1.6Absolute Value Functions
Graph f(x) = |x| and its transformations, write piecewise definitions, and solve absolute value equations and inequalities.
Absolute value functions model distance and error. They appear in optimization problems, error analysis, and piecewise definitions — and solving absolute value equations is a key algebraic skill for standardized tests.
Essential Question
Why does the absolute value function produce a V-shaped graph, and how does its piecewise definition help us solve absolute value equations?
Lesson Overview
The absolute value of a number is its distance from zero on the number line — always non-negative. The parent function f(x) = |x| produces a V-shaped graph with vertex at the origin, slopes of +1 (right arm) and −1 (left arm), domain (−∞, ∞), and range [0, ∞). Its piecewise definition is f(x) = x if x ≥ 0, and f(x) = −x if x < 0. The general transformed form is g(x) = a|x − h| + k: h shifts horizontally, k shifts vertically, |a| stretches or compresses, and a < 0 reflects over the x-axis (opening downward). To solve |expression| = c (c ≥ 0), split into two cases: expression = c and expression = −c. To solve |expression| < c, write −c < expression < c. To solve |expression| > c, write expression < −c OR expression > c.
General Form
g(x) = a|x − h| + k
- Vertex: (h, k)
- h: horizontal shift (right if h > 0)
- k: vertical shift (up if k > 0)
- |a| > 1: steeper arms (stretch)
- 0 < |a| < 1: shallower arms (compression)
- a < 0: opens downward (∧ shape)
Solving Absolute Value Equations
|A| = c (c ≥ 0)
A = c OR A = −c
|A| < c
−c < A < c
(AND — between)
|A| > c
A < −c OR A > c
(OR — outside)
Parent Function: f(x) = |x|
Transformed: f(x) = |x| vs g(x) = 2|x − 2| − 1
Shift right 2
|x − 2|: vertex moves from x=0 to x=2.
Vertical stretch ×2
Coefficient 2: arms are steeper (slope ±2 instead of ±1).
Shift down 1
−1 outside: vertex moves from y=0 to y=−1.
Piecewise Definition of |x − h| + k
General Piecewise Form
f(x) = |x − h| + k
= (x − h) + k, if x ≥ h
= −(x − h) + k, if x < h
The vertex is at x = h. For x ≥ h, use the positive branch; for x < h, use the negative branch.
Example: f(x) = |x − 3| + 1
f(x) =
(x − 3) + 1 = x − 2, x ≥ 3
−(x − 3) + 1 = −x + 4, x < 3
Vertex at (3, 1). Check: f(3) = |3−3|+1 = 1. ✓
Worked Examples
Graph g(x) = 2|x − 3| − 4 and identify the vertex, domain, and range.
Compare to a|x − h| + k: a = 2, h = 3, k = −4.
Vertex: (h, k) = (3, −4).
a = 2 > 0: opens upward. Arms have slope ±2.
Plot vertex (3, −4). Right arm: slope +2 → (4, −2), (5, 2). Left arm: slope −2 → (2, −2), (1, 2).
Domain: (−∞, ∞). Range: [−4, ∞) since vertex is the minimum.
Write the piecewise definition for f(x) = |2x − 6| + 1.
Factor inside: |2x − 6| = |2(x − 3)| = 2|x − 3|. Vertex at x = 3.
For x ≥ 3: 2x − 6 ≥ 0, so |2x − 6| = 2x − 6. f(x) = (2x − 6) + 1 = 2x − 5.
For x < 3: 2x − 6 < 0, so |2x − 6| = −(2x − 6) = −2x + 6. f(x) = (−2x + 6) + 1 = −2x + 7.
Solve |3x − 9| = 12.
Set up two cases: 3x − 9 = 12 OR 3x − 9 = −12.
Case 1: 3x − 9 = 12 → 3x = 21 → x = 7.
Case 2: 3x − 9 = −12 → 3x = −3 → x = −1.
Check: |3(7)−9| = |12| = 12 ✓. |3(−1)−9| = |−12| = 12 ✓.
Solve |2x + 1| < 7.
|A| < c means −c < A < c. So: −7 < 2x + 1 < 7.
Subtract 1 from all parts: −8 < 2x < 6.
Divide by 2: −4 < x < 3.
Solve |x − 5| ≥ 3.
|A| ≥ c means A ≤ −c OR A ≥ c. So: x − 5 ≤ −3 OR x − 5 ≥ 3.
Case 1: x − 5 ≤ −3 → x ≤ 2.
Case 2: x − 5 ≥ 3 → x ≥ 8.
Guided Practice
Identify the vertex, direction of opening, and range of g(x) = −3|x + 2| + 5.
Hint: a = −3 (negative → opens downward). h = −2 (x + 2 means h = −2). k = 5. Vertex is (h, k). Range: since it opens down, the vertex is the maximum.
Write the piecewise definition for f(x) = |x + 4| − 2.
Hint: Vertex at x = −4. For x ≥ −4, |x+4| = x+4. For x < −4, |x+4| = −(x+4). Add −2 to each branch.
Solve |4x − 8| = 20.
Hint: Set up two cases: 4x − 8 = 20 and 4x − 8 = −20. Solve each for x.
Solve |3x + 6| ≤ 9.
Hint: |A| ≤ c means −c ≤ A ≤ c. Write −9 ≤ 3x + 6 ≤ 9, then solve the compound inequality.
Write the equation of an absolute value function with vertex (−1, 4) that opens downward with slope magnitude 2.
Hint: Use g(x) = a|x − h| + k. Vertex gives h = −1, k = 4. Opens downward means a < 0. Slope magnitude 2 means |a| = 2.
Key Vocabulary
Absolute Value
The distance of a number from zero. |x| ≥ 0 always. |x| = x if x ≥ 0; |x| = −x if x < 0.
V-Shape / Vertex
The characteristic shape of f(x) = |x|. The vertex is the turning point — the minimum (opens up) or maximum (opens down) of the graph.
Piecewise Definition
Writing |x − h| as two linear pieces: (x − h) for x ≥ h and −(x − h) for x < h. Removes the absolute value symbol.
General Form g(x) = a|x−h|+k
a: stretch/reflection. h: horizontal shift (vertex x-coord). k: vertical shift (vertex y-coord). Vertex is always (h, k).
Absolute Value Equation
|A| = c splits into A = c OR A = −c (two solutions when c > 0, one when c = 0, none when c < 0).
Absolute Value Inequality
|A| < c gives −c < A < c (AND). |A| > c gives A < −c OR A > c (OR). Think: less than = between; greater than = outside.
Interactive Practice — 5 Questions
What is the vertex of g(x) = |x − 5| + 3?
Which piecewise definition matches f(x) = |x − 2|?
Solve |2x − 4| = 10.
What is the range of g(x) = −2|x + 1| + 6?
The solution to |x − 3| < 5 is:
Independent Practice
Independent Practice
Identify the vertex, direction of opening, and range of g(x) = 4|x − 1| − 7.
Write the piecewise definition for f(x) = |3x − 9| + 2.
Solve |5x + 10| = 25. Check both solutions.
Solve |2x − 6| > 8. Write the solution in interval notation.
Write the equation of an absolute value function with vertex (3, −2), opening upward, with slope magnitude 1/2.
Common Mistakes
Solving |A| = c as only one equation: A = c, missing the second case A = −c.
|A| = c always produces TWO equations (when c > 0): A = c AND A = −c. Both must be solved.
Writing |A| > c as −c < A < c (using the 'between' rule for 'greater than').
|A| > c means A < −c OR A > c (outside). |A| < c means −c < A < c (between). Less than = between; greater than = outside.
Confusing the vertex: thinking g(x) = |x + 3| has vertex at (3, 0).
g(x) = |x − h| + k has vertex at (h, k). Since x + 3 = x − (−3), h = −3. Vertex is (−3, 0).
Forgetting that |A| = c has NO solution when c < 0.
Absolute value is always ≥ 0. If c < 0, |A| = c has no solution. If c = 0, there is exactly one solution.
Math Tips
Vertex of g(x) = a|x − h| + k is always (h, k). The sign of a tells you direction: a > 0 opens up (∨), a < 0 opens down (∧).
Memory trick for inequalities: "Less than" absolute value → AND (between). "Greater than" absolute value → OR (outside). Think: |x| < 3 is a short interval; |x| > 3 is two rays going out.
To write a piecewise definition: find the vertex x = h. For x ≥ h, drop the absolute value. For x < h, negate the expression inside.
The slope of each arm of g(x) = a|x − h| + k is ±|a|. Larger |a| = steeper V. Smaller |a| = shallower V.
|A| = c with c < 0 has NO solution. Always check the right-hand side before solving.