Absolute Value Function- Graphing and Solving Equations
What Absolute Value Actually Means
Absolute value is the distance a number sits from zero on a number line. That's it. It doesn't care about direction, sign, or any of that. -5 and 5 both have an absolute value of 5 because both sit exactly five units from zero.
We write absolute value using those vertical bars: |x|. So |−3| = 3 and |7| = 7.
This simple concept becomes a function when you plug it into an equation. The absolute value function outputs the positive version of whatever you give it.
The Absolute Value Function Graph
The basic absolute value function is f(x) = |x|. When you graph this, you get a V shape that opens upward.
Key features of the graph:
- The vertex is at the origin (0, 0)
- The graph is symmetric across the y-axis
- The left arm slopes down at a 45-degree angle
- The right arm slopes up at a 45-degree angle
How to Graph |x| From Scratch
You only need two points to sketch this function:
- Pick x = 1, then f(1) = |1| = 1. Plot (1, 1)
- Pick x = -1, then f(-1) = |-1| = 1. Plot (-1, 1)
- Connect both points to (0, 0)
That's your V shape. It really is that simple.
Transformations Change Everything
Most absolute value problems won't give you a clean f(x) = |x|. You'll see things like f(x) = |x - 2| + 3. Here's how transformations work:
| Change | Effect on Graph | Example |
|---|---|---|
| |x - h| | Shifts right by h units | |x - 2| moves right 2 |
| |x| + k | Shifts up by k units | |x| + 3 moves up 3 |
| -|x| | Flips the V upside down | Opens downward |
| a|x| | Stretches or compresses vertically | 2|x| is steeper |
For f(x) = |x - 2| + 3, you take the basic V, slide it right 2 units, then lift it up 3 units. The vertex ends up at (2, 3).
Solving Absolute Value Equations
This is where most people get stuck. The trick is understanding what absolute value actually does.
When you see |x| = 5, you're really asking: "What numbers are exactly 5 units from zero?" Two numbers satisfy this — 5 and -5.
That's the core rule: If |x| = a (where a > 0), then x = a or x = -a
Step-by-Step: Solving |2x - 3| = 7
Follow this process every time:
- Isolate the absolute value expression first. In this case, it's already isolated.
- Split into two cases.
Case 1: 2x - 3 = 7
Case 2: 2x - 3 = -7 - Solve each equation separately.
Case 1: 2x = 10, so x = 5
Case 2: 2x = -4, so x = -2 - Check both answers.
|2(5) - 3| = |7| = 7 ✓
|2(-2) - 3| = |-7| = 7 ✓
Both solutions check out. Your answer is x = 5 or x = -2.
Solving |x + 4| = 3
This one's even simpler:
- x + 4 = 3 → x = -1
- x + 4 = -3 → x = -7
Check: |(-1) + 4| = |3| = 3 ✓ and |(-7) + 4| = |-3| = 3 ✓
Answer: x = -1 or x = -7
When the Right Side is Negative
Here's the part that trips up almost everyone. If you have |x| = -3, there is no solution.
Absolute value always outputs a positive number (or zero). It cannot equal a negative number. Period. End of discussion.
Same deal with |x - 5| < -2. That inequality has no solution because absolute values can't be negative.
Solving Absolute Value Inequalities
Inequalities require a different approach. You need to split them based on whether they're "less than" or "greater than" problems.
|x| < a (Less Than)
|x| < 4 means x is between -4 and 4.
Solution: -4 < x < 4
|x| > a (Greater Than)
|x| > 4 means x is either less than -4 or greater than 4.
Solution: x < -4 or x > 4
Example: |2x + 1| ≤ 7
Split into a compound inequality:
-7 ≤ 2x + 1 ≤ 7
Subtract 1: -8 ≤ 2x ≤ 6
Divide by 2: -4 ≤ x ≤ 3
Done. That's your solution set.
Common Mistakes to Avoid
- Forgetting to check both solutions. Always verify both answers in the original equation.
- Solving |x| = a as just x = a. You always need two equations.
- Dropping absolute value bars too early. Isolate them completely before splitting cases.
- Assuming a solution exists when it doesn't. Negative right side means no solution.
Quick Reference
| Equation Type | Solution Method | Example |
|---|---|---|
| |x| = a | x = a or x = -a | |x| = 5 → x = 5, -5 |
| |x| < a | -a < x < a | |x| < 3 → -3 < x < 3 |
| |x| > a | x < -a or x > a | |x| > 3 → x < -3 or x > 3 |
| |x| = 0 | x = 0 only | |x| = 0 → x = 0 |
| |x| = -a | No solution | |x| = -5 → ∅ |
Getting Started With Practice
You need to drill this until it becomes automatic. Here's a practice set:
- Solve: |x - 3| = 8
- Solve: |5x + 2| = 17
- Solve: |x + 1| < 5
- Graph: f(x) = |x + 2| - 4
- Solve: |3x - 7| > 2
Work through each one without looking at answers. Check your work. If you got any wrong, figure out exactly where your thinking went off track.
The goal isn't to memorize steps. It's to understand why the process works. Once that clicks, you'll solve these problems in seconds.