Absolute Value Function

Absolute Value as Distance from Zero

The parent absolute value function maps each real number to its distance from zero on the number line. A distance cannot be negative, so $f(x)=|x|$ 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.

Definition of the Absolute Value Function

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

Components of absolute value functions:

• The symbol $|x|$ is read as absolute value of $x$.
• The function result is always non-negative ($|x| \geq 0$)
• This function is even: $|-x| = |x|$

Properties of Absolute Value Functions

Interpreting absolute value as distance gives the following properties:

Basic properties:

Triangle inequality properties:

Graphs of Absolute Value Functions

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

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

Transformations of Absolute Value Functions

Absolute value functions can be transformed in various ways. A vertical stretch or a vertical shift moves the whole V, while a reflection or a horizontal shift changes its orientation. Each change is applied to the parent function $f(x) = |x|$.

Vertical Translation

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

Horizontal Translation

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

Stretching and Compression

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

• If $|a| > 1$: the graph becomes steeper
• If $0 < |a| < 1$: the graph becomes gentler
• If $a < 0$: the graph is also reflected across the $x$-axis

The following comparison lists these changes:

General Form of Absolute Value Functions

The parameters in this form are:

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

Transformation table:

Absolute Value Equations and Inequalities

Solving absolute value equations:

To solve $|x| = a$ with $a \geq 0$:

Example: Solve $|x - 3| = 5$

Solving absolute value inequalities:

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

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

Exercises

Each problem gives an absolute-value equation or inequality, so split it into the two cases the absolute value allows before you solve.

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

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

3. Solve the inequality $|x - 2| < 4$

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

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

Worked Solutions

1. Calculating function values for various inputs:

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

   $$\begin{aligned} f(-1) &= |2(-1) - 6| \\ &= |-2 - 6| \\ &= |-8| \\ &= 8 \end{aligned}$$

   $$\begin{aligned} f(0) &= |2(0) - 6| \\ &= |0 - 6| \\ &= |-6| \\ &= 6 \end{aligned}$$

   $$\begin{aligned} f(3) &= |2(3) - 6| \\ &= |6 - 6| \\ &= |0| \\ &= 0 \end{aligned}$$

   $$\begin{aligned} f(5) &= |2(5) - 6| \\ &= |10 - 6| \\ &= |4| \\ &= 4 \end{aligned}$$

2. Solving absolute value equations:

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

   $$3x + 1 = 7 \quad \text{or} \quad 3x + 1 = -7$$

   $$3x = 6 \quad \text{or} \quad 3x = -8$$

   $$x = 2 \quad \text{or} \quad x = -\frac{8}{3}$$

3. Solving absolute value inequalities:

   For $|x - 2| < 4$, we use the property that $|a| < b$ is equivalent to $-b < a < b$:

   $$-4 < x - 2 < 4$$

   $$-4 + 2 < x < 4 + 2$$

   $$-2 < x < 6$$

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

4. Determining the vertex:

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

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

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

5. Calculating position based on distance:

   Given that the distance between cities $A$ and $B$ is $150 \text{ km}$, with city $A$ at coordinate $-50 \text{ km}$. Let city $B$ be at coordinate $x_B$:

   $$|-50 - x_B| = 150$$

   $$\begin{aligned} &-50 - x_B = 150 \\ &\text{or} \\ &-50 - x_B = -150 \end{aligned}$$

   $$\begin{aligned} &x_B = -50 - 150 = -200 \\ &\text{or} \\ &x_B = -50 + 150 = 100 \end{aligned}$$

   So city $B$ can be located at coordinate $100 \text{ km}$ or $-200 \text{ km}$.