Absolute-Value Equations and Graphs

7.1: Absolute-Value Equations and Graphs. Original editorial SAT preparation with three worked examples, ten flashcards, and thirty practice questions. Educator-review draft.

Lesson 1 of 83 parts Practice Flashcards

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Distance equations

Objective: Solve an absolute-value equation as two signed cases.

Prerequisite skills: equations, distance, and piecewise reasoning. The absolute value |u| measures distance from zero. If |u|=k with k>0, then u=k or u=−k. Do not drop either branch; if k is negative there are no real solutions. Common error: keeping only the positive branch. Restrictions and interpretation: Absolute value is nonnegative; a positive right side usually gives two branches, zero one, and a negative none.

Formulas

$|x − h| = k ⇒ x = h ± k, k ≥ 0$

Worked examples

Try itWhat is the greater solution of |x − 24| = 22?Show answer

Use x−24=22 or x−24=−22. Solutions are 46 and 2; the greater is 46. Therefore the requested answer is 46.

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Distance equations

Objective: Solve an absolute-value equation as two signed cases.

Formulas

$|x − h| = k ⇒ x = h ± k, k ≥ 0$

Worked examples

Try itWhat is the greater solution of |x − 24| = 22?Show answer

Use x−24=22 or x−24=−22. Solutions are 46 and 2; the greater is 46. Therefore the requested answer is 46.

Absolute-value graph vertex

Objective: Read the vertex of a shifted absolute-value graph.

The graph y=a|x−h|+k has vertex (h,k). For a>0 it opens upward with minimum k; for a<0 it opens downward. The slopes of the two arms have opposite signs. Common error: reversing the horizontal shift or confusing the corner with an intercept. Restrictions and interpretation: Absolute value is nonnegative; a positive right side usually gives two branches, zero one, and a negative none.

Formulas

y = a|x − h| + k has vertex (h,k)

Worked examples

Try itWhat is the minimum value of f(x) = 2|x − 24| + 28?Show answer

Since |x−24|≥0, the term 2|x−24| is at least zero. It vanishes at x=24; the minimum output is 28. Therefore the requested answer is 28.

Nested linear absolute equations

Objective: Isolate absolute value before splitting into cases.

Remove outside additions and multipliers first. After isolating the absolute value, check its right side is nonnegative, solve both linear branches, and substitute into the original equation. Common error: splitting into cases before removing the outside constant. Restrictions and interpretation: Absolute value is nonnegative; a positive right side usually gives two branches, zero one, and a negative none. Desmos and method choice: Use Desmos for arithmetic or a graph/table check when that reduces work. Enter parentheses explicitly and verify the requested quantity and units. Use the algebraic steps shown here for an exact result; a rounded graph display is not proof of exact equality.

Formulas

$a|bx + c| + d = e ⇒ |bx+c| = (e−d)/a, a ≠ 0$

Worked examples

Try itWhat is the sum of the real solutions of 2|3x − 6| + 5 = 57?Show answer

Subtract 5 and divide by 2: |3x−6|=26. The branches give x=(6+26)/3 and x=(6−26)/3. Their sum is 12/3=4. Both return the same original absolute value. Therefore the requested answer is 4.

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