The Modulus Function and Combinations of Graph Transformations

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

The four single transformations, derived rather than listed

Spec 1.4 names four transformations of $y = f(x)$ : $y = af(x)$ , $y = f(x) + a$ , $y = f(x+a)$ , $y = f(ax)$ . Below IAL level these are usually handed over as four rules to memorise, with a diagram to match each. Here they are four consequences of one idea: a transformation either changes what happens to the OUTPUT after $f$ has done its work, or it changes what gets fed into $f$ as the INPUT before it does any work at all — and every rule below falls out of tracking a single point $(p, q)$ on $y = f(x)$ through each change.

Output changes — $y = af(x)$ and $y = f(x) + a$ . These act on $f(p) = q$ directly, once it already exists. $y = af(x)$ multiplies it: the point becomes $(p, aq)$ — a vertical stretch by scale factor $a$ from the x-axis (a scale factor between $0$ and $1$ squashes the curve towards the axis; a negative $a$ multiplies by a negative number, which is a reflection combined with a stretch). $y = f(x) + a$ adds to it: the point becomes $(p, q+a)$ — a vertical translation, by $a$ units up if $a>0$ . Both act on the y-coordinate only, exactly matching the order they are written in: multiply, or add.

Input changes — $y = f(x+a)$ and $y = f(ax)$ . These change what value has to be substituted before you can reach the same output $q$ . For $y = f(x+a)$ : you need $x + a = p$ , i.e. $x = p - a$ — the point moves to $(p-a, q)$ , which is $a$ units in the OPPOSITE direction to the sign in the formula. This is the one worth checking by substitution every time rather than trusting instinct: plug $x = p-a$ into $f(x+a)$ and you get $f((p-a)+a) = f(p) = q$ , confirmed. For $y = f(ax)$ : you need $ax = p$ , i.e. $x = p/a$ — the point moves to $(p/a, q)$ , a horizontal stretch by scale factor $1/a$ (dividing, not multiplying — a large $a$ compresses the graph horizontally, since it now takes a smaller x to reach the same argument).

The pattern worth keeping, stated once rather than as four separate facts: an operation applied OUTSIDE $f(\cdot)$ acts on the y-coordinate in the direction the formula reads; an operation applied INSIDE the brackets, on $x$ itself, acts on the x-coordinate in the OPPOSITE direction to how it reads — because it is describing what $x$ has to become before it produces the output you are looking for, which is a backwards question by nature.

Combining two transformations — and the one combination the spec explicitly excludes

The spec's own guidance gives two examples of a combined transformation: $y = 2f(3x)$ and $y = f(-x) + 1$ (spec 1.4 guidance). Look at what each actually combines. $2f(3x)$ combines one INPUT change ( $x \to 3x$ ) with one OUTPUT change ( $\times 2$ ). $f(-x) + 1$ combines one INPUT change ( $x \to -x$ , a reflection in the y-axis, the $a=-1$ case of $f(ax)$ ) with one OUTPUT change ( $+1$ ). Every combined transformation this paper asks for has exactly this shape: one thing done to $x$ before $f$ is applied, and one (or two, e.g. $y = af(x) + c$ ) done to the result afterwards.

What is missing from that list is a combination of TWO input changes in the same bracket — $y = f(ax + b)$ , both a stretch and a translation applied to $x$ before $f$ ever sees it. The spec's own guidance states this directly and is worth quoting exactly: it names the four base forms above and then adds, verbatim, "the graph of y = f(ax + b) will not be required." A lesson on this topic could derive how to sequence two input changes correctly — it is not difficult, only easy to get backwards — but there is no version of the WMA13 paper that will ask for it, so this lesson does not spend time teaching it. Recognise the excluded shape on sight (an $x$ -coefficient and a constant added TOGETHER inside the same $f(\cdot)$ ) and treat any exam-style question with that shape as a red flag, not a target to prepare for.

Mechanism

Why the input change and the output change can be done in either order

Take a concrete combination — $y = 3f(x+2) - 1$ — and track a single point $(p, q)$ on $y = f(x)$ through it two different ways, to see whether the order in which the two moves are made actually matters. Route one, input first: apply the translation $x \to x+2$ first, which by the rule above moves $(p, q)$ to $(p - 2, q)$ ; then apply the two output changes in the order they are written, $\times 3$ then $-1$ : $(p-2, q) \to (p-2, 3q) \to (p-2, 3q-1)$ . Route two, output first: apply $\times 3$ then $-1$ to the ORIGINAL point first, giving $(p, 3q-1)$ on $y = 3f(x)-1$ ; then apply the same translation to that: $(p-2, 3q-1)$ . Both routes land on exactly the same point, $(p-2, 3q-1)$ — and substituting directly into the formula confirms it independently: $3f((p-2)+2) - 1 = 3f(p) - 1 = 3q - 1$ . The reason this works is not a coincidence of this particular example. The input change touches only the x-coordinate; the two output changes touch only the y-coordinate. Operations that act on different coordinates cannot interfere with each other, so it makes no difference which is imagined as happening "first" — the combined picture is the same either way. Contrast this with the TWO output changes done together, $\times 3$ then $-1$ : swap THEIR order and the result changes. $3(f(x) - 1) = 3f(x) - 3$ , not $3f(x) - 1$ — a different function entirely. Two operations on the same coordinate generally do not commute; an input change and an output change, touching different coordinates, always do. This is also the real reason $y = f(ax+b)$ is excluded from this spec: it is two input changes sharing one coordinate, which — like the two output changes above — do not commute, and getting the order and the exact translation amount right together is a genuinely harder skill than anything this paper's guidance actually lists.

What reflecting the output vs reflecting the input actually does — with real numbers, before any letters

In plain terms

Take one concrete rule: subtract 3 from whatever number you're given. Run it on eight numbers in a row, from −2 up to 5, and write down what comes out each time: −2 gives −5, −1 gives −4, 0 gives −3, 1 gives −2, 2 gives −1, 3 gives 0, 4 gives 1, 5 gives 2. Five of those results came out negative (from inputs −2 through 2) and three came out zero or positive (from inputs 3, 4, 5). Now do two different things to that same list of eight results, one at a time, and watch how differently they behave. Thing one: fix up the OUTPUTS after the subtracting is already done. Go down the list and, wherever a result came out negative, flip its sign to positive; leave every result that was already zero or positive completely alone. −5 becomes 5. −4 becomes 4. −3 becomes 3. −2 becomes 2. −1 becomes 1. 0 stays 0 — it wasn't negative, there's nothing to flip. 1 stays 1. 2 stays 2. New list: 5, 4, 3, 2, 1, 0, 1, 2. Notice the last three entries, from inputs 3, 4 and 5, are completely untouched — they were already non-negative, and this operation does nothing at all to a number that's already non-negative. Thing two: instead, fix up the INPUTS before subtracting even happens. For each of the same eight starting numbers, strip off any minus sign first, THEN subtract 3. −2 loses its sign and becomes 2, then 2 − 3 = −1. −1 becomes 1, then 1 − 3 = −2. 0 stays 0 (nothing to strip), then 0 − 3 = −3. 1 stays 1 (it had no minus sign to begin with), then 1 − 3 = −2. 2 stays 2, then 2 − 3 = −1. So for inputs −2 through 2, the new results are −1, −2, −3, −2, −1. Compare the first two entries against the last two: input −2 produced −1, and input 2 — further down the very same list — also produces −1. Input −1 produced −2, and input 1 also produces −2. Every negative starting number has been silently swapped for its positive twin before the subtracting rule ever runs on it — the negative-input half of this list isn't new information at all, it's a copy of the positive-input half, reversed left to right. Those are the two genuinely different things a modulus sign can do to the same rule and the same numbers: fix the OUTPUT afterward, which only touches results that came out negative and only changes their sign — or fix the INPUT beforehand, which throws away every negative starting number and replaces its result with whatever the positive twin already gave.

Name what you were doing to that subtract-3 rule: call it $f$ , so $f(x) = x - 3$ . Thing one — fixing the output afterward — is $y = |f(x)|$ : take $f(x)$ , and if it's negative, flip its sign; if it's already $\geq 0$ , leave it alone. That's exactly why the last three rows of the first table (inputs 3, 4, 5) never moved: $f(x) \geq 0$ there already, so $|f(x)| = f(x)$ with nothing left to do. Thing two — fixing the input beforehand — is $y = f(|x|)$ : strip the sign off $x$ first (that's $|x|$ ), then apply $f$ to whatever's left. That's why input $-2$ and input $2$ landed on the same output, $-1$ : $f(|-2|) = f(2) = -1$ and $f(|2|) = f(2) = -1$ as well — they're computing the literal same thing. Every negative $x$ gets routed to the calculation its positive twin, $|x|$ , already produces.

Formally

Both are the piecewise modulus definition — $|t| = t$ for $t \geq 0$ , $|t| = -t$ for $t < 0$ — applied in two different places. $y = |f(x)|$ applies it to $f$ 's OUTPUT: wherever $f(x) \geq 0$ the curve is left unchanged, and wherever $f(x) < 0$ it is reflected in the x-axis via $-f(x)$ , selectively, one region at a time — a curve produced this way can never dip below the x-axis anywhere. $y = f(|x|)$ applies it to $x$ itself, BEFORE $f$ runs: for $x \geq 0$ , $|x| = x$ so $f(|x|) = f(x)$ , the right-hand half is untouched; for $x < 0$ , $|x| = -x$ , so at $x = -a$ , $f(|x|) = f(a)$ — the entire left-hand half is discarded and rebuilt as a mirror image of the right-hand half, not merely adjusted.

Mechanism

Where $y = |f(x)|$ and $y = f(|x|)$ actually come from

Both are direct consequences of the from the first prequestion — $|t| = t$ when $t \geq 0$ , and $|t| = -t$ when $t < 0$ — applied in two different places. $y = |f(x)|$ applies it to the OUTPUT of $f$ . At any x-value where $f(x)$ is already $\geq 0$ , $|f(x)| = f(x)$ : nothing changes, the curve is left exactly where it was. At any x-value where $f(x) < 0$ , $|f(x)| = -f(x)$ — and $-f(x)$ is precisely the reflection-in-the-x-axis rule from the third prequestion, applied only over that stretch of x. So $y = |f(x)|$ is $y = f(x)$ with every part that dipped below the x-axis reflected back up above it, and every part already on or above the axis left untouched — a single mechanism, applied selectively based on the sign of the original output, not two different rules to remember for "positive bits" and "negative bits." A curve produced this way can never dip below the x-axis anywhere, because $|f(x)| \geq 0$ for every real $x$ by the same definition that opened this lesson. $y = f(|x|)$ applies the definition to the INPUT instead. For $x \geq 0$ , $|x| = x$ , so $f(|x|) = f(x)$ : the right-hand half of the curve, $x \geq 0$ , is completely unchanged. For $x < 0$ , $|x| = -x$ , which is a positive number — specifically, it is the same positive value that $|x|$ would be at $+x$ 's reflection. Concretely: at $x = -a$ for some $a > 0$ , $f(|-a|) = f(a)$ — the height of the curve at $-a$ is defined to equal the height at $+a$ . The left-hand half of $y = f(x)$ , whatever it looked like, is discarded entirely and replaced by a mirror image of the right-hand half, reflected in the y-axis. These are genuinely different operations — one reflects part of the curve in the x-axis based on where the ORIGINAL output was negative; the other discards the whole left half and rebuilds it as a mirror of the right half, based on nothing but the sign of $x$ itself — and an exam question can ask for either one given the same starting curve $y = f(x)$ , with two different correct pictures as the answer.

Diagram — y = |2x − 1| and y = x + 5 — the spec's own worked example, solved by sketch

x-axis: x · y-axis: y

y = |2x − 1|: A V-shape: the line y = 2x − 1 for x ≥ 1/2 (where 2x − 1 ≥ 0, so the modulus does nothing), reflected in the x-axis for x < 1/2 (where 2x − 1 < 0) to give the rising line y = 1 − 2x on that side instead. The vertex sits at (1/2, 0), the one point where 2x − 1 is exactly zero and there is nothing to reflect.
y = x + 5: A straight line of gradient 1, y-intercept (0, 5) — shallower than either arm of the V, which is exactly why it only crosses the V twice rather than not at all or more than twice.

Intersection at x = 6: On the right arm (x ≥ 1/2, where |2x−1| = 2x−1): solving 2x − 1 = x + 5 gives x = 6, y = 11. Substituting back confirms it: |2(6) − 1| = 11 = 6 + 5.
Intersection at x = −4/3: On the left arm (x < 1/2, where |2x−1| = 1−2x): solving 1 − 2x = x + 5 gives x = −4/3, y = 11/3. Confirmed: |2(−4/3) − 1| = |−11/3| = 11/3 = −4/3 + 5.
The outside region solves |2x − 1| > x + 5: Between the two intersection points the line sits above the V (check x = 0: |−1| = 1 < 5 = 0 + 5). Outside them — x < −4/3 or x > 6 — the V is above the line instead, which is the solution set to the strict inequality: neither critical value is included, since at both of them the two sides are exactly equal, not one strictly greater than the other.

Diagram — One curve, two different modulus transformations: f(x) = x² − 4x

x-axis: x · y-axis: y

y = f(x) = x² − 4x: An upward parabola through the origin, crossing the x-axis again at (4, 0), with its minimum turning point at (2, −4) — read off directly from x(x−4) for the roots and (x−2)² − 4 for the vertex. Negative for the whole stretch strictly between its two roots, 0 < x < 4, and non-negative everywhere else.
y = |f(x)|: Identical to y = f(x) outside [0, 4], where f was already non-negative. Between the roots, the dip down to (2, −4) is reflected up to a peak at (2, 4) instead — the only part of the curve that changes, and it changes by reflection in the x-axis, not by being redrawn from scratch.
y = f(|x|): Identical to y = f(x) for x ≥ 0 (roots at 0 and 4, minimum at (2, −4)) — completely unchanged. For x < 0 the right-hand half is mirrored across the y-axis: a new minimum appears at (−2, −4), and a new x-intercept appears at (−4, 0) that has no counterpart anywhere on the original curve for negative x. The point (0, 0) — a straightforward crossing on y = f(x), since f changes sign there with a non-zero gradient — becomes a sharp peak on y = f(|x|) instead, because the curve now approaches it from both sides as a mirror image of the same falling branch.

(0, 0) and (4, 0) — shared by all three curves: Roots of f(x) itself are untouched by either modulus operation, because |0| = 0: wherever the original curve meets the x-axis, so do both transformed versions.
(2, 4) — only on y = |f(x)|: The reflected image of the original minimum (2, −4). Nowhere else does this point appear.
(−4, 0) and (−2, −4) — only on y = f(|x|): These exist only because of the mirroring for x < 0; the original y = f(x) has no feature at x = −4 or x = −2 at all (f(−4) = 32, f(−2) = 12 — nothing special happens there on the original curve).

Common error: Sketching y = f(|x|) by keeping the right-hand half and simply continuing the original left-hand half of y = f(x) unchanged, rather than replacing it with a mirror image of the right-hand half.

Correct: The left-hand half of y = f(|x|) has nothing to do with the original left-hand half of y = f(x) — it is discarded completely and rebuilt as a reflection of the RIGHT-hand half in the y-axis. An examiner report on exactly this transformation records the gap directly: "very few students understood that... the negative x part of the graph is a reflection of the positive x part... so [the second solution] is also a solution."

examiner-report · Jun 2023 · Q6(d)

Where the exam hides two equations inside |f(x)| = k

An equation of the form $|f(x)| = k$ , for a positive constant $k$ , is not one equation — it is two, and both have to be tried. $|f(x)| = k$ means $f(x) = k$ OR $f(x) = -k$ , because those are exactly the two values whose modulus is $k$ . This is the same case-split idea as $|ax+b| = cx+d$ above, just with the roles of the two sides swapped: there the modulus expression sat on the left and both branches were solved as ordinary linear or quadratic equations; here it is $f$ itself inside the modulus, and each branch is $f(x)$ equal to a plain constant.

The second branch is not always solvable, and knowing why not is itself useful, not just a technicality to check off. $f(x) = -k$ can only have a real solution if $-k$ is actually a value $f$ can take — if $-k$ is below every value in $f$ 's range (for instance, more negative than $f$ 's minimum, on a curve with a minimum), the branch has no real roots at all, and that is not an error, it is the correct outcome of a genuinely unsolvable equation. Working out the discriminant of that branch's equation and getting a negative number is not a sign that something went wrong; it is the mark-earning way of showing the branch was tried and correctly found empty.

