A modulus folds a graph. Where it goes decides which way.

Put the modulus round the whole function, or round the $x$ inside it, and you get two different pictures. Then the fold explains why modulus equations throw up answers that don't work.

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In $|f(x)|$ the modulus acts on the output. Every negative $y$-value becomes positive, so each part of the curve below the $x$-axis folds up over it. Nothing is left below the axis.

In $f(|x|)$ the modulus acts on the input: $x$ and $-x$ give the same output. So the right half is kept and copied onto the left, and the graph is symmetrical about the $y$-axis.

Solve $|x - 2| = 2x + 1$. The modulus means $x - 2$ is either $2x + 1$ or $-(2x + 1)$, so there are two cases to solve. The graph says to expect one answer: the line meets the V once, at $x = \tfrac{1}{3}$.

The other case gives $x = -3$, but there $2x + 1 = -5$, and a modulus is never negative. On the graph, the line meets the part of $y = x - 2$ that was folded away, below the axis. Reject it.

The same picture answers inequalities. The V is below the line everywhere to the right of $x = \tfrac{1}{3}$, so

Fold the output, mirror the input, check every answer.

Modulus inequalities, combining transformations, and twelve graded exam questions with full worked solutions are waiting.

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