10.1 Absolute Value Equations and Functions

Quick Answer: An equation of form |ax + b| = c (with c ≥ 0) splits into two linear cases, an inequality of form |x| < c becomes a compound AND while |x| > c becomes an OR, and the graph y = a|x - h| + k is a V shape with vertex (h, k).

Absolute value measures a number's distance from zero, so |x| is always non-negative. This single fact drives every rule in this section. Florida's B.E.S.T. benchmark MA.912.AR.4.1 asks you to solve absolute value equations and inequalities in one variable; MA.912.AR.4.3 asks you to graph absolute value functions and pull out vertex, axis of symmetry, domain, and range. Both show up in the Non-Linear Relationships reporting category (31-38% of the exam).

Solving Absolute Value Equations

For an equation of form |expression| = c:

• If c > 0, split into two cases: expression = c OR expression = -c. Solve each, then check both candidates in the original equation.
• If c = 0, the expression itself equals zero (one solution).
• If c < 0, there is no solution, because an absolute value can never equal a negative number.

Worked example: |2x - 5| = 7. Split into 2x - 5 = 7 giving x = 6, and 2x - 5 = -7 giving x = -1. Both check: |2(6)-5| = 7 and |2(-1)-5| = |-7| = 7. Solution set: {-1, 6}.

When the variable appears on both sides, isolate a single absolute value first, then split. For |3x + 1| = |x - 5|, use |A| = |B| → A = B OR A = -B: 3x + 1 = x - 5 gives x = -3, and 3x + 1 = -(x - 5) = -x + 5 gives 4x = 4 so x = 1. Both satisfy the original.

Solving Absolute Value Inequalities

The inequality sign controls the structure. For c > 0:

• |x| < c becomes the compound inequality -c < x < c (an AND statement, one interval).
• |x| > c becomes x < -c OR x > c (two disjoint intervals).
• |x| ≤ c and |x| ≥ c follow the same pattern with inclusive endpoints.

The trap cases are when the bound is negative or zero:

• |x| < -4 has no solution (absolute value cannot be less than a negative).
• |x| > -4 is true for every real x, so the solution is all real numbers.
• |x| ≤ 0 is satisfied only by x = 0.

Worked example: |3 - 2x| ≤ 5. Rewrite as -5 ≤ 3 - 2x ≤ 5. Subtract 3: -8 ≤ -2x ≤ 2. Divide by -2 and reverse both inequalities: 4 ≥ x ≥ -1, or -1 ≤ x ≤ 4. In interval notation: [-1, 4].

Graphing Absolute Value Functions

The vertex form y = a|x - h| + k produces a V-shaped graph with these key features:

Worked example: graph y = -2|x + 3| + 4. Rewrite as y = -2|x - (-3)| + 4, so vertex is (-3, 4). Because a = -2 is negative, the V opens downward with range (-∞, 4]. The axis of symmetry is x = -3. y-intercept: y = -2|0 + 3| + 4 = -6 + 4 = -2, so (0, -2). x-intercepts: -2|x + 3| + 4 = 0 → |x + 3| = 2 → x + 3 = 2 (x = -1) or x + 3 = -2 (x = -5). So x-intercepts at (-5, 0) and (-1, 0).

Transformations and the Parameter a

The value of a controls both direction (sign) and vertical stretch (magnitude). A larger |a| makes the V narrower; a smaller |a| (between -1 and 1, excluding 0) makes it wider. The horizontal shift h moves the vertex left when h is negative inside the bars; remember x - h means the graph moves right h units.

A common EOC trap: y = |x - 5| has vertex (5, 0), not (-5, 0). The form uses x - h, so h = 5. Another: y = 2|x + 1| - 3 has vertex (-1, -3) and range [-3, ∞) because a = 2 > 0.

Classifying from a Graph or Table

An absolute value function is recognizable from its V shape on a graph. From a table, look for outputs that decrease to a minimum then increase symmetrically (or increase to a maximum then decrease), with equal y-values at x-values equally spaced from the vertex. The first differences change sign around the vertex, and the second differences are constant in magnitude but the sign flips.

EOC Trap Summary

1. Forgetting the negative branch when splitting — always write both equations.
2. Answering |x - 4| = -6 with a numeric solution instead of "no solution."
3. Dividing an inequality by a negative without reversing the inequality symbols (both sides of a compound).
4. Misreading the sign of h in vertex form, placing the vertex on the wrong side of the y-axis.
5. Reporting range as all reals when the V opens upward and has a minimum.

Absolute value items are quick wins if you mechanically isolate the bars, check the sign of the constant, split correctly, and verify each candidate in the original equation.