Algebra Dilations and Reflections of Absolute Value Functions

Algebra
Name ________________________________________
CHAPTER 7
FUNCTIONS AND TRANSFORMATIONS
Period ______ Date ____________________________
Dilations and Reflections of Absolute Value Functions: 𝒇(𝒙) = |𝒙|
In questions 1 through 6,
a) quickly sketch the graph of the parent function, 𝑓(π‘₯) = |π‘₯|
b) identify the vertex of each new function
c) graph the new function
d) make a table of values for the new function
e) write a verbal description of the transformation from the parent graph, 𝑓(π‘₯) = |π‘₯|, to the new function
1.
𝑦 = 2|π‘₯|
2.
𝑦 = βˆ’3|π‘₯|
Vertex: (
,
)
Vertex: (
,
)
x
y
x
Description of the transformation:
1
3.
𝑦 = 3 |π‘₯ + 3|
Vertex: (
,
)
x
y
y
Description of the transformation:
3
4.
𝑦 = βˆ’ 4 |π‘₯ βˆ’ 2|
Vertex: (
,
)
x
y
Description of the transformation:
Description of the transformation:
5.
6.
𝑦 = βˆ’ |π‘₯ βˆ’ 4| + 3
4
Vertex: (
,
)
𝑦 = 2|π‘₯ + 1| βˆ’ 6
Vertex: (
x
,
)
y
Description of the transformation:
1
x
y
Description of the transformation:
In questions 7 through 12,
a) write a verbal description of the transformation from the parent graph, 𝑓(π‘₯) = |π‘₯|, to the new function
b) write an absolute value function for the graph
7. Description:
8. Description:
9. Description:
Equation:
10. Description:
Equation:
Equation:
11. Description:
Equation:
Equation:
12. Description:
Equation:
13. The graph of 𝑓(π‘₯) = |π‘₯| is given below. Sketch the new graph completing the transformations in the
order given.
a) First translate the absolute value function up 2 b) First vertically stretch the function by a factor
spaces. Then vertically stretch the translated
of 3 (multiply the function by 3). Then
function by a factor of 3 (multiply the function
translate the stretched absolute value function
by 3).
up 2 spaces.
14.
Are the graphs in 13(a) and 13(b) the same graph? Why or why not?
15.
Write equations for each of the two new functions you graphed in question 13(a) and (b).
Equation for 13(a):
Equation for 13(b):
In problems 16-23, given the graphs shown below, find the value of each expression.
Evaluate:
16. 𝑓(3) =
17. (𝑔)(βˆ’2) =
18.
𝑓(π‘₯)
𝑔(π‘₯)
𝑓(2) + 𝑔(βˆ’4) =
19.
𝑔(1) βˆ’ 𝑓(1) =
Solve:
20. 𝑓(π‘₯) = βˆ’4;
π‘₯ =?
21.
𝑔(π‘₯) = 0;
22.
π‘₯ =?
23.
𝑓(π‘₯) = 𝑔(π‘₯);
𝑓(π‘₯) = βˆ’4;
π‘₯ =?
π‘₯ =?
Algebra
ANSWERS
CHAPTER 7
FUNCTIONS AND TRANSFORMATIONS
Dilations and Reflections of Absolute Value Functions: 𝒇(𝒙) = |𝒙|
1.
𝑦 = |π‘₯| βˆ’ 2
Vertex: (0, βˆ’2)
x
-2
-1
0
1
2
y
0
-1
-2
-1
0
2.
𝑦 = |π‘₯| + 4
Vertex: (0, 4)
x
-2
-1
0
1
2
y
6
5
4
6
0
Description of the transformation:
The graph of 𝑓(π‘₯) = |π‘₯| was shifted down two
units.
Description of the transformation:
The graph of 𝑓(π‘₯) = |π‘₯| was shifted up four units.
3.
𝑦 = |π‘₯ + 3|
Vertex: (βˆ’3, 0)
4.
𝑦 = |π‘₯ βˆ’ 2|
Vertex: (2, 0)
x
-5
-4
-3
-2
-1
y
2
1
0
1
2
x
-5
-4
-3
-2
-1
y
2
1
0
1
2
Description of the transformation:
The graph of 𝑓(π‘₯) = |π‘₯| was shifted left three
units.
Description of the transformation:
The graph of 𝑓(π‘₯) = |π‘₯| was shifted right two units.
5.
𝑦 = |π‘₯ βˆ’ 1| + 3
Vertex: (1, 3)
6.
𝑦 = |π‘₯ + 5| βˆ’ 2
Vertex: (βˆ’5, βˆ’2)
x
-1
0
1
2
3
y
5
4
3
4
5
Description of the transformation:
The graph of 𝑓(π‘₯) = |π‘₯| was shifted right one unit
and up three units.
x
-7
-6
-5
-4
-3
y
0
-1
-2
-1
0
Description of the transformation:
The graph of 𝑓(π‘₯) = |π‘₯| was shifted left five units
and down two units.
7.
𝑦 = |π‘₯ + 2| + 1
Vertex: (βˆ’2, 1)
x
-4
-3
-2
-1
0
8.
𝑦 = |π‘₯ βˆ’ 3| βˆ’ 1
Vertex: (3, βˆ’ 1)
y
3
2
1
2
3
x
1
2
3
4
5
y
1
0
-1
0
1
Description of the transformation:
The graph of 𝑓(π‘₯) = |π‘₯| was shifted left two units
and up one unit.
Description of the transformation:
The graph of 𝑓(π‘₯) = |π‘₯| was shifted right three
units and down one unit.
9.
Vertex: (0, βˆ’3)
10. Vertex: (βˆ’2, 0)
Equation:
𝑦 = |π‘₯| βˆ’ 3
Equation:
𝑦 = |π‘₯ + 2|
11. Vertex: (3, 1)
12. Vertex: (βˆ’4, βˆ’2)
Equation:
𝑦 = |π‘₯ βˆ’ 3| + 1
Equation:
𝑦 = |π‘₯ + 4| βˆ’ 2
13. The cost in dollars of trampoline jumping at Sky High Sports in Valencia is a function of the number of
hours you jump. The relationship between both variables can be modeled as 𝐢(𝑑) = 3.50𝑑 + 6.00.
a) Find the value of 𝐢(4) Write a sentence to interpret your answer in the context of the problem.
𝐢(4) = 20. This means that for four hours on trampoline jumping, you must pay $20.
b) Solve the equation 𝐢(𝑑) = 30.50. Write a sentence to interpret your answer in the context of the
problem.
If 𝐢(𝑑) = 30.50, then 𝑑 = 7. This means that if you are willing to spend $30.50 on trampoline
jumping, you will be able to do so for seven hours.
In problems 14-21, given the graphs shown below, find the value of each expression.
14. 𝑓(2) = 3
15. 𝑔(2) = 1
16.
𝑔(3) = 5
17.
𝑓(0) = β€’1
18.
𝑔(0) + 𝑓(βˆ’1) = β€’5
19.
20.
If 𝑓(π‘₯) = 3, then π‘₯ =?
π‘₯=2
21.
If 𝑔(π‘₯) = βˆ’2, then π‘₯ =?
π‘₯=0
If 𝑔(π‘₯) = 𝑓(π‘₯), then π‘₯ =?
π‘₯=3