Worked, in full

Solve |3x − 2| > 2x + 1

01
Split into the two cases the definition of $|3x-2|$ actually contains. For $x \geq \frac{2}{3}$ (where $3x-2 \geq 0$ ): $|3x-2| = 3x-2$ , so the inequality is $3x - 2 > 2x + 1$ . For $x < \frac{2}{3}$ (where $3x-2 < 0$ ): $|3x-2| = -(3x-2) = 2 - 3x$ , so the inequality is $2 - 3x > 2x + 1$ .
Earns: M1 — attempts to split the modulus into its two cases (or, equally validly, squares both sides, which reaches the same combined quadratic inequality). Either route is credited; what earns the mark is committing to a method that actually handles both possibilities, not just the visible one.
02
Solve each linear inequality on its own, ignoring for now whether the answer belongs in that case's region. Case one: $3x - 2 > 2x + 1 \Rightarrow x > 3$ . Case two: $2 - 3x > 2x + 1 \Rightarrow 1 > 5x \Rightarrow x < \frac{1}{5}$ .
Earns: A1 — both critical values correct, $x = 3$ and $x = \frac{1}{5}$ . On a real WMA13 linear modulus-inequality question this is a single accuracy mark covering BOTH branches together, not a second method mark — verified mark-scheme wording for exactly this step: "A1: Correct critical values x... which may be part of an incorrect inequality" (Jan 2022, Q7(b)). The mark is earned by the two numbers being right; which way either inequality sign happens to be pointing at this stage does not matter yet — sorting that out is stage 3's job, not this one's.
03
Now reconcile each solution with the region its own case assumed. Case one required $x \geq \frac{2}{3}$ , and found $x > 3$ — entirely inside that region (every $x > 3$ does satisfy $x \geq \frac{2}{3}$ ), so the whole of $x > 3$ survives. Case two required $x < \frac{2}{3}$ , and found $x < \frac{1}{5}$ — also entirely inside its own region ( $\frac{1}{5} < \frac{2}{3}$ ), so the whole of $x < \frac{1}{5}$ survives too.
Earns: dM1 — selects the outside region for these critical values, dependent on the method mark from stage 1. Verified real mark-scheme language for exactly this step, from a linear modulus-inequality question of this same shape: "dM1: Selects outside region for their critical values... It is dependent upon having attempted to solve one correct equation" (Jan 2022, Q7(b)) — a single 'd', not a doubly-dependent 'dd', because a plain modulus inequality like this one produces only ONE prior method mark for this step to depend on. (WMA13 does use a doubly-dependent ddM1 for a region-selection step that looks superficially like this one — but only on a harder question that combines the modulus with a second technique, so that TWO separate method marks, not one, sit in front of it; that is not what this question is.) This is the step most commonly skipped or done by instinct rather than by checking each case's own restriction — which works here because both results happened to sit entirely inside their case, but will not always.
04
State the combined answer: $x < \frac{1}{5}$ or $x > 3$ , and check it with one point inside the stated region and one point in the excluded middle. Inside, at $x = 0$ : $|3(0)-2| = 2$ and $2(0)+1 = 1$ , and $2 > 1$ is true, confirming $x=0$ (which satisfies $x < \frac{1}{5}$ ) is a genuine solution. In the excluded middle, at $x = 1$ : $|3(1)-2| = 1$ and $2(1)+1 = 3$ , and $1 > 3$ is false, confirming $x=1$ is correctly excluded, exactly as $\frac{1}{5} < 1 < 3$ predicts.
Earns: A1 — correct final answer, both bounds strict. The two-point check costs one line and confirms both the shape of the answer (outside, not inside, the two critical values) and that neither bound was accidentally reversed.

Source — Mark scheme, Jan 2022

"dM1: Selects outside region for their critical values... It is dependent upon having attempted to solve one correct equation"

Complete it yourself

Complete the chain — solve |4x + 3| = 2x + 9

01
Split into the two cases. For $x \geq -\frac{3}{4}$ (where $4x+3 \geq 0$ ): $4x + 3 = 2x + 9$ . For $x < -\frac{3}{4}$ (where $4x + 3 < 0$ ): $-(4x+3) = 2x + 9$ , i.e. $-4x - 3 = 2x + 9$ .
02
Solve the first case: $4x + 3 = 2x + 9 \Rightarrow 2x = 6 \Rightarrow x = 3$ . Check against that case's own region: $3 \geq -\frac{3}{4}$ ✓, so $x=3$ is a genuine solution.

Marked, line by line

$f(x) = x^2 - 4x$ , $x \in \mathbb{R}$ . (a) Write down the coordinates of the points where the curve $y = f(x)$ crosses the x-axis. (2) (b) Sketch $y = |f(x)|$ , stating the coordinates of the points where the curve meets the x-axis and the coordinates of its maximum point. (3) (c) Sketch $y = f(|x|)$ , stating the coordinates of ALL the points where the curve meets the x-axis. (3) (d) Solve the equation $|f(x)| = 3$ , giving your answers as exact values. (4) — VERIDIAN-original question, inspired by the structure of real WMA13 items combining a "state the coordinates" transformation part with an equation-solving part on the same function (spec 1.3 and 1.4 in one question, echoing the pilot's own combination of adjacent spec points). Not a reproduction of any past-paper question, and the per-line mark allocations are modelled on verified mark-scheme conventions rather than transcribed from a real scheme, which for an original question does not exist.

