ABSOLUTE VALUE
What is Absolute Value?
Absolute Value is the distance from zero on the real number line.
Absolute Value is never negative. It is not about the value of the number
but rather the distance from zero on the real number line, left or right.
Absolute Value only tells you how far, not in which direction. Absolute
Value is also referred to as the Modulus.
The absolute value of
some number “x” is the distance of “x” from zero. It is denoted as
│x│. The absolute value notation is bars ││.
Do not use parentheses ( ) or brackets [ ] because they do not mean the same thing.
EXAMPLES:
│0│= 0
│1│= 1
│2│= 2
│3│= 3
│4│= 4
│5│= 5
│9│= 9
│79│= 79
│319│= 319
│999│= 999
│1,000,000│= 1,000,000
│-1│= 1
│-2│= 2
│-3│= 3
│-4│= 4
│-5│= 5
│-9│= 9
│-25│= 25
│-51│= 51
│-409│= 409
│-901│= 901
│-1,000│= 1,000
WORKING WITH ABSOLUTE VALUES
Given: -│-9│= -(+9) = -9
First you find the absolute value of -9, and then you indicate the negative value outside of the bars. If no value is indicated outside of the bars, that indicates a positive value.
Notice: Given -(-9) = +9 negative x negative = positive. Bars and parentheses do not work the same.
The negative of any absolute value is a negative number.
*Remember Order of Operations – PEMDAS
Review the following simplifications:
-│-47│= -(+47) = -47
-│-76│= -(+76) = -76
-│-321│= -(+321) = -321
│7 – 4│= 3
│5 + 6│= 11
│10 – 6│= 4
│20 + 30│= 50
│0 – 5│= │-5│= 5
│4 – 9│= │-5│= 5
│1– 9│= │-8│= 8
│0 – 27│= │-27│= 27
│-1 + (-3)│= │-4│= 4
│-1 – (-3)│= │(-1) + 3│= │2│= 2
│-5 + (-3)│= │-8│= 8
│-5 – (-3)│= │(-5) + 3│= │-2│= 2
│5 + 3(-5)│= │5 + (-15)│= │-10│= 10
│-6 + 4(-12)│= │-6 + (-48)│= │-54│= 54
│52│= │25│= 25
│(-5)2│= │25│= 25
│-52│= │-25│= 25
-│52│= -│25│= -25
-│(-5)2│= -│25│= -25
-│-52│= -│-25│= -25
│5│2 = 52 = 25
│(-5)│2 = 52 = 25
│-5│2 = 52 = 25
-│5│2 = -(52) = -25
-│(-5)│2 = -(52) = -25
-│-5│2 = -(52) = -25
(-│-5│)2 = (-5)2 = 25
(-│-5│)3 = (-5)3 = -125
Note: When you multiply an even number of negative signs, the answer (product) is positive. Negative x Negative = Positive.
Note: When you multiply an odd number of negative signs, the answer (product) is negative. Negative x Negative x Negative = Negative.
INEQUALITIES
12 is less than 15. (12 < 15)
-12 is greater than -15. (-12 > -15)
│12│is less than │15│ (12 < 15)
│-12│is less than │-15│ (12 < 15)
-│12│is greater than -│15│ (-12 > -15)
Note: Equality │3 + 6│equals │9 - 18│ │9│equals │-9│ 9 = 9
VARIABLES
│m│= m
-│m│= -m
I cannot tell whether a variable is positive or negative from an absolute value. When I take the absolute value of any number the answer is always positive. So if I am asked what is the value of “m”, I cannot answer this question, but I can say the following:
If the variable “m” is greater than zero, then “m” is a positive number. The sign does not change when I take the absolute value.
If the variable “m” is less than zero, then “m” is a negative number. The sign changes from negative to positive when I take the absolute value.
If the variable “m” is zero, then “m” is neither positive nor negative, it is simply zero.
Given │m│ then m = m, if m ≥ 0 │7│ m = 7, because m ≥ 0
Given │m│ then m = -m, if m < 0 │-3│ m = (-3) , because m < 0
Both of these absolute values are positive. I am talking about the value of “m” inside the absolute value bars.
GRAPH OF THE ABSOLUTE VALUE FUNCTION
The Absolute Value Function is y = │x│.
For every value of “x”, y
is positive.
This graph is the shape
of a “V”, starting from the origin (0, 0) and going out (in both directions)
and up at 45 degree angles.
Whatever “x” is, “y” is the positive value of “x”.
SOLVING ABSOLUTE VALUE EQUATIONS
Example A: │x + 5│= 9
To solve this equation I must remove the absolute value bars and write two equations that take into account the possibility of “x + 5” being positive or negative.
