Equations, inequalities, the V-shaped graph, and transformations — with the AND/OR trick for inequalities.
|x| = x if x ≥ 0
|x| = −x if x < 0
Absolute value gives the distance from 0 on the number line. Distance is always non-negative.
|7| = 7
Already positive
|−7| = 7
−(−7) = 7
|0| = 0
Zero is its own absolute value
Rule: |A| = k
If k > 0: solve A = k AND A = −k (two solutions)
If k = 0: solve A = 0 (one solution)
If k < 0: no solution (|A| can never be negative)
Example 1: |2x − 3| = 7
Case 1: 2x − 3 = 7 → 2x = 10 → x = 5
Case 2: 2x − 3 = −7 → 2x = −4 → x = −2
Solutions: x = 5 or x = −2
Example 2: |x + 1| + 3 = 3
Isolate |·| first: |x + 1| = 0
k = 0 → only one case: x + 1 = 0
Solution: x = −1
Example 3: |3x − 1| = −4
k = −4 < 0 → absolute value cannot equal a negative number
No solution ∅
−k < A < k
Solution is BETWEEN −k and k
Graph: closed segment on number line
Example: |x − 2| ≤ 5
−5 ≤ x − 2 ≤ 5
−3 ≤ x ≤ 7
[−3, 7]
A < −k OR A > k
Solution is OUTSIDE −k and k
Graph: two rays going outward
Example: |2x + 1| > 3
2x + 1 < −3 OR 2x + 1 > 3
x < −2 OR x > 1
(−∞, −2) ∪ (1, ∞)
Memory Trick
Less than (<) → goes inward → AND. Greater than (>) → goes outward → OR. Think: less than = "and" is in the middle (−k < A < k has "A" sandwiched). Greater than = OR breaks apart.
| Parameter | Effect | Example |
|---|---|---|
| h | Horizontal shift (right h units) | y = |x − 3|: vertex at (3, 0) |
| k | Vertical shift (up k units) | y = |x| + 2: vertex at (0, 2) |
| a > 1 | Vertical stretch (narrower V) | y = 3|x|: steeper sides |
| 0 < a < 1 | Vertical compression (wider V) | y = 0.5|x|: flatter sides |
| a < 0 | Reflection over x-axis (upside-down V) | y = −|x|: opens downward |
Worked Example: Describe y = −2|x + 4| − 1
h = −4 (shift LEFT 4, since x − h = x − (−4) = x + 4)
k = −1 (shift DOWN 1)
a = −2 (reflect over x-axis, stretch by factor 2)
Vertex at (−4, −1). Opens downward. Sides have slope ±2.
Absolute value distributes over multiplication.
Absolute value distributes over division (b ≠ 0).
Triangle inequality — absolute value of sum ≤ sum of absolute values.
Always true. Squaring removes the absolute value sign.
Square root of a square always gives absolute value, not just a.
f(−x) = |−x| = |x| = f(x). Symmetric about the y-axis.
Split into two cases: 2x − 3 = 7 OR 2x − 3 = −7. Solve each: x = 5 or x = −2. Always check both solutions in the original equation to verify they are valid (watch for extraneous solutions when the absolute value equation is more complex).
For |x| < a (less than): the solution is AND — write −a < x < a. The values are between −a and a. For |x| > a (greater than): the solution is OR — write x < −a OR x > a. The values are outside the interval. Memory trick: Less than = AND (the 'and' is between the two signs); Greater than = OR (opens outward on both sides).
The graph of y = |x| is a V-shape with vertex at the origin (0,0). The left side has slope −1 and the right side has slope +1. It is symmetric about the y-axis (an even function). Transformations: y = a|x − h| + k shifts the vertex to (h, k), stretches by |a|, and reflects if a < 0.
Interactive problems with step-by-step solutions and private tutoring — free to try.
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