12 marks available

(a) — 2 marks

01
$f(x) = x(x-4)$
Method mark for factorising to find the roots.
M1
02
$x = 0$ or $x = 4$ , so the curve crosses the x-axis at $(0,0)$ and $(4,0)$ .
Accuracy mark, both coordinates, correct answer only.
A1

(b) — 3 marks

101
$|f(x)|$ meets the x-axis at exactly the points where $f(x) = 0$ , since $|0| = 0$ and nowhere else can a non-negative quantity be zero: $(0,0)$ and $(4,0)$ , using the roots from (a).
Independent mark, follow through the candidate's own roots from (a) — the roots of |f(x)| are always identical to the roots of f(x), whatever those turned out to be.
B1 ft
102
Between the roots, $0 < x < 4$ , $f(x) < 0$ (an upward parabola with two real roots is negative strictly between them) — this is the section that gets reflected in the x-axis.
Method mark for correctly identifying which section of the curve requires reflecting.
M1
103
$f(2) = 4 - 8 = -4$ , so the original minimum is $(2,-4)$ ; reflected, the maximum of $|f(x)|$ is $(2, 4)$ .
Accuracy mark, correct maximum point, correct answer only.
A1

(c) — 3 marks

201
For $x \geq 0$ , $f(|x|) = f(x)$ , so the roots of $f(x)$ at $x \geq 0$ carry over unchanged: from (a), both $(0,0)$ and $(4,0)$ already satisfy $x \geq 0$ , so both remain roots of $f(|x|)$ .
Independent mark, follow through the candidate's own non-negative roots from (a).
B1 ft
202
For $x < 0$ , $f(|x|)$ mirrors the $x>0$ branch: the root at $x=4$ has a mirror image at $x=-4$ , giving an additional root $(-4, 0)$ with no counterpart on the original curve.
Independent mark for the mirrored root — the one most commonly missed, since it requires recognising that the left half of the sketch is not simply the original left half of y = f(x) left in place.
B1
203
The complete set of x-intercepts is $(-4,0)$ , $(0,0)$ and $(4,0)$ , and the curve is symmetric about the y-axis, with a second minimum at $(-2,-4)$ mirroring $(2,-4)$ .
Independent mark for stating the symmetry and the complete, correctly-mirrored picture.
B1

(d) — 4 marks

301
$|f(x)| = 3$ means $f(x) = 3$ or $f(x) = -3$ . Case one: $x^2 - 4x - 3 = 0$ .
Method mark for splitting into the two cases and forming the first quadratic.
M1
302
$x = \frac{4 \pm \sqrt{16+12}}{2} = \frac{4 \pm \sqrt{28}}{2} = 2 \pm \sqrt{7}$
Accuracy mark, exact surd form as the question requires, correct answer only.
A1
303
Case two: $x^2 - 4x + 3 = 0$ .
Method mark for forming the second case's quadratic.
M1
304
$(x-1)(x-3) = 0$ , so $x = 1$ or $x = 3$ . Full solution set: $x = 1, 3, 2-\sqrt{7}, 2+\sqrt{7}$ .
Accuracy mark, both roots of the second case, correct answer only. Both cases here do have real solutions — unlike the reference-card's warning example, $-3$ is within f's range, since f's minimum is $-4$ , which is below $-3$ .
A1

In your own words

In one sentence: why does solving |f(x)| = k need you to try f(x) = −k as well as f(x) = k, and why might that second equation sometimes have no real solutions at all?

