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MathChapter 3: Functions and Linear Equations

Notes

Absolute Value Equations Notes

Interpret absolute value as distance, solve its equation branches, and understand parent and transformed V-shaped graphs.

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Absolute value measures distance from zero, so it is never negative. This distance meaning explains why a positive target usually creates two branches, a zero target creates one, and a negative target creates none.

Absolute value as distance

Equal distance from zeroPoints negative four and four are each four units from zero, demonstrating that both have absolute value four.-6-5-4-3-2-10123456-4044 units left of zero4 units right of zero
Equal distance from zero
Solution count for an absolute-value target
EquationConditionSolutions
\(|u|=a\)\(a>0\)\(u=a\) or \(u=-a\)
\(|u|=0\)zero distance\(u=0\)
\(|u|=a\)\(a<0\)no solution

Solve absolute-value equations

Isolate → classify target → split → verify

  1. Isolate the absolute value

    Remove outside coefficients and constants first.

  2. Inspect the target

    Negative means no solution; zero means one branch; positive means two branches.

  3. Write branches

    For \(|E|=a>0\), solve \(E=a\) and \(E=-a\).

  4. Verify

    Substitute both candidates into the original equation.

Worked example

Solve two branches

Solve \(3|2x-5|-4=17\).

  1. Isolate

    Add \(4\) and divide by \(3\): \(|2x-5|=7\).

  2. Positive branch

    \(2x-5=7\), so \(x=6\).

  3. Negative branch

    \(2x-5=-7\), so \(x=-1\).

  4. Verify

    Both inputs make \(|2x-5|=7\).

\(x=-1\) or \(x=6\).

The parent absolute-value function

Piecewise definition
\[|x|=\begin{cases}-x,&x<0\\x,&x\ge0\end{cases}\]

Negative inputs are negated to produce positive distance; nonnegative inputs remain unchanged.

Values of the parent absolute-value function
x|x|
\(-3\)\(3\)
\(-2\)\(2\)
\(-1\)\(1\)
\(0\)\(0\)
\(1\)\(1\)
\(2\)\(2\)
\(3\)\(3\)
Graph of y = 1|x - 0| + 0V-shaped absolute-value graph with vertex at 0, 0, opening upward.-6-5-4-3-2-1123456-6-5-4-3-2-1123456xyvertex (0, 0)
Graph of y = 1|x - 0| + 0

The left branch has slope \(-1\) and the right branch has slope \(1\). They meet at vertex \((0,0)\), producing a V rather than a single line.

Transformed absolute-value graphs

Transformation form
\[y=a|x-h|+k\]

The vertex is \((h,k)\). The graph opens upward for \(a>0\), downward for \(a<0\), and becomes steeper as \(|a|\) grows.

Graph of y = -2|x - 1| + 3V-shaped absolute-value graph with vertex at 1, 3, opening downward.-6-5-4-3-2-1123456-6-5-4-3-2-1123456xyvertex (1, 3)
Graph of y = -2|x - 1| + 3
Mini check

Check your understanding

How many solutions does \(5|x-2|+1=1\) have?

  1. None
  2. One
  3. Two
  4. Infinitely many
Show answer and explanation

Answer: One

Subtract \(1\) and divide by \(5\): \(|x-2|=0\), so \(x=2\) only.

Key takeaways

What to remember

  • Absolute value is nonnegative distance from zero.
  • Isolate the absolute-value expression before splitting.
  • Positive targets usually give two branches, zero gives one, and negative gives none.
  • The parent graph is a V with vertex \((0,0)\).
  • For \(y=a|x-h|+k\), the vertex is \((h,k)\) and the sign of \(a\) controls opening direction.
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Put these notes into practice

Apply the ideas with SAT-style questions, then reinforce key details with flashcards.