Unit 6: Absolute Value

Absolute Value means ...

... only how far a number is from zero:

"6" is 6 away from zero,

and "-6" is also 6 away from zero.

So the absolute value of 6 is 6,

and the absolute value of -6 is also 6

More Examples:

 The absolute value of -9 is 9

 The absolute value of 3 is 3
 The absolute value of 0 is 0
 The absolute value of -156 is 156

No Negatives!

So in practice "absolute value" means to remove any negative sign in front of a number, and to think of all
numbers as positive (or zero).

Absolute Value Symbol

To show that you want the absolute value of something, you put "|" marks either side (they are called
"bars" and are found on the right side of your keyboard), like these examples:

|-5| = 5 |7| = 7

Sometimes absolute value is also written as "abs()", so abs(-1) = 1 is the same as |-1| = 1

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Examples

|8-3| = 5 |3-8| = 5
(8-3 = 5) (3-8 = -5, and |-5| = 5)

|-3×6| = 18 -|-12| = -12

(|-12| = 12 and then the
(-3×6 = -18, and |-18| = 18)
first minus gets you -12)

So, when a number is positive or zero we leave it alone, when it is negative we change it to positive.

Useful Properties

Here are some properties of absolute values that can be useful:

|a| ≥ 0 always!

That makes sense ... |a| can never be less than zero.

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 Absolute Value Function Investigations
Goal: Explore parameter changes to the absolute value parent function.
Parent Function: 𝒚 = |𝒙|
General Function: 𝒚 = 𝒂 |𝒙 − 𝒉| + 𝒌
1. Graph the parent function. This is the most basic form of the function. For absolute value it is
𝑦 = |𝑥|.

2. Changing h:

Graph 𝑦 = |𝑥 + 4|. What happened?

________________________________________________

Graph 𝑦 = |𝑥 − 4|. What happened?

________________________________________________
What is the rule when h is changed?
________________________________________________
________________________________________________

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Page 4 of 23
6. Prediction: Put your calculator down. Using the rules you came up with, draw the graph of
𝑦 = −2|𝑥 − 3| + 4. Now check the answer by graphing the function on your calculator. Did you get it
right?____________

Page 5 of 23
Graphing Absolute Value Equations with two variables

The equation should be in the form 𝒚 = 𝒂|𝒙 − 𝒉| + 𝒌

1. (h, k) is the vertex (Remember h has the opposite sign)
2. a is the slope of the two lines. If a is negative the graph reflects (flips) and opens down.

Example: graph y=|x+2| - 5

The vertex is at (-2, -5)

then shift it down to make

Start with |x| then shift it left to make it|x+2|
it|x+2|-5

WOW! THE SOLUTIONS TO THE EQUATION ARE WHERE THE GRAPH

CROSSES THE X-AXIS (in this case 3 and -7)

If the equation had been y= - 3|x+2| - 5 the graph would:

1. Flip over: the vertex is still at (-2, -5).

2. Have a slope of 3 instead of 1 (it got steeper).

(since the graph doesn’t cross the x axis, there is no solution)

__________________________________________________________________________________________

Page 6 of 23
Algebra 1: Unit 6 Name___________________________________ ID: 1
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Goal: Graphing Absolute Value Functions Date________________ Period____

Graph each equation.
1) y = x + 3 2) y = x - 2
y y
6 6
5 5
4 4
3 3
2 2
1 1
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x
-1 -1
-2 -2
-3 -3
-4 -4
-5 -5
-6 -6

3) y = x - 3 4) y = x + 2
y y
6 6
5 5
4 4
3 3
2 2
1 1
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x
-1 -1
-2 -2
-3 -3
-4 -4
-5 -5
-6 -6

5) y = x - 3 6) y = x + 4
y y
6 6
5 5
4 4
3 3
2 2
1 1
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x
-1 -1
-2 -2
-3 -3
-4 -4
-5 -5
-6 -6

Page 7 of 23 Worksheet by Kuta Software LLC

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7) y = x - 3 - 4 8) y = x + 2 - 2
y y
6 6
5 5
4 4
3 3
2 2
1 1
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x
-1 -1
-2 -2
-3 -3
-4 -4
-5 -5
-6 -6

9) y = x + 1 - 2 10) y = x - 4 + 1
y y
6 6
5 5
4 4
3 3
2 2
1 1
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x
-1 -1
-2 -2
-3 -3
-4 -4
-5 -5
-6 -6

11) y = x + 4 + 2 12) y = x - 2 + 4
y y
6 6
5 5
4 4
3 3
2 2
1 1
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x
-1 -1
-2 -2
-3 -3
-4 -4
-5 -5
-6 -6

Page 8 of 23 Worksheet by Kuta Software LLC

-2-
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13) y = - x + 2 + 3 14) y = - x - 4 - 3
y y
6 6
5 5
4 4
3 3
2 2
1 1
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x
-1 -1
-2 -2
-3 -3
-4 -4
-5 -5
-6 -6

15) y = - x + 3 + 2 16) y = - x - 1 + 1
y y
6 6
5 5
4 4
3 3
2 2
1 1
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x
-1 -1
-2 -2
-3 -3
-4 -4
-5 -5
-6 -6

17) y = - x + 1 - 1 18) y = - x + 1 + 4
y y
6 6
5 5
4 4
3 3
2 2
1 1
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x
-1 -1
-2 -2
-3 -3
-4 -4
-5 -5
-6 -6

Page 9 of 23 Worksheet by Kuta Software LLC

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19) y = 3 x - 2 + 2 1
y 20) y = × x-3 +2
8 2
y
6 8

4 6

2 4

2
-8 -6 -4 -2 2 4 6 8 x
-2 -8 -6 -4 -2 2 4 6 8 x
-4 -2

-6 -4

-8 -6

-8

21) y = 2 x + 4 - 3 1
y 22) y = × x+3 -1
8 3
y
6 8

4 6

2 4

2
-8 -6 -4 -2 2 4 6 8 x
-2 -8 -6 -4 -2 2 4 6 8 x
-4 -2

-6 -4

-8 -6

-8

23) y = 4 x + 4 - 2 1
y 24) y = - × x + 1 - 2
8 2
y
6 8

4 6

2 4

2
-8 -6 -4 -2 2 4 6 8 x
-2 -8 -6 -4 -2 2 4 6 8 x
-4 -2

-6 -4

-8 -6

-8

Page 10 of 23 Worksheet by Kuta Software LLC

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 Absolute Value Functions

Solving Equations having one variable

(Absolute value equations have 2 solutions)

1. Get the absolute value alone on one side of the equation.

2. Using the fact that |u| = a is the same as u = ±a and vice versa:

a. Rewrite the equation in two parts.

b. Solve each equation.

Example: solve 5|x+2|=25

Divide both sides by 5 to get: |x+2|=5

Using "|u| = a is the same as u = ±a":

this: |x+2|=5

is the same as this: x+2 = ±5

Which will have two solutions:

x+2 = -5 x+2 = +5

x = -7 x=3
_____________________________________________________________________________________

Page 11 of 23
Algebra 1: Unit 6 Name___________________________________ ID: 1
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Goal: Solving Absolute Value Equations Date________________ Period____

Solve each equation.
1) n = 1 2) r = 3

3) n = 7 4) x = 9

5) -1 + k = 6 6) x - 5 = 5

7) x + 7 = 15 n
8) = -3
10

9) 10r = 100 10) -7a = 63

Page 12 of 23 Worksheet by Kuta Software LLC

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11) x - 9 = 9 12) x + 9 = 9

13) p + 10 = 2 14) 5 + 4k = 33

15) 2b + 10 = 24 16) 9v - 1 = -64

17) -4 + 7n = 11 18) -3v + 7 = 11

19) 7 - 5r = -28 20) 2r - 4 = 2

Page 13 of 23 Worksheet by Kuta Software LLC

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Algebra 1: Unit 6 Name___________________________________ ID: 1
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Goal: Solving Hard Absolute Value Equations Date________________ Period____

Solve each equation.
1) -7 + p = 3 2) m - 10 = -13

3) m - 7 = -7 4) -4 x = -32

v 6) x - 7 = -1
5) =4
8

7) -7 + n + 8 = 12 v+5
8) =3
10

x x
9) -9 + = -10 10) -2 + =1
8 3

x 12) x + 1 + 1 = 8
11) - 2 = -1
10

Page 14 of 23 Worksheet by Kuta Software LLC

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 n 14) x + 7 - 2 = -2
13) 2 =2
9

15) 4 - -8n = -60 16) -2 - m - 1 = -5

x 18) -2 + 7 x - 6 = 40
17) 2 +3=5
9

19) - -4v - 8 = -44 p

20) 7 - 4 = 10
3

21) 8 -7 - n - 1 = 55 r
22) 2 +2=4
9

23) 2 1 + p - 9 = -21 24) 5 6 + a - 2 = 68

25) 3 -10b - 3 = 117 26) 5 + 4 b + 4 = 29

Page 15 of 23 Worksheet by Kuta Software LLC

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Goal: Solve absolute value equations by graphing a system of equation
This method is similar to solving systems of equations by substitution – but in reverse!

1. Let each side of the original equation be a separate equation equal to y. There are now two new
equations.
2. Graph the two new equations.
3. The solutions are where the two graphs intercept.

Example:
Original equation: |𝑥 + 3| = 2
Split it into the two new equations: 𝑦 = |𝑥 + 3| 𝑎𝑛𝑑 𝑦 = 2

Graph the two new equations:

Solutions are x ={-5, -1}

_______________________________________________________________________________________
To solve an absolute value equation as a system on a graphing calculator:
a. Graph each side of the equation.
b. Using calc /intersect find where the two graphs cross.
c. These are your answers.

Page 16 of 23
Algebra 1: Unit 6 Name___________________________________ ID: 1
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Goal: Solve by graphing a system Date________________ Period____

Solve each equation by graphing a system.
1) a + 3 = 2 2) m - 1 = 3

n 4) 3k = 9
3) =1
2

5) m - 3 = 1 6) r + 1 = 1

7) 1 + v = 2 8) 1 + x = 3

9) 3 b - 2 + 3 = 6 10) 3 - 4 n - 1 = -1

11) 4 + m + 1 = 5 12) 2 n + 2 + 3 = 5

Page 17 of 23 Worksheet by Kuta Software LLC

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NAME________________________________________________ PERIOD ____________
1) Solutions:__________________ 2) Solutions:__________________

3) Solutions:__________________ 4) Solutions:__________________

5) Solutions:__________________ 6) Solutions:__________________

Page 18 of 23
7) Solutions:__________________ 8) Solutions:__________________

9) Solutions:__________________ 10) Solutions:__________________

11) Solutions:__________________ 12) Solutions:__________________

Page 19 of 23
Regents Exam Questions A2.A.1: Absolute Value Equations 2 Name: ________________________
www.jmap.org

A2.A.1: Absolute Value Equations 2: Solve absolute value equations and inequalities involving
linear expressions in one variable

1 What is the solution set of the equation 2x  1  9?

2 What is the solution set of the equation 2x  1  9?

3 What is the solution set for the equation 3x  2  5?

4 What is the solution set of the equation 4x  3  17?

5 What is the solution set for the equation 3  2x  5?

6 What is the solution set of the equation 2  3x  5?

7 The solution set of the equation 2x  1  4  8

8 What is the solution set of the equation x  6   4  10?

9 What is the solution set for the equation 3x  1  x  5?

10 What is the solution set for the equation 2x  x   2?

11 What is the solution set of the equation 4a  6  4a  10?

Page 20 of 23
Regents Exam Questions A2.A.1: Absolute Value Equations 3 Name: ________________________
www.jmap.org

A2.A.1: Absolute Value Equations 3: Solve absolute value equations and inequalities involving
linear expressions in one variable

1 Solve for all values of x: |2x − 5| = 3

2 Solve for all values of x: |2x − 5| = 7

3 Solve for all values of x: |3x − 2| = 6

4 Solve for all values of x: |3x − 2| = 16

5 Find the solution set for |2x − 1| = 3.

6 What is the solution set of the equation |2x − 1| = 5?

7 Solve for all values of x: |2x + 3| = 7

8 Find all values of x that satisfy the equation |2x + 3| = 11?

9 Find the solution set of |2x + 1| = 9.

10 Find the complete solution set of |2x − 4| = 8

11 Solve for all values of x: |2x + 5| = 4

12 Solve for all values of x: |3x + 5| = 7

Page 21 of 23
Regents Exam Questions A2.A.1: Absolute Value Equations 3 Name: ________________________
www.jmap.org

13 What is the solution set of the equation |2x + 5| − 4 = 3?

14 Solve for x: |5 − 2x| = 7?

15 Solve for all values of x: |3x − 1| = 5

16 Solve for the negative value of x: |3x − 1| = 19

17 What is the negative value of x that satisfies the equation |3x + 1| = 8?

18 Find the negative root of the equation |3x − 2| = 4.

19 Find the negative member of the solution set for |2x − 4| = 6

20 What is the negative root of the equation |2x + 3| = 11?

21 Solve for the negative value of x: |2x + 5| + 1 = 13

22 Solve for the negative value of x: |2x − 3| + 1 = 17

23 Solve for the positive value of x: |2x − 3| = 11

Page 22 of 23
Integrated Algebra Practice: Absolute Value Equations #2 Page 1
www.jmap.org
NAME:_____________________________

P.I. A2.A.1: Solve absolute value equations involving 4. Use the problem solving strategy of Draw a
linear expressions in one variable Diagram to solve the following problem. The
ideal diameter of the inside opening of a pipe
1. When Spheres-R-Us ships bags of golf balls, is 3.85 cm. This opening can vary by at most
each bag must be within 7 balls of 600. 0.015 cm. Find the maximum and minimum
Which equation is an absolute value equation diameters of this inside opening.
representing the high and low values, and
what are those values?
[A] m −600 = 7; 593, 607

[B] 600− 7 = m; 7, 14 5. Use the problem solving strategy Guess and

Test to determine the values of a and b for
[C] m −7 = 600; 593, 607
which the equation x + a = b will have
[D] 600− m = 7; 7, 14 exactly one solution.

2. The ideal surface area of a rubber ball shipped 6. a. Use a graphing calculator to graph
from the factory is 54.25 square inches. The y = 4 x − 1 and y = 2 .
company inspector insists that the balls b. Use the CALC feature of the calculator to
shipped vary no more than 0.5 square inches identify the x-coordinates of the points where
of this total. Find the minimum and maximum the graphs intersect.
values for the surface area of rubber balls c. What absolute value equation have you
shipped. solved?

3. One poll reported the 48% of city residents 7. Use each of the symbols
were against building a new highway. The , x , 4, 2, − 1, − , and = once to write an
polling service stated that this poll was
absolute value equation. Then solve your
accurate to within 3%. Use an absolute value
equation.
equation to find the minimum and maximum
percents of city residents who are against
building a new highway.

8. Write a problem about a poll that is accurate

to within 3.5% that can be solved using an
absolute value equation. Then solve your
problem.

Page 23 of 23
Notes: