Graphing Piecewise Functions with Absolute Values

What You're Actually Working With

Piecewise functions with absolute values sound complicated, but they're just regular piecewise functions wearing a disguise. The absolute value symbols force parts of the function to bounce above the x-axis, which changes how the graph looks.

If you can't graph basic piecewise functions yet, fix that first. This article assumes you know what piecewise notation means and can plot simple linear functions.

The Core Idea You Need to Get

Absolute value always outputs a non-negative number. |x| equals x when x ≥ 0, and -x when x < 0. That's it. That's the whole concept.

When absolute value meets piecewise notation, you're usually dealing with one of two scenarios:

• The absolute value wraps around an expression that changes sign within your domain
• You need to break the function into cases to remove the absolute value properly

How to Rewrite These Functions (The Method That Actually Works)

Here's the process you should follow every time:

1. Find where the expression inside the absolute value equals zero. This is your boundary point.
2. Split your piecewise function at that boundary. Everything below the zero point gets one treatment, everything above gets another.
3. Apply the definition: If |expression| = expression when expression ≥ 0, and |expression| = -expression when expression < 0.
4. Graph each piece on its assigned interval.

A Real Example That Makes This Clear

Let's graph f(x) = |x - 2| + 3

Step 1: The expression inside absolute value is x - 2. Set it equal to zero: x - 2 = 0, so x = 2.

Step 2: Break into two cases.

Case 1: When x - 2 ≥ 0 (meaning x ≥ 2), |x - 2| = x - 2

So f(x) = (x - 2) + 3 = x + 1

Case 2: When x - 2 < 0 (meaning x < 2), |x - 2| = -(x - 2) = -x + 2

So f(x) = (-x + 2) + 3 = -x + 5

The rewritten function is:

f(x) = x + 1 when x ≥ 2
f(x) = -x + 5 when x < 2

Graph the line y = x + 1 starting at x = 2, and y = -x + 5 to the left of x = 2. The vertex lands at (2, 3). You'll get a V-shape that opens upward.

Comparing Your Options for Solving These

Common Mistakes That Will Sink You

Forgetting the negative sign when rewriting. When x - 2 < 0, |x - 2| = -(x - 2). Students often write x - 2 instead of -x + 2. This is the #1 error.

Using the wrong inequality direction. When x - 2 ≥ 0, the expression equals itself. When x - 2 < 0, it equals the negative. The sign flips.

Not checking the boundary point. At x = 2 in our example, both pieces give f(2) = 3. The function is continuous there. Always verify your boundary values match.

Graphing the wrong region. If your piece is defined for x ≥ 2, you graph that piece starting at x = 2 and extending right. Not left.

A Messier Example

Graph f(x) = |x² - 4|

Set x² - 4 = 0. This gives x² = 4, so x = -2 and x = 2.

You now have three regions: x < -2, -2 ≤ x < 2, and x ≥ 2.

Region 1 (x < -2): x² - 4 > 0, so |x² - 4| = x² - 4

Region 2 (-2 ≤ x < 2): x² - 4 < 0, so |x² - 4| = -(x² - 4) = 4 - x²

Region 3 (x ≥ 2): x² - 4 > 0, so |x² - 4| = x² - 4

This gives you a graph that bounces off the x-axis at x = -2 and x = 2, with the middle section flipping below the axis to above it. You get a W-shape with zeros at those points.

Quick Reference: Signs to Watch For

• |expression| where expression ≥ 0 → use expression
• |expression| where expression < 0 → use -(expression)
• Multiple absolute values → multiple boundary points
• Nested absolute values → work from inside out

Getting Started: Practice Problem

Try this one on your own before looking at the answer:

Graph f(x) = |x + 1| - 2

Once you can do this without breaking a sweat, you're ready for anything with absolute values and piecewise notation.