Absolute Value Functions

13 min read

Quick Reference

For lecture, the load-bearing ideas are the piecewise definition (the rule splits at the point where the inside expression is zero), the distance-from-zero meaning of the output, the V shape of the graph and where its corner sits, and the move from a graph to a solution set for an absolute value equation or inequality.

Before You Start: Prerequisite Check

Try This First

A number line runs left and right with zero in the middle. Pick the point at $x = -4$ and the point at $x = 6$. Ignore which side of zero each one is on, and ask only one question: how far is each point from zero?

Before any symbols, answer with the number line in mind.

• The point at $-4$ is a distance of how many units from zero?
• The point at $6$ is a distance of how many units from zero?
• Now do the same for a few more points: $-1$, $0$, $3$, $-3$. Write the distance next to each.

A Function That Reports Distance

A function is a process that takes an input and gives back an output. The absolute value function takes any number and gives back its distance from zero on the number line. That output is never negative, because a distance is never negative.

Watch how the output changes as the input moves. Start at a large negative input and slide it toward zero: the distance shrinks. The output decreases all the way down to $0$ at the input $0$. Keep sliding the input past zero into the positives, and the distance grows again. The output decreases, touches zero, then increases. That single change of direction at the input $0$ is the whole personality of this function, and it is why the graph has a corner instead of a smooth turn.

Prerequisite Hub

Builds on:

Unlocks:

Nothing later depends on this yet.

The Official Definition

Absolute value function:

The absolute value function is defined piecewise as $f(x) = |x|$, equal to $x$ if $x$ is greater than or equal to $0$, and equal to $-x$ if $x$ is less than $0$. It gives the distance of the input from zero on the number line, output without regard to sign.

Source: OpenStax Precalculus 2e, Section 1.6 Absolute Value Functions

Written as a piecewise rule, this is

$$f(x) = |x| = \begin{cases} x & \text{if } x \geq 0, \\ -x & \text{if } x < 0. \end{cases}$$

Read the second piece carefully. When $x$ is negative, the rule outputs $-x$, and $-x$ is positive because the negative of a negative number is positive. So $f(-4) = -(-4) = 4$. The $-x$ piece is not making the answer negative; it is flipping a negative input back up to its positive distance.

Two Representations of the Same Idea

Translate between these on purpose. Reading down the table, the outputs fall to $0$ and then rise again, which is exactly the V you see on the graph. The corner in the graph is the input where the piecewise rule switches from the $-x$ piece to the $x$ piece.

The V-Shaped Graph

The graph of $f(x) = |x|$ is two straight rays joined at a corner.

        y
        |
   \    |    /
    \   |   /
     \  |  /
      \ | /
       \|/
--------+--------  x
        0

• For inputs to the right of zero, $f(x) = x$, a line of slope $1$ rising to the right.
• For inputs to the left of zero, $f(x) = -x$, a line of slope $-1$ rising to the left.
• The two rays meet at the corner $(0, 0)$, the lowest point of the graph.

The general absolute value function $f(x) = a\lvert x - h \rvert + k$ is this same V moved and reshaped:

The corner always sits where the inside expression equals zero, that is, at $x = h$. Finding that input is the first move in graphing or solving anything with an absolute value.

Worked Examples

Example 1: Evaluate the Function (Mind the Negative Piece)

Let $f(x) = |x|$. Find $f(6)$, $f(0)$, and $f(-9)$.

Predict first. Each output is a distance from zero, so expect $6$, $0$, and $9$, all nonnegative.

Apply the piecewise rule.

$$f(6) = 6 \quad (\text{since } 6 \geq 0, \text{ use the } x \text{ piece})$$ $$f(0) = 0 \quad (\text{since } 0 \geq 0, \text{ use the } x \text{ piece})$$ $$f(-9) = -(-9) = 9 \quad (\text{since } -9 < 0, \text{ use the } -x \text{ piece})$$

The results $6$, $0$, and $9$ match the prediction. The third output came from the $-x$ piece, which turned the negative input back into its positive distance.

Example 2: Graph a Shifted Absolute Value Function

Graph $g(x) = \lvert x - 2 \rvert + 1$ by locating its corner and one point on each ray.

Find the corner. The inside $x - 2$ equals zero when $x = 2$. At that input, $g(2) = \lvert 0 \rvert + 1 = 1$. The corner is $(2, 1)$.

Build a short table around the corner.

