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Functions and Their Modeling

Absolute Value Function

Understanding Absolute Value Functions

An absolute value function is a function that produces positive or zero values from any input, regardless of the original sign of the input. Geometrically, absolute value can be understood as the distance of a number from the zero point on the number line.

Mathematical Definition

For any real number xxx, the absolute value function is defined as:

f(x)=∣x∣={x,if x≥0−x,if x<0f(x) = |x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}f(x)=∣x∣={x,−x,​if x≥0if x<0​

Components of absolute value functions:

  • The symbol ∣x∣|x|∣x∣ is read as "absolute value of 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∣

Properties of Absolute Value Functions

Absolute value functions have several important properties that need to be understood:

Basic properties:

∣x∣≥0 for all x∈R|x| \geq 0 \text{ for all } x \in \mathbb{R}∣x∣≥0 for all x∈R
∣−x∣=∣x∣|-x| = |x|∣−x∣=∣x∣
∣x∣2=x2|x|^2 = x^2∣x∣2=x2
∣xy∣=∣x∣⋅∣y∣|xy| = |x| \cdot |y|∣xy∣=∣x∣⋅∣y∣
∣xy∣=∣x∣∣y∣, with y≠0\left|\frac{x}{y}\right| = \frac{|x|}{|y|}, \text{ with } y \neq 0​yx​​=∣y∣∣x∣​, with y=0

Triangle inequality properties:

∣x+y∣≤∣x∣+∣y∣|x + y| \leq |x| + |y|∣x+y∣≤∣x∣+∣y∣
∣∣x∣−∣y∣∣≤∣x−y∣||x| - |y|| \leq |x - y|∣∣x∣−∣y∣∣≤∣x−y∣

Graphs of Absolute Value Functions

The following is a visualization of the basic absolute value function:

Graph of Function f(x)=∣x∣f(x) = |x|f(x)=∣x∣
The graph shows the characteristic shape of an absolute value function that forms the letter V.

Value table for function f(x)=∣x∣f(x) = |x|f(x)=∣x∣:

xxx−4-4−4−3-3−3−2-2−2−1-1−1000111222333444
f(x)=xf(x) = x f(x)=xxxxxxxxxxxxxxxxxxxxxxxxxxxx

Transformations of Absolute Value Functions

Absolute value functions can be transformed in various ways:

Vertical Translation

The function f(x)=∣x∣+kf(x) = |x| + kf(x)=∣x∣+k shifts the graph upward (if k>0k > 0k>0) or downward (if k<0k < 0k<0).

Vertical Translation
Comparison of f(x)=∣x∣f(x) = |x|f(x)=∣x∣ with g(x)=∣x∣+2g(x) = |x| + 2g(x)=∣x∣+2 and h(x)=∣x∣−2h(x) = |x| - 2h(x)=∣x∣−2.

Horizontal Translation

The function f(x)=∣x−h∣f(x) = |x - h|f(x)=∣x−h∣ shifts the graph to the right (if h>0h > 0h>0) or to the left (if h<0h < 0h<0).

Horizontal Translation
Comparison of f(x)=∣x∣f(x) = |x|f(x)=∣x∣ with g(x)=∣x−3∣g(x) = |x - 3|g(x)=∣x−3∣ and h(x)=∣x+3∣h(x) = |x + 3|h(x)=∣x+3∣.

Stretching and Compression

The function f(x)=a∣x∣f(x) = a|x|f(x)=a∣x∣ changes the slope of the graph:

  • If a>1a > 1a>1: the graph becomes steeper
  • If 0<a<10 < a < 10<a<1: the graph becomes gentler
  • If a<0a < 0a<0: the graph is inverted (reflection across the x-axis)

To make it easier to understand, let's look at the following example:

Stretching and Compression
Comparison of f(x)=∣x∣f(x) = |x|f(x)=∣x∣ with g(x)=2∣x∣g(x) = 2|x|g(x)=2∣x∣ and h(x)=0.5∣x∣h(x) = 0.5|x|h(x)=0.5∣x∣.

General Form of Absolute Value Functions

The general form of an absolute value function is:

f(x)=a∣x−h∣+kf(x) = a|x - h| + kf(x)=a∣x−h∣+k

where:

  • aaa: stretching/compression factor and reflection
  • hhh: horizontal translation
  • kkk: vertical translation
  • The vertex is located at (h,k)(h, k)(h,k)

Transformation table:

ParameterValueEffect on Graph
a>1a > 1a>1Positive > 1Graph becomes steeper
0<a<10 < a < 10<a<1Positive < 1Graph becomes gentler
a<0a < 0a<0NegativeGraph is inverted
h>0h > 0h>0PositiveShift to the right
h<0h < 0h<0NegativeShift to the left
k>0k > 0k>0PositiveShift upward
k<0k < 0k<0NegativeShift downward

Absolute Value Equations and Inequalities

Solving absolute value equations:

To solve ∣x∣=a|x| = a∣x∣=a with a≥0a \geq 0a≥0:

∣x∣=a⇒x=a or x=−a|x| = a \Rightarrow x = a \text{ or } x = -a∣x∣=a⇒x=a or x=−a

Example: Solve ∣x−3∣=5|x - 3| = 5∣x−3∣=5

∣x−3∣=5|x - 3| = 5∣x−3∣=5
x−3=5 or x−3=−5x - 3 = 5 \text{ or } x - 3 = -5x−3=5 or x−3=−5
x=8 or x=−2x = 8 \text{ or } x = -2x=8 or x=−2

Solving absolute value inequalities:

For ∣x∣<a|x| < a∣x∣<a with a>0a > 0a>0:

∣x∣<a⇒−a<x<a|x| < a \Rightarrow -a < x < a∣x∣<a⇒−a<x<a

For ∣x∣>a|x| > a∣x∣>a with a>0a > 0a>0:

∣x∣>a⇒x<−a or x>a|x| > a \Rightarrow x < -a \text{ or } x > a∣x∣>a⇒x<−a or x>a

Exercises

  1. Determine the value of f(x)=∣2x−6∣f(x) = |2x - 6|f(x)=∣2x−6∣ for x=−1,0,3,5x = -1, 0, 3, 5x=−1,0,3,5

  2. Solve the equation ∣3x+1∣=7|3x + 1| = 7∣3x+1∣=7

  3. Solve the inequality ∣x−2∣<4|x - 2| < 4∣x−2∣<4

  4. Determine the vertex of the function f(x)=2∣x−3∣+1f(x) = 2|x - 3| + 1f(x)=2∣x−3∣+1

  5. The distance between two cities is 150 km. If city A is located at coordinate -50 km, where is city B located?

Answer Key

  1. Calculating function values for various inputs:

    Substitute each value of x into the function f(x)=∣2x−6∣f(x) = |2x - 6|f(x)=∣2x−6∣:

    f(−1)=∣2(−1)−6∣=∣−2−6∣=∣−8∣=8f(-1) = |2(-1) - 6| = |-2 - 6| = |-8| = 8f(−1)=∣2(−1)−6∣=∣−2−6∣=∣−8∣=8
    f(0)=∣2(0)−6∣=∣0−6∣=∣−6∣=6f(0) = |2(0) - 6| = |0 - 6| = |-6| = 6f(0)=∣2(0)−6∣=∣0−6∣=∣−6∣=6
    f(3)=∣2(3)−6∣=∣6−6∣=∣0∣=0f(3) = |2(3) - 6| = |6 - 6| = |0| = 0f(3)=∣2(3)−6∣=∣6−6∣=∣0∣=0
    f(5)=∣2(5)−6∣=∣10−6∣=∣4∣=4f(5) = |2(5) - 6| = |10 - 6| = |4| = 4f(5)=∣2(5)−6∣=∣10−6∣=∣4∣=4

  2. Solving absolute value equations:

    For the equation ∣3x+1∣=7|3x + 1| = 7∣3x+1∣=7, we use the definition of absolute value which produces two possibilities:

    3x+1=7or3x+1=−73x + 1 = 7 \quad \text{or} \quad 3x + 1 = -73x+1=7or3x+1=−7
    3x=6or3x=−83x = 6 \quad \text{or} \quad 3x = -83x=6or3x=−8
    x=2orx=−83x = 2 \quad \text{or} \quad x = -\frac{8}{3}x=2orx=−38​

  3. 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:

    −4<x−2<4-4 < x - 2 < 4−4<x−2<4
    −4+2<x<4+2-4 + 2 < x < 4 + 2−4+2<x<4+2
    −2<x<6-2 < x < 6−2<x<6

    So the solution set is x∈(−2,6)x \in (-2, 6)x∈(−2,6).

  4. Determining the vertex:

    From the function f(x)=2∣x−3∣+1f(x) = 2|x - 3| + 1f(x)=2∣x−3∣+1, we can identify the parameters:

    • a=2a = 2a=2 (stretching factor)
    • h=3h = 3h=3 (horizontal translation)
    • k=1k = 1k=1 (vertical translation)

    The vertex is located at (h,k)=(3,1)(h, k) = (3, 1)(h,k)=(3,1).

  5. Calculating position based on distance:

    Given that the distance between cities A and B is 150 km, with city A at coordinate -50 km. Let city B be at coordinate xBx_BxB​:

    ∣−50−xB∣=150|-50 - x_B| = 150∣−50−xB​∣=150
    −50−xB=150or−50−xB=−150-50 - x_B = 150 \quad \text{or} \quad -50 - x_B = -150−50−xB​=150or−50−xB​=−150
    xB=−50−150=−200orxB=−50+150=100x_B = -50 - 150 = -200 \quad \text{or} \quad x_B = -50 + 150 = 100xB​=−50−150=−200orxB​=−50+150=100

    So city B can be located at coordinate 100 km or -200 km.

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  • Absolute Value FunctionMaster absolute value functions with interactive graphs, transformations, and step-by-step solutions. Learn properties, equations, and real applications.
On this page
  • Understanding Absolute Value Functions
    • Mathematical Definition
  • Properties of Absolute Value Functions
  • Graphs of Absolute Value Functions
  • Transformations of Absolute Value Functions
    • Vertical Translation
    • Horizontal Translation
    • Stretching and Compression
  • General Form of Absolute Value Functions
  • Absolute Value Equations and Inequalities
  • Exercises
    • Answer Key
  • Comments
  • Report
  • Source code