Mastering Absolute Value Functions: Dive Deep into Graphing

$x$	$y = 2 ∣ x - 2 ∣ + 3$	$y$
$- 4$	$y = 2 ∣ - 4 - 2 ∣ + 3$	$15$
$- 2$	$y = 2 ∣ - 2 - 2 ∣ + 3$	$11$
$0$	$y = 2 ∣ 0 - 2 ∣ + 3$	$7$
$2$	$y = 2 ∣ 2 - 2 ∣ + 3$	$3$
$4$	$y = 2 ∣ 4 - 2 ∣ + 3$	$7$
$6$	$y = 2 ∣ 6 - 2 ∣ + 3$	$11$
$8$	$y = 2 ∣ 8 - 2 ∣ + 3$	$15$

x

y = 2 ∣ x - 2 ∣ + 3

y

- 4

y = 2 ∣ - 4 - 2 ∣ + 3

15

- 2

y = 2 ∣ - 2 - 2 ∣ + 3

11

0

y = 2 ∣ 0 - 2 ∣ + 3

7

2

y = 2 ∣ 2 - 2 ∣ + 3

3

4

y = 2 ∣ 4 - 2 ∣ + 3

7

6

y = 2 ∣ 6 - 2 ∣ + 3

11

8

y = 2 ∣ 8 - 2 ∣ + 3

15

$t$	$s = - 2 ∣ t - 10 ∣ + 40$	$s$
$0$	$s = - 2 ∣ 0 - 10 ∣ + 40$	$20$
$5$	$s = - 2 ∣ 5 - 10 ∣ + 40$	$30$
$10$	$s = - 2 ∣ 10 - 10 ∣ + 40$	$40$
$15$	$s = - 2 ∣ 15 - 10 ∣ + 40$	$30$
$20$	$s = - 2 ∣ 20 - 10 ∣ + 40$	$20$
$25$	$s = - 2 ∣ 25 - 10 ∣ + 40$	$10$
$30$	$s = - 2 ∣ 30 - 10 ∣ + 40$	$0$

t

s = - 2 ∣ t - 10 ∣ + 40

s

0

s = - 2 ∣ 0 - 10 ∣ + 40

20

5

s = - 2 ∣ 5 - 10 ∣ + 40

30

10

s = - 2 ∣ 10 - 10 ∣ + 40

40

15

s = - 2 ∣ 15 - 10 ∣ + 40

30

20

s = - 2 ∣ 20 - 10 ∣ + 40

20

25

s = - 2 ∣ 25 - 10 ∣ + 40

10

30

s = - 2 ∣ 30 - 10 ∣ + 40

0

$x$	$y = \frac{3}{2} ∣ x - 2 ∣$	$y$
$- 2$	$y = \frac{3}{2} ∣ - 2 - 2 ∣$	$6$
$0$	$y = \frac{3}{2} ∣ 0 - 2 ∣$	$3$
$2$	$y = \frac{3}{2} ∣ 2 - 2 ∣$	$0$
$4$	$y = \frac{3}{2} ∣ 4 - 2 ∣$	$3$
$6$	$y = \frac{3}{2} ∣ 6 - 2 ∣$	$6$

x

y = \frac{3}{2} ∣ x - 2 ∣

y

- 2

y = \frac{3}{2} ∣ - 2 - 2 ∣

6

0

y = \frac{3}{2} ∣ 0 - 2 ∣

3

2

y = \frac{3}{2} ∣ 2 - 2 ∣

0

4

y = \frac{3}{2} ∣ 4 - 2 ∣

3

6

y = \frac{3}{2} ∣ 6 - 2 ∣

6

$x$	$y = - 2 ∣ x - 1 ∣ + 4$	$y$
$- 5$	$y = - 2 ∣ - 5 - 1 ∣ + 4$	$- 8$
$- 3$	$y = - 2 ∣ - 3 - 1 ∣ + 4$	$- 4$
$- 1$	$y = - 2 ∣ - 1 - 1 ∣ + 4$	$0$
$1$	$y = - 2 ∣ 1 - 1 ∣ + 4$	$4$
$3$	$y = - 2 ∣ 3 - 1 ∣ + 4$	$0$
$5$	$y = - 2 ∣ 5 - 1 ∣ + 4$	$- 4$
$7$	$y = - 2 ∣ 7 - 1 ∣ + 4$	$- 8$

x

y = - 2 ∣ x - 1 ∣ + 4

y

- 5

y = - 2 ∣ - 5 - 1 ∣ + 4

- 8

- 3

y = - 2 ∣ - 3 - 1 ∣ + 4

- 4

- 1

y = - 2 ∣ - 1 - 1 ∣ + 4

0

1

y = - 2 ∣ 1 - 1 ∣ + 4

4

3

y = - 2 ∣ 3 - 1 ∣ + 4

0

5

y = - 2 ∣ 5 - 1 ∣ + 4

- 4

7

y = - 2 ∣ 7 - 1 ∣ + 4

- 8

$x$	$h = 71 - \frac{7 1}{5 6} ∣ x - 56 ∣$	$h$
$0$	$h = 71 - \frac{7 1}{5 6} ∣ 0 - 56 ∣$	$0$
$20$	$h = 71 - \frac{7 1}{5 6} ∣ 20 - 56 ∣$	$\approx 25.36$
$40$	$h = 71 - \frac{7 1}{5 6} ∣ 40 - 56 ∣$	$\approx 50.71$
$56$	$h = 71 - \frac{7 1}{5 6} ∣ 56 - 56 ∣$	$71$
$70$	$h = 71 - \frac{7 1}{5 6} ∣ 60 - 56 ∣$	$\approx 53.25$
$90$	$h = 71 - \frac{7 1}{5 6} ∣ 80 - 56 ∣$	$\approx 27.89$
$110$	$h = 71 - \frac{7 1}{5 6} ∣ 100 - 56 ∣$	$\approx 2.54$

x

h = 71 - \frac{7 1}{5 6} ∣ x - 56 ∣

h

0

h = 71 - \frac{7 1}{5 6} ∣ 0 - 56 ∣

0

20

h = 71 - \frac{7 1}{5 6} ∣ 20 - 56 ∣

\approx 25.36

40

h = 71 - \frac{7 1}{5 6} ∣ 40 - 56 ∣

\approx 50.71

56

h = 71 - \frac{7 1}{5 6} ∣ 56 - 56 ∣

71

70

h = 71 - \frac{7 1}{5 6} ∣ 60 - 56 ∣

\approx 53.25

90

h = 71 - \frac{7 1}{5 6} ∣ 80 - 56 ∣

\approx 27.89

110

h = 71 - \frac{7 1}{5 6} ∣ 100 - 56 ∣

\approx 2.54

$x$	$y = 12 ∣ x - 5 ∣ - 3$	$y$
$0$	$y = 12 ∣ 0 - 5 ∣ - 3$	$57$
$2$	$y = 12 ∣ 2 - 5 ∣ - 3$	$33$
$4$	$y = 12 ∣ 4 - 5 ∣ - 3$	$9$
$5$	$y = 12 ∣ 5 - 5 ∣ - 3$	$- 3$
$6$	$y = 12 ∣ 6 - 5 ∣ - 3$	$9$
$8$	$y = 12 ∣ 8 - 5 ∣ - 3$	$33$
$10$	$y = 12 ∣ 10 - 5 ∣ - 3$	$57$

x

y = 12 ∣ x - 5 ∣ - 3

y

0

y = 12 ∣ 0 - 5 ∣ - 3

57

2

y = 12 ∣ 2 - 5 ∣ - 3

33

4

y = 12 ∣ 4 - 5 ∣ - 3

9

5

y = 12 ∣ 5 - 5 ∣ - 3

- 3

6

y = 12 ∣ 6 - 5 ∣ - 3

9

8

y = 12 ∣ 8 - 5 ∣ - 3

33

10

y = 12 ∣ 10 - 5 ∣ - 3

57

Consider the absolute value function.

f (x) = ∣ x ∣ + 3

Now consider the following graphs.

Which of these graphs is the graph of

f (x) ?

Let's consider the given absolute value function. f(x)=|x|+3 We can graph an absolute value function by making a table of values. To find a range of points for our graph, we will create this table for values of x from -3 to 3.

x	\|x\|+3	f(x)=\|x\|+3
-3	\| -3\|+3	6
-2	\| -2\|+3	5
-1	\| -1\|+3	4
0	\| 0\|+3	3
1	\| 1\|+3	4
2	\| 2\|+3	5
3	\| 3\|+3	6

Next, we will plot these ordered pairs on the coordinate plane.

Recall that the graphs of absolute value functions follow a common pattern — that is, they all form a V-shape. Let's now draw a V-shaped graph that passes through these points.

Considering the available options, we can now see that the graph of f(x) corresponds to choice C. This is the correct answer.

Consider the absolute value equation.

3 g (x) + 2 = 6 ∣ x ∣ - 1

Now consider the following graphs.

Which of these graphs is the graph of

g (x) ?

Let's consider the absolute value equation. 3g(x)+2=6|x|-1 To draw the graph of g(x), we should first isolate g(x).

We can now graph g(x) by making a table of values. We will create this table by using values of x from -4 to 4.

x	2\|x\|-1	g(x)
-4	2\| -4\|-1	7
-2	2\| -2\|-1	3
0	2\| 0\|-1	-1
2	2\| 2\|-1	3
4	2\| 4\|-1	7

Next, we will plot these ordered pairs on the coordinate plane.

Finally, we can draw a V-shaped graph that passes through these points.

Therefore, the graph of g(x) corresponds to choice A.

The graph of which function has the same

y -

intercept as the graph of

f (x) = ∣ x - 2 ∣ + 3 ?

To determine which of the absolute value functions has the same y-intercept as f(x)=|x-2|+3, we will draw the graphs of the functions and compare their y-intercepts. We will start by graphing f(x). To do so, we will make a table of values for the x-values between -1 and 5.

x	\|x-2\|+3	f(x)
-1	\| -1-2\|+3	6
0	\| 0-2\|+3	5
1	\| 1-2\|+3	4
2	\| 2-2\|+3	3
3	\| 3-2\|+3	4
4	\| 4-2\|+3	5
5	\| 5-2\|+3	6

Now we will plot and connect the obtained points. Keep in mind that the graph of an absolute value function is V-shaped.

We can see that the y-intercept of the graph is 5. We can graph g(x)=|3x-2|+3 and h(x)=3|x-2|+3 by following the same procedure.

We can see that the graphs of g(x) and f(x) intersect the y-axis at the same point. Therefore, they have the same y-intercept.

Another way to identify the y-intercepts of the graphs is to substitute 0 for x and simplify the right-hand side of the functions. Let's begin with f(x).

The graph of f(x) intercepts the y-axis at (0,5). By applying the same operations, we can find the y-intercepts of the other functions.

	x=0	Simplify	y-intercept
f(x)=\|x-2\|+3	f( 0)=\| 0-2\|+3	f(0)=5	(0,5)
g(x)=\|3x-2\|+3	g( 0)=\|3( 0)-2\|+3	g(0)=5	(0,5)
h(x)=3\|x-2\|+3	h( 0)=3\| 0-2\|+3	h(0)=9	(0,9)

As we can see, f(x) and g(x) have the same y-intercept.

The vertex of an absolute value function is the point at which its graphs changes direction. Consider the absolute value function.

f (x) = - ∣ x - 2 ∣ + 3

What are the coordinates of the vertex of

f (x) ?

Let's recall the absolute value function. f(x)=-|x-2|+3 We can graph an absolute value function by making a table of values. We will create this table for values of x between -1 and 5.

x	-\|x-2\|+3	f(x)
-1	-\| -1-2\|+3	0
0	-\| 0-2\|+3	1
1	-\| 1-2\|+3	2
2	-\| 2-2\|+3	3
3	-\| 3-2\|+3	2
4	-\| 4-2\|+3	1
5	-\| 5-2\|+3	0

Next, we will plot these ordered pairs on the coordinate plane.

We can now draw a V-shaped graph that passes through these points.