The table is symmetric about $x = 2$: the values $2$ and $3$ appear on both sides. Plotting these points gives a V opening upward with its corner at $(2, 1)$, the basic V shifted right $2$ and up $1$. The slope is $+1$ to the right of the corner and $-1$ to the left.

Example 3: Solve an Absolute Value Equation

Solve $\lvert x - 5 \rvert = 3$.

Predict first. The equation asks which inputs sit a distance of $3$ from the point $5$ on the number line. Two such points exist, one on each side of $5$, so expect two solutions placed symmetrically around $5$.

Split into the two distance cases. A distance of $3$ from $5$ means the inside expression is either $3$ or $-3$:

$$x - 5 = 3 \quad \text{or} \quad x - 5 = -3.$$

Solve each:

$$x = 8 \quad \text{or} \quad x = 2.$$

Check against the prediction. The points $8$ and $2$ are each exactly $3$ units from $5$ ($8 - 5 = 3$ and $5 - 2 = 3$), matching the two symmetric solutions predicted. Verify in the original equation: $\lvert 8 - 5 \rvert = 3$ and $\lvert 2 - 5 \rvert = \lvert {-3} \rvert = 3$. Both hold.

Example 4: Solve an Absolute Value Inequality

Solve $\lvert x + 1 \rvert \leq 4$.

Read it as distance. The expression $\lvert x + 1 \rvert$ is the distance from $x$ to $-1$ (because $x + 1 = x - (-1)$). The inequality asks for every input within a distance of $4$ from $-1$.

One way to see this: a compound inequality. A distance of at most $4$ means the inside is between $-4$ and $4$:

$$-4 \leq x + 1 \leq 4.$$

Subtract $1$ from all three parts:

$$-5 \leq x \leq 3.$$

Another way to see this: read it off the number line. Start at $-1$ and step $4$ units in each direction. Stepping left reaches $-5$; stepping right reaches $3$. Every point in between is within distance $4$, so the solution is the closed interval from $-5$ to $3$.

Solution: $[-5, 3]$. Both routes agree. Which one do you prefer, and why? The compound inequality is reliable when the algebra is messy; the number-line reading is faster when the distance meaning is clear.

Common Misconceptions

Practice Problems

Mastery Checklist

Novice (Level 1-2):

• I can evaluate $f(x) = |x|$ at positive, negative, and zero inputs, and explain why the output is never negative.
• I can state the piecewise definition and say which piece applies to a given input.
• I can locate the corner of $y = a\lvert x - h \rvert + k$ by finding where the inside equals zero.

Competent (Level 3-4):

• I can graph a shifted absolute value function from its corner and a short table.
• I can solve an absolute value equation by splitting it into two distance cases.
• I can solve “less than” and “greater than” absolute value inequalities and write the result in interval notation.

Proficient (Level 5):

• I can recover a function $f(x) = a\lvert x - h \rvert + k$ from its graph using the corner and one extra point.
• I can convert between the absolute value form and the piecewise form of the same V.
• I can explain, using distance from a point, why an absolute value equation has two solutions and why each inequality type gives the interval shape it does.

Mental Model

The distance reporter with a turning point.

Picture a single sensor sitting at zero on the number line. Hand it any input, and it reports only the distance from itself to that input, never which side the input was on. That is why two opposite inputs share an output and why no output is ever negative.

The turning point is the second half of the picture. As the input slides in from far left, the reported distance shrinks to zero, then grows again as the input moves right. That single change of direction is the corner of the V. Shifting the sensor to the point $h$ and raising the floor to height $k$ moves the corner to $(h, k)$ without changing the V shape, which is exactly what the constants in $a\lvert x - h \rvert + k$ do.

Connections

Looking back:

• Absolute Value gives the distance-from-zero meaning that the function reports for every input.
• Transformation of Functions gives the shift, stretch, and reflection rules that turn the basic V into $a\lvert x - h \rvert + k$.

Looking ahead:

• The piecewise structure here is the first example of a function defined by cases, a pattern that returns whenever a rule changes behavior at a breakpoint.
• Reading solutions of $\lvert \text{expression} \rvert = c$ and $\lvert \text{expression} \rvert < c$ as distances prepares the distance reasoning used later for tolerance conditions and for the precise definition of a limit.

Real-world connections:

• A tolerance specification such as “the part must measure within $0.5$ mm of $20$ mm” is the inequality $\lvert \text{length} - 20 \rvert \leq 0.5$.
• The size of an error, regardless of whether a measurement came in high or low, is an absolute value of the difference from the target.

Resources

Last updated: 2026-06-16