Named traps

positive-x-branch-not-mirrored-for-f-of-modulus-x: Confirmed directly on a real f(|x|) reflective-symmetry question: "very few students understood that... the negative x part of the graph is a reflection of the positive x part... so [the second solution] is also a solution" (Jun 2023, Q6(d)). The mechanism is the one this lesson derives rather than states: for x < 0, f(|x|) equals f evaluated at the corresponding positive value, so the entire left-hand branch is a rebuild, not a survival of the curve's original left half.
only-one-branch-of-modulus-equation-solved: A confirmed, recurring pattern across at least three series (Jan 2022 Q7, Oct 2020 Q4(c), Jun 2023 Q6(c)): candidates solve only one of the two cases a modulus equation or inequality actually splits into, or find both critical values correctly but then fail to select the correct combined region at the end. Jan 2022's own examiner report calls the specific question this pattern showed up on "very demanding by many." The fix is structural, not a matter of care: always write out both cases before solving either.
critical-values-correct-but-wrong-region-selected: Confirmed by the mark scheme's own design, not just examiner commentary: Pearson attaches a dedicated, separately-earned mark to the single step of choosing which region satisfies a modulus inequality — distinct from, and dependent on, the earlier mark(s) that found the critical values in the first place. Verbatim from a plain linear modulus-inequality question of exactly the shape this lesson teaches: "dM1: Selects outside region for their critical values... It is dependent upon having attempted to solve one correct equation" (Jan 2022, Q7(b)). On a harder WMA13 question that combines the modulus with a second technique — so that TWO separate method marks, not one, precede this same step — the dependency escalates and the mark scheme uses a doubly-dependent ddM1 instead, verified verbatim elsewhere in this paper's mark schemes: "ddM1: Chooses the outside region for their values... It is dependent on both previous method marks" (Oct 2023, Q9(c)). Either way, the region-selection step is never folded into the mark that found the values — it is always its own separate checkpoint, exactly because getting the two boundary values right and then picking the wrong side of them is common enough to need one.
combined-transformation-only-partially-applied: Confirmed on a real combined-transformation coordinate question: common errors included giving "an error of (−8, −3) or (−8, −6)" and, separately, that "a common error was to subtract 1 from the x coordinate to give (−5, −9)" (Jan 2024, Q1). Both are the same underlying failure: applying only one of the two combined moves correctly, or applying the right move to the wrong coordinate. The check from the mechanism above is the direct fix — track one point through both moves independently and confirm each one only touched the coordinate it was supposed to.
calculator-used-with-no-algebraic-method-shown: WMA13 carries an explicit no-calculator-methods rubric on specific questions, verified verbatim from the January 2023 question paper: "In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable." The facts bank confirms this same requirement recurred specifically on the January 2024 modulus simultaneous-equation question (Q8) — the exact topic this lesson covers. Treat the rubric as a cue that the algebra itself, not just the final numbers, is what earns the marks on that question.

Beyond the spec

Spec 1.3 only asks for sketching y = |f(x)| and y = f(|x|) and solving equations/inequalities built from them (nothing in it names "even function" or requires the term). But the mechanism block above already proved, case by case, that the left half of y = f(|x|) is always a mirror of the right half — and that proof is really one instance of a single, nameable property. Seeing the general property makes the specific trap this lesson keeps returning to (mirroring the left branch of f(|x|)) something you can recognise on sight for ANY function, not something re-derived from scratch each time a new f(x) shows up.

A function $g$ is called even if $g(-x) = g(x)$ for every $x$ in its domain — geometrically, its graph is symmetric about the y-axis. $y = f(|x|)$ is for every choice of $f$ , with no exceptions and no extra checking required: $g(x) = f(|x|)$ gives $g(-x) = f(|-x|) = f(|x|) = g(x)$ , using only the fact that $|-x| = |x|$ for every real $x$ . That is the entire mechanism from earlier in this lesson, restated as one line of algebra instead of a case-by-case argument about which half survives — the left branch isn't merely SIMILAR to a mirror of the right branch, $f(|x|)$ is structurally forced to be y-axis-symmetric before a single number is substituted in, whatever the original $f$ happened to look like. (For contrast, $y = |f(x)|$ has no such guarantee — reflecting the negative parts of $f$ upward does not generally produce a y-axis-symmetric curve, which is exactly why the two transformations in this lesson need two separate mechanisms, not one.)

Retrieval — with feedback on every choice

Question 1

1 mark

The curve $y = f(x)$ has a minimum turning point at $(2, -4)$ . What is the corresponding point on $y = |f(x)|$ ?

Question 2

1 mark

The curve $y = f(x)$ crosses the x-axis at exactly one point, $(6, 0)$ , and nowhere else. Where does $y = f(|x|)$ cross the x-axis?

Question 3

3 marks

Solve $|x + 2| = 3x - 6$ .

Question 4

2 marks

The curve $y = f(x)$ passes through $(4, 10)$ . What are the coordinates of the corresponding point on $y = 3f(x - 2)$ ?

Question 5

1 mark

Which of these transformations of $y = f(x)$ will NOT be required on the WMA13 paper?

Reference — not a study method, a lookup

|t| = t if t ≥ 0, −t if t < 0. Never negative. Two branches, always both checked.
y=af(x): (p,q)→(p,aq). y=f(x)+a: →(p,q+a). y=f(x+a): →(p−a,q). y=f(ax): →(p/a,q).
y = |f(x)|: reflects parts below the x-axis upward; parts already ≥0 unchanged.
y = f(|x|): keeps x ≥ 0; discards x < 0 and rebuilds it as a mirror of x ≥ 0.
Combined: one input change + one/two output changes. f(ax+b) is NOT required.
|A| = B: solve A=B and A=−B, check each root against the case it came from.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme or examiner report in this lesson was independently verified against the primary Pearson document, not carried over from prior course material. Every question in this lesson — prequestion, worked chain, chain drill, marked solution and MCQ alike — is VERIDIAN-original wording, inspired by confirmed real question types and (for the diagram block quoting it) the specification's own published guidance example, never a reproduction of a real Pearson past-paper question; and because the questions are original, the per-line mark allocations attached to them are modelled on verified mark-scheme conventions (what M, A and B marks mean, when follow-through applies, what a doubly-dependent ddM1 mark is for) rather than transcribed from a real mark scheme, which for an original question does not exist.

