Download Graphing absolute values on a numberline

SOLVING AND GRAPHING ABSOLUTE
VALUES ON A NUMBER LINE
Absolute Value Inequality Graph and Solution
THE GEOMETRIC DEFINTION OF ABS VALUE
The absolute value of a number measures its
distance to the origin on the real number line.
For example: 𝒙 =5 Translates into English: we
are looking for those real numbers x whose
distance from the origin is 5 units.
X=5 or x=-5
WHAT ABOUT THE SOLUTIONS TO INEQUALITIES
𝑥 <5
Translate into English: we are looking for those
real numbers x whose distance from the origin is
less than 5 units.
we are talking about values in the interval
between -5 and 5:
−𝟓 < 𝒙 < 𝟓
WHAT ABOUT THE SOLUTIONS TO INEQUALITIES
𝑥 ≥2
In English: which numbers, x, are at least 2 units
away from the origin?
On the left side, real numbers less than or equal
to -2 qualify, on the right all real numbers greater
than or equal to 2:
𝒙 ≤ −𝟐 𝑶𝑹 𝒙 ≥ 𝟐
QUICK SUMMERY
If
If
𝒙 < 𝒂 𝒕𝒉𝒆𝒏
-a<x<a
𝒙 > 𝒂 𝒕𝒉𝒆𝒏
x<-a OR x>a
LOCATOR POINT
Think of zero on the number line as the locator
point of absolute value.
 Do you remember how parent graphs shift
horizontally?
 Do you remember how the parent graph shrink
or stretch?
 The same concepts apply to graphing abs
values on a number line.

LET'S FIND THE SOLUTIONS TO THE INEQUALITY:
Let's find the solutions to the inequality:
𝒙−𝟐 ≤𝟏
 The locator point has been translated from the
zero to 2.
 So the equation above means any points that
are 1 unit or less away from 2.

1≤𝑥≤3
WHAT ABOUT THE EXAMPLE
𝑥+1 ≥3
The locator point has shifted 1 unit left. You are
looking for points that are 3 or more units away
from -1.
𝒙 ≤ −𝟒 𝑶𝑹 𝒙 ≥ 𝟐
WHAT ABOUT THE EXAMPLE
2𝑥 ≥ 6
 The locator point has not been shifted.
 The distance from zero has been reduced by a
factor of 2. So you are looking for points that
are more than 3 (i.e 6 ÷ 2) units away from
zero.

𝒙 ≤ −𝟑 𝑶𝑹 𝒙 ≥ 𝟑
WITH A LITTLE BIT OF TWEAKING, OUR METHOD
CAN ALSO HANDLE INEQUALITIES SUCH AS
𝟐𝒙 − 𝟓 < 𝟖
Shift 5 units right
Shrink the distance of 8 by a factor of 2
Start at 5 and move 4 units in each
direction: 4+5=9 and 5+-4=1
1<𝑥<9
TEST POINT
Choose a value of x between -1 and 9.
 Let’s say x=0
 𝟐(𝟎) − 𝟓 < 𝟖
 −𝟓 < 𝟖
 5<8 is true

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