(x + 5) = 9 or -(x + 5) = 9
x = 4 -x + -5 = 9
│4 + 5│= 9 -x = 14
x = -14
│-14 + 5│= │-9│= 9
The solutions to this equation are x = 4, -14. In other words, there are two solutions to this equation, 4 and -14.
Practice 1: │r – 17│= 23
To solve this equation I must remove the absolute value bars and write two equations that take into account the possibility of “r - 17” being positive or negative.
(r – 17) = 23 or -(r – 17) = 23
r = 23 + 17 = 40 -r + 17 = 23
│40 – 17│= 23 -r = 23 – 17 = 6
r = -6
│-6 – 17│= │-23│= 23
The solutions to this equation are r = 40, -6. In other words, there are two solutions to this equation, 40 and -6.
Example B: │3y – 4│– 8 = 6
To solve this equation I must first isolate the absolute value.
│3y – 4│= 6 + 8
Next I remove the absolute value bars and write two equations that take into account the possibility of “3y - 4” being positive or negative.
(3y – 4) = 14 or -(3y – 4) = 14
3y = 18 -3y + 4 = 14
y = 6 -3y = 10
│3(6) – 4│= 14 y = 10/-3
│18 – 4│= 14 │3(10/-3) – 4│= 14
│14│= 14 │-10 – 4│= 14
│-14│= 14
The solutions to this equation are y = 6, -10/3. In other words, there are two solutions to this equation, 6 and -10/3.
Practice 2: │7 – y│+ 2 = 6
To solve this equation I must first isolate the absolute value.
│7 – y│ = 4
Next I remove the absolute value bars and write two equations that take into account the possibility of “7 – y” being positive or negative.
(7 – y) = 4 or -(7 – y) = 4
-y = -3 -7 + y = 4
y = 3 y = 11
│7 – 3│+ 2 = 6 │7 – 11│+ 2 = 6
│4│+ 2 = 6 │-4│+ 2 = 6
4 + 2 = 6 4 + 2 = 6
The solutions to this equation are y = 3, 11. In other words, there are two solutions to this equation, 3 and 11.
SOLVING ABSOLUTE VALUE INEQUALITIES
Example C – Absolute Value Inequalities:
Given │a│ < b Therefore -b < a < b
│3x + 3│ < 15 Therefore -15 < 3x + 3 < 15
To solve this equation I must first isolate the variable “x”.
-15 < 3x + 3 3x + 3 < 15
-18 < 3x 3x < 12
-6 < x x < 4
The answers to this equation are all numbers falling between -6 and 4.
We can check this by
looking at the number line and choosing some values to plug into the original
equation.
│3(0) + 3│ < 15 │0 + 3│ < 15 │3│ < 15 Correct
│3(5) + 3│ < 15 │15 + 3│ < 15 │18│ < 15 Incorrect
│3(-7) + 3│ < 15 │-21 + 3│ < 15 │-18│ < 15 Incorrect
Example D – Absolute Value Inequalities:
Given │a│ > b Therefore a > b and a < -b
│3x + 3│ > 15 Therefore 3x + 3 > 15 and 3x + 3 < -15
To solve this equation I must first isolate the variable “x”.
3x + 3 > 15 3x + 3 < -15
3x > 12 3x < -18
x > 4 x < -6
The answers to this equation are all numbers less than -6 and greater than 4.
We can check this by
looking at the number line and choosing some values to plug into the original
equation.
│3(0) + 3│ > 15 │0 + 3│ > 15 │3│ > 15 Incorrect
│3(5) + 3│ > 15 │15 + 3│ > 15 │18│ > 15 Correct
│3(-10) + 3│ > 15 │-30 + 3│ > 15 │-27│ > 15 Correct
Practice 3 – Absolute Value Inequalities:
Given │a│ < b Therefore -b < a < b
│2x – 7│ < 9 Therefore -9 < 2x – 7 < 9
To solve this equation I must first isolate the variable “x”.
-9 < 2x – 7 2x – 7 < 9
-2 < 2x 2x < 16
-1 < x x < 8
The answers to this equation are all numbers falling between -1 and 8.
We can check this by looking at the number line and choosing some values to plug into the original equation.
│2(0) – 7│ < 9 │0 – 7│ < 9 │-7│ < 9 Correct
│2(10) – 7│ < 9 │20 – 7│ < 9 │13│ < 9 Incorrect
│2(-3) – 7│ < 9 │-6 – 7│ < 9 │-13│ < 9 Incorrect
Practice 4 – Absolute Value Inequalities:
Given │a│ > b Therefore a > b and a < -b
│2x – 7│ > 9 Therefore 2x – 7 > 9 and 2x – 7< -9
To solve this equation I must first isolate the variable “x”.
2x – 7 > 9 2x – 7< -9
2x > 16 2x < -2
x > 8 x < -1
The answers to this equation are all numbers less than -1 and greater than 8.