Recall that the vertex of an absolute value function is the point at which the graph changes direction. By looking at the above graph, we can see that the coordinates of the vertex are (2,3).

The vertex of an absolute value function can be also identified by its equation. Let's recall the vertex form of an absolute value function. f(x)=a|x-h|+k In this form, (h,k) is the vertex of the function. With this information, we can immediately identify the vertex of f(x)=-|x-2|+3 as (2,3).

At a given cell phone store, the number of phones sold

s,

in thousands, increases and then decreases throughout the course of a year as described by the following absolute value function.

s (t) = - \frac{1}{2} ∣ t - 4 ∣ + 8

In this function,

t

represents time in months. What is the greatest number of phones sold in a month?

Let's begin by recalling the absolute value function. s(t)=- 1/2|t-4|+8 We can graph an absolute value function by making a table of values. Note that the number of months cannot be negative. Therefore, we will create this table of values with x-values from 0 to 12.

t	- 1/2\|t-4\|+8	s(t)
0	-1/2\| 0-4\|+8	6
2	-1/2\| 2-4\|+8	7
4	-1/2\| 4-4\|+8	8
6	-1/2\| 6-4\|+8	7
8	-1/2\| 8-4\|+8	6
10	-1/2\| 10-4\|+8	5
12	-1/2\| 12-4\|+8	4

Next, we will plot these ordered pairs on the coordinate plane.

We can now draw a V-shaped graph that passes through these points.

We can see that the number of phones sold increases until the fourth month. Then, it starts to decrease. This means that the greatest number of phones sold in one month is represented by the vertex of the function.

We see that the vertex is at (4,8). Therefore, in the 4^(th) month, 8000 phones were sold. This is the greatest number of phones sold in a month.

x	\|x\|+3	f(x)=\|x\|+3
-3	\| -3\|+3	6
-2	\| -2\|+3	5
-1	\| -1\|+3	4
0	\| 0\|+3	3
1	\| 1\|+3	4
2	\| 2\|+3	5
3	\| 3\|+3	6

x	2\|x\|-1	g(x)
-4	2\| -4\|-1	7
-2	2\| -2\|-1	3
0	2\| 0\|-1	-1
2	2\| 2\|-1	3
4	2\| 4\|-1	7

x	\|x-2\|+3	f(x)
-1	\| -1-2\|+3	6
0	\| 0-2\|+3	5
1	\| 1-2\|+3	4
2	\| 2-2\|+3	3
3	\| 3-2\|+3	4
4	\| 4-2\|+3	5
5	\| 5-2\|+3	6

x	-\|x-2\|+3	f(x)
-1	-\| -1-2\|+3	0
0	-\| 0-2\|+3	1
1	-\| 1-2\|+3	2
2	-\| 2-2\|+3	3
3	-\| 3-2\|+3	2
4	-\| 4-2\|+3	1
5	-\| 5-2\|+3	0

t	- 1/2\|t-4\|+8	s(t)
0	-1/2\| 0-4\|+8	6
2	-1/2\| 2-4\|+8	7
4	-1/2\| 4-4\|+8	8
6	-1/2\| 6-4\|+8	7
8	-1/2\| 8-4\|+8	6
10	-1/2\| 10-4\|+8	5
12	-1/2\| 12-4\|+8	4

Graphing Absolute Value Equations

Catch-Up and Review

Modeling a Real-World Situation With an Absolute Value Function

Absolute Value Functions and Their Graphs

Absolute Value Function

Graphing an Absolute Value Function Using a Table of Values

Modeling Shoes Sold With an Absolute Value Function

Answer

Hint

Solution

Modeling the Path of a Billiard-Ball With an Absolute Value Function

Answer

Hint

Solution

Characteristics of Absolute Value Graphs

Vertex Form of an Absolute Value Function

Identifying the Characteristics of an Absolute Value Graph

Identifying the Characteristics of an Absolute Value Graph

Hint

Solution

Interpreting the Characteristics of an Absolute Value Graph

Hint

Solution

Interpreting an Absolute Value Graph in a Context

Hint

Solution

Graphing Absolute Value Equations

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