For AI agents: use /llms.txt for the Nakafa content index.
The parent absolute value function maps each real number to its distance from zero on the number line. A distance cannot be negative, so f ( x ) = ∣ x ∣ f(x)=|x| f ( x ) = ∣ x ∣ always has a non-negative output. Transformed absolute value functions can shift or reflect this parent graph, so their outputs need not remain non-negative.
For any real number x x x , the absolute value function is defined as:
Components of absolute value functions:
The symbol ∣ x ∣ |x| ∣ x ∣ is read as absolute value of x x x .
The function result is always non-negative (∣ x ∣ ≥ 0 |x| \geq 0 ∣ x ∣ ≥ 0 )
This function is even: ∣ − x ∣ = ∣ x ∣ |-x| = |x| ∣ − x ∣ = ∣ x ∣
Interpreting absolute value as distance gives the following properties:
Triangle inequality properties:
The following is a visualization of the basic absolute value function:
Value table for function f ( x ) = ∣ x ∣ f(x) = |x| f ( x ) = ∣ x ∣ :
x x x − 4 -4 − 4 − 3 -3 − 3 − 2 -2 − 2 − 1 -1 − 1 0 0 0 1 1 1 2 2 2 3 3 3 4 4 4 f ( x ) = ∣ x ∣ f(x) = \lvert x \rvert f ( x ) = ∣ x ∣ 4 4 4 3 3 3 2 2 2
Absolute value functions can be transformed in various ways:
The function f ( x ) = ∣ x ∣ + k f(x) = |x| + k f ( x ) = ∣ x ∣ + k shifts the graph upward (if k > 0 k > 0 k > 0 ) or downward (if k < 0 k < 0 k < 0 ).
The function f ( x ) = ∣ x − h ∣ f(x) = |x - h| f ( x ) = ∣ x − h ∣ shifts the graph to the right (if h > 0 h > 0 h > 0 ) or to the left (if h < 0 h < 0 h < 0 ).
The function f ( x ) = a ∣ x ∣ f(x) = a|x| f ( x ) = a ∣ x ∣ changes the slope of the graph:
If ∣ a ∣ > 1 |a| > 1 ∣ a ∣ > 1 : the graph becomes steeper
If 0 < ∣ a ∣ < 1 0 < |a| < 1 0 < ∣ a ∣ < 1 : the graph becomes gentler
If a < 0 a < 0 a < 0 : the graph is also reflected across the x x x -axis
The following comparison lists these changes:
The general form of an absolute value function is:
The parameters in this form are:
a a a : stretching/compression factor and reflection
h h h : horizontal translation
k k k : vertical translation
The vertex is located at ( h , k ) (h, k) ( h , k )
Parameter Value Effect on Graph ∣ a ∣ > 1 \lvert a \rvert > 1 ∣ a ∣ > 1 Magnitude > 1 > 1 > 1 Graph becomes steeper 0 < ∣ a ∣ < 1 0 < \lvert a \rvert < 1 0 < ∣ a ∣ < 1 Magnitude < 1 < 1 < 1 Graph becomes gentler a < 0 a < 0 a < 0 Negative Graph is also reflected across the x x x -axis h > 0 h > 0 h > 0 Positive Shift to the right h < 0 h < 0 h < 0 Negative Shift to the left k > 0 k > 0 k > 0 Positive Shift upward k < 0 k < 0 k < 0 Negative Shift downward
Solving absolute value equations:
To solve ∣ x ∣ = a |x| = a ∣ x ∣ = a with a ≥ 0 a \geq 0 a ≥ 0 :
Example: Solve ∣ x − 3 ∣ = 5 |x - 3| = 5 ∣ x − 3∣ = 5
Solving absolute value inequalities:
For ∣ x ∣ < a |x| < a ∣ x ∣ < a with a > 0 a > 0 a > 0 :
For ∣ x ∣ > a |x| > a ∣ x ∣ > a with a > 0 a > 0 a > 0 :
Determine the value of f ( x ) = ∣ 2 x − 6 ∣ f(x) = |2x - 6| f ( x ) = ∣2 x − 6∣ for x = − 1 , 0 , 3 , 5 x = -1, 0, 3, 5 x = − 1 , 0 , 3 , 5
Solve the equation ∣ 3 x + 1 ∣ = 7 |3x + 1| = 7 ∣3 x + 1∣ = 7
Solve the inequality ∣ x − 2 ∣ < 4 |x - 2| < 4 ∣ x − 2∣ < 4
Determine the vertex of the function f ( x ) = 2 ∣ x − 3 ∣ + 1 f(x) = 2|x - 3| + 1 f ( x ) = 2∣ x − 3∣ + 1
The distance between two cities is 150 km 150 \text{ km} 150 km . If city A A A is located at coordinate − 50 km -50 \text{ km} − 50 km , where is city B B B located?
Calculating function values for various inputs:
Substitute each value of x x x into the function f ( x ) = ∣ 2 x − 6 ∣ f(x) = |2x - 6| f ( x ) = ∣2 x − 6∣ :
Solving absolute value equations:
For the equation ∣ 3 x + 1 ∣ = 7 |3x + 1| = 7 ∣3 x + 1∣ = 7 , we use the definition of absolute value which produces two possibilities:
Solving absolute value inequalities:
For ∣ x − 2 ∣ < 4 |x - 2| < 4 ∣ x − 2∣ < 4 , we use the property that ∣ a ∣ < b |a| < b ∣ a ∣ < b is equivalent to − b < a < b -b < a < b − b < a < b :
Determining the vertex:
From the function f ( x ) = 2 ∣ x − 3 ∣ + 1 f(x) = 2|x - 3| + 1 f ( x ) = 2∣ x − 3∣ + 1 , we can identify the parameters:
a = 2 a = 2 a = 2 (stretching factor)
h = 3 h = 3 h = 3 (horizontal translation)
Calculating position based on distance:
Given that the distance between cities A A A and B B B is 150 km 150 \text{ km} 150 km , with city A A A at coordinate − 50 km -50 \text{ km} − 50 km . Let city B B B be at coordinate :
Published: May 18, 2025 . Updated: Sep 1, 2026 .
So the solution set is x ∈ ( − 2 , 6 ) x \in (-2, 6) x ∈ ( − 2 , 6 ) .
k = 1 k = 1 k = 1 (vertical translation)The vertex is located at ( h , k ) = ( 3 , 1 ) (h, k) = (3, 1) ( h , k ) = ( 3 , 1 ) .
So city B B B can be located at coordinate 100 km 100 \text{ km} 100 km or − 200 km -200 \text{ km} − 200 km .