We can check this by
looking at the number line and choosing some values to plug into the original
equation.
│2(0) – 7│ > 9 │0 – 7│ > 9 │-7│ > 9 Incorrect
│2(-2) – 7│ > 9 │-4 – 7│ > 9 │-11│ > 9 Correct
│2(10) – 7│ > 9 │20 – 7│ > 9 │13│ > 9 Correct
Note: The
argument of the absolute value is whatever appears within the absolute
value bars.
Given │a│ = b, “a” is called the argument.
I have learned far too many math concepts to keep special case equations in my head. Instead I choose to understand concepts, as opposed to memorizing countless equations. When dealing with absolute value inequalities, I change the inequality to an equality and solve for the variable. Then I choose three values: one between the two solutions and two outside of each of the solutions. Then I know which way to set the inequalities. Let me show you how!
Example E – Absolute Value Inequalities:
│a│ < b I change this to an equality. │a│ = b
│2x – 7│ < 9 I change this to an equality. │2x – 7│ = 9
First I remove the absolute value bars and write two equations that take into account the possibility of “2x – 7” being positive or negative.
To solve these equations I must first isolate the variable “x”.
2x – 7 = 9 -(2x – 7) = 9
2x = 16 -2x + 7 = 9
x = 8 -2x = 2
x = -1
Now I look at the number
line and choose values inside and outside the range of numbers between -1 and
8.
I am choosing the numbers 0, -3 and 10. I then solve the original equation using these values.
│2x – 7│ < 9 │2(0) – 7│ < 9 │- 7│ < 9 7 < 9
This is correct. The answers to this inequality fall between the values of -1 and 8.
│2x – 7│ < 9 │2(-3) – 7│ < 9 │-6 – 7│ < 9 │-13│ < 9 13 < 9
This is not correct. So the answers do not fall outside of the range of numbers between -1 and 8.
│2x – 7│ < 9 │2(10) – 7│ < 9 │20 – 7│ < 9 │13│ < 9 13 < 9
This is not correct. So the answers do not fall outside of the range of numbers between -1 and 8.
Note: I did not use the values of “x” that I found by re-writing the equation as an equality (-1, 8) because using those values would be equal to 9, and therefore would not tell me in which direction the answers to the inequality are located.
Practice 5 – Absolute Value Inequalities:
│a│ > b I change this to an equality. │a│ = b
│2x – 7│ > 9 I change this to an equality. │2x – 7│ = 9
To solve this equation I must first isolate the variable “x”.
2x – 7 = 9 -(2x – 7) = 9
2x = 16 -2x + 7 = 9
x = 8 -2x = 2
x = -1
Now I look at the number
line and choose values inside and outside the range of numbers between -1 and
8.
I am choosing the numbers 0, -3 and 10. I then solve the original equation using these values.
│2x – 7│ > 9 │2(0) – 7│ > 9 │- 7│ > 9 7 > 9
This is not correct. The answers to this inequality do not fall between the numbers -1 and 8.
│2x – 7│ > 9 │2(-3) – 7│ > 9 │-6 – 7│ > 9 │-13│ > 9 13 > 9
This is correct. The answers to this inequality fall outside of the range of numbers between -1 and 8.
│2x – 7│ > 9 │2(10) – 7│ > 9 │20 – 7│ > 9 │13│ > 9 13 > 9
This is correct. The answers to this inequality fall outside of the range of numbers between -1 and 8.
Note: I did not use the values of “x” that I found by re-writing the equation as an equality (-1, 8) because using those values would be equal to 9, and therefore would not tell me in which direction the answers to the inequality are located.
GRAPHING ABSOLUTE VALUE EQUATIONS
The Absolute Value Function is y = │x│.
For every value of “x”, y is positive.
This graph is the shape of a “V”, starting from the origin (0, 0) and going out (in both directions) and up at 45 degree angles. Whatever “x” is, “y” is the positive value of “x”. This graph is often referred to as the parent graph (in other words the basic/generic graph).
For the absolute value function y = -│x│,
For every value of “x”, y is negative.
This graph is the shape of a “V”, starting from the origin (0, 0) and going out (in both directions) and down at 45 degree angles. Whatever “x” is, “y” is the negative value of “x”. The parent graph is flipped.
For the absolute value function y = │x│ + 3,
You move the absolute value function graph y = │x│ (the parent graph) up the y-axis three units. The +3 tells you the vertex of the graph. When x = 0, y = 3.
If you were given the graph above and asked to write an equation, normally you are also given the parent graph y = │x│. You know the equation of the parent graph. Now you need to indicate the change in position up the y-axis. That number is added outside the second absolute value bar. y = │x│ + 3
For the absolute value function y = │x│ – 5,
You move the absolute value function graph y = │x│ (the parent graph) down the y-axis five units. The -5 tells you the vertex of the graph. When x = 0, y = -5.
