x| - 3

• m(x) = |x² - 2x - 3|

Multiple Choice

2h. For what values of x will n(x) = q(x)?

• x ≤ -1 or x ≥ 3

• -1 ≤ x ≤ 3

• x ≥ 0

• x ≤ 0

Drag and Drop

2i. The graph of y = n(x) has reflection symmetry over the line ​

Placeholder

= 0. This is because the ​

Placeholder

values of q are negative for x < 0. The function ​

Placeholder

is the absolute value of the input values of ​

Placeholder

, so the graph of n is the reflection over the ​

Placeholder

of the part of the graph of q that is to the right of the x-axis. The graphs of n and q are the same for values of x that are greater than or equal to 0.

Drag these tiles and drop them in the correct blank above

x

y

input

output

n

q

x-axis

y-axis

Random Question of the Day Time

https://wheelofnames.com/4ke-epz We’ll spin the wheel as a class and spend a minute or so
discussing our answers.

Problem 3

Labelling

Put the correct formulas under each part. Also explain your reasoning on the handout.

Drag labels to their correct position on the image

a(f(x)) = f(|x|)

f(a(x)) = f(|x|)

f(a(x)) = |f(x)|

a(f(x)) = |f(x)|