Answers26-27 AISD - Algebra 2A-IC HallReflections and Dilations of Absolute Value Functions

Reflections and Dilations of Absolute Value Functions Answers

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1
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Question text not available

Question illustration
A
f(x) = 1.3|x| – 2
B
f(x) = 3|x – 2|
C
f(x) = |x – 2|
Option C
D
f(x) = |x| + 2
Option D
2
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The function shown is reflected across the y-axis to create a new function.

Question illustration
A
Both the domain and range change.
B
Both the range and domain stay the same.
C
The domain stays the same, but the range changes.
D
The range stays the same, but the domain changes.
3

Which graph represents the function f(x) = –|x + 3|?

A
Option
B
Option
C
Option
D
Option
4

Which graph represents the function f(x) = –|x| – 2?

A
Option
B
Option
C
Option
D
Option
5

The absolute value function, f(x) = –|x| – 3, is shown.

Question illustration
A
all real numbers
B
all real numbers less than or equal to 0
C
all real numbers greater than or equal to –3
D
all real numbers less than or equal to –3
6

The absolute value function, f(x) = |x + 2|, is shown.

Question illustration
A
all real numbers
B
all real numbers greater than or equal to 0
C
all real numbers greater than or equal to –2
D
all real numbers less than or equal to –2
7

Which graph represents the function f(x) = |x|?

Question illustration
A
Option
B
Option
C
Option
D
Option
8

Question text not available

Question illustration
A
Both the domain and range of the transformed function are the same as those of the parent function.
B
Neither the domain nor the range of the transformed function are the same as those of the parent function.
C
The range of the transformed function is the same as the parent function, but the domains of the functions are different.
D
The domain of the transformed function is the same as the parent function, but the ranges of the functions are different.
9

Question text not available

Question illustration
A
f(x) = |x| – 3
B
f(x) = |x + 2|
C
f(x) = 0.5|x|
D
f(x) = 4|x|
10

Question text not available

Question illustration
A
f(x) = –|x|
B
f(x) = |–x|
C
f(x) = |x| + 1
D
f(x) = |x – 1|

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