If you were given the graph above and asked to write an equation, normally you are also given the parent graph y = │x│. You know the equation of the parent graph. Now you need to indicate the change in position down the y-axis. That number is added outside the second absolute value bar. y = │x│ + (-5) = y = │x│ – 5
For the absolute value function y = │x + 4│,
You move the absolute
value function graph y = │x│ (the parent graph) left on the x-axis four
units. The value of x inside the absolute value bars must be -4 for the absolute
value term to equal zero. This is the vertex of the graph. When x =
-4, y = 0.
If you were given the graph above and asked to write an equation, normally you are also given the parent graph y = │x│. You know the equation of the parent graph. Now you need to indicate the change in position left on the x-axis. Determine what number needs to be added to the location on the x-axis for the absolute value to equal zero. Since “x” is -4, +4 must be added to “x”. You add the +4 inside the absolute value bars. y = │x + 4│
For the absolute value function y = │x – 3│,
You move the absolute
value function graph y = │x│ (the parent graph) right on the x-axis three
units. The value of x inside the absolute value bars must be 3 for the
absolute value term to equal zero. This is the vertex of the graph.
When x = 3, y = 0.
If you were given the graph above and asked to write an equation, normally you are also given the parent graph y = │x│. You know the equation of the parent graph. Now you need to indicate the change in position right on the x-axis. Determine what number needs to be added to the location on the x-axis for the absolute value to equal zero. Since “x” is 3, -3 must be added to “x”. You add the -3 inside the absolute value bars. y = │x – 3│
For the absolute value function y = -│x + 3│ + 4,
You flip the absolute
value function graph y = │x│(the parent graph) because of the negative sign in
front of the absolute value bars, move left on the x-axis three units, and up
four units. The value of x inside the absolute value bars must be -3 for
the absolute value term to equal zero. The +4 tell you the y-coordinate
of the vertex. When x = -3, y = 4. This is the vertex of the graph
(-3, 4).
If you were given the graph above and asked to write an equation, normally you are also given the parent graph y = │x│. You know the equation of the parent graph. First indicate the flipped graph with a negative sign in front of the absolute value bars. Now you need to indicate the change in position of the vertex up the y-axis by adding this value outside the second absolute value bar. y = -│x│ + 4
Now you need to indicate
the change in position of the vertex left on the x-axis. Determine what
number needs to be added to the location on the x-axis for the absolute value
to equal zero. Since “x” is -3, +3 must be added to “x”. You add
the +3 inside the absolute value bars.
y = -│x + 3│ + 4
For the absolute value functions y = 2│x│ or y = │2x│
The absolute value function graph y = │x│ (the purple graph) becomes steeper (thinner) (the green graph). The larger the number, the thinner the graph. To accurately graph these equations, use a negative and a positive value for “x” and solve for “y” to get two sets of coordinate points. You only need two points to draw a straight line. You already know the vertex point (0, 0).
If you were given the
graph above and asked to write an equation, normally you are given the parent
graph y = │x│ (the purple graph). You know the equation of the parent
graph. Now you need to determine how steep is the green graph. That
number goes in front of the “x” or outside the first absolute value bar.
The right green line goes through the point (1, 2), so the slope of this line
is 2/1, which equals 2!
y = 2│x│ or y = │2x│
For the absolute value functions y = ½│x│ or y = │(½)x│
The absolute value
function graph y = │x│ becomes less steep (wider). The smaller the number
(the fraction), the wider the graph. To accurately graph these equations,
use a negative and a positive value for “x” and solve for “y” to get two sets
of coordinate points. You only need two points to draw a straight line.
You already know the vertex point (0, 0).
If you were given the graph above and asked to write an equation, normally you are given the parent graph y = │x│ (the purple graph). You know the equation of the parent graph. Now you need to determine how steep is the green graph. That number goes in front of the “x” or outside the first absolute value bar. The left green line goes through the point (-2, 1), so the slope of this line is -1/2. Be careful here. The negative would indicate the parent graph was flipped. The slope is 1/2 not -1/2! y = ½│x│ or y = │(½)x│
For the absolute value functions y = │x│2 or y = │x2│
We cannot graph this by adjusting the graph of y = │x│.
This graph is a parabola. This is now a quadratic equation.
The standard form of a quadratic equation is ax2 + bx + c = 0, if a ≠ 0.
If a = 0, then the equation becomes a linear equation bx + c = 0.
To learn how to
graph parabolas, please listen to the lesson on quadratic equations.
Hopefully I was successful in teaching you to work with Absolute Values.
Please go to the website blog and let me know.
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