Question 11 mark

The curve $y = f(x)$ has a minimum turning point at $(2, -4)$ . What is the corresponding point on $y = |f(x)|$ ?

$(2, 4)$ — a maximum
Correct. Since $f(2) = -4 < 0$ , the modulus reflects that specific output in the x-axis: $|-4| = 4$ . Reflecting a minimum in the x-axis produces a maximum at the same x-coordinate.
B $(2, -4)$ — unchanged
This is the value $|f(x)|$ would give only if $f(2)$ were already $\geq 0$ . It is $-4$ , strictly negative, so the modulus does change it — to $4$ , not $-4$ .
C $(-2, 4)$
This reflects the x-coordinate as well, which $y = |f(x)|$ never does — it acts only on the output of $f$ , after $x$ has already been substituted in.
D $(2, 0)$
The modulus does not zero out a negative value — it makes it positive. $|-4| = 4$ , not $0$ ; only a value that started at exactly $0$ stays at $0$ .

Traps tested: Modulus reflection not applied to negative output · X coordinate changed when only y should be · Modulus confused with setting to zero

Question 21 mark

The curve $y = f(x)$ crosses the x-axis at exactly one point, $(6, 0)$ , and nowhere else. Where does $y = f(|x|)$ cross the x-axis?

$(6, 0)$ and $(-6, 0)$
Correct. The root at $x=6$ is on the $x \geq 0$ side, so it survives unchanged; and it gains a mirror image at $x = -6$ , since $f(|-6|) = f(6) = 0$ . Two roots, symmetric about the y-axis.
B $(6, 0)$ only
This treats $f(|x|)$ as identical to $f(x)$ for negative x, which it never is — the left-hand side is rebuilt as a mirror of the right-hand side, not left as the original curve's own (unknown, unspecified) left half.
C $(6, 0)$ and $(0, 0)$
There is no reason for $f(|x|)$ to cross the axis at the origin unless $f(0) = 0$ , which was never stated — the only given root is at $x=6$ , and mirroring a root at $x=6$ produces a root at $x=-6$ , not at $x=0$ .
D $(-6, 0)$ only
This keeps the mirrored root and drops the original one. The right-hand half of $y = f(|x|)$ , $x \geq 0$ , is identical to $y = f(x)$ there, so the original root at $x=6$ is retained, not replaced.

Traps tested: Positive x branch not mirrored for f of modulus x · Mirrored root placed at origin instead of reflected position · Original root dropped instead of retained

Question 33 marks

Solve $|x + 2| = 3x - 6$ .

$x = 4$ only
Correct. Case $x \geq -2$ : $x+2 = 3x-6 \Rightarrow x=4$ , which satisfies $x \geq -2$ ✓. Case $x < -2$ : $-(x+2) = 3x-6 \Rightarrow x = 1$ — but $1 < -2$ is FALSE, so this is not a valid solution to that case, even though the algebra inside it was done correctly. It is an extraneous root: check directly, $|1+2| = 3$ but $3(1)-6 = -3$ , and $3 \neq -3$ .
B $x = 4$ or $x = 1$
$x=1$ solves the equation you get by assuming $x < -2$ , but $1$ is not less than $-2$ — the assumption that produced it is false for this value, so the equation it was derived from does not apply here. Solving a case's equation correctly is not the same as the result belonging to that case; both have to hold.
C $x = 1$ only
This is the case-two result on its own, without checking it against case two's own restriction ( $x < -2$ ) — which it fails — and without noticing case one's own valid result, $x=4$ .
DNo solution
There is a genuine solution: $x=4$ satisfies both the equation and case one's restriction, and direct substitution confirms it ( $|4+2|=6=3(4)-6$ ).

Traps tested: Case restriction not checked after solving · Wrong branch selected · Valid solution missed entirely

Question 42 marks

The curve $y = f(x)$ passes through $(4, 10)$ . What are the coordinates of the corresponding point on $y = 3f(x - 2)$ ?

$(6, 30)$
Correct. $f(x-2)$ needs $x - 2 = 4$ , i.e. $x = 6$ — a shift RIGHT by 2 (the input change), giving $(6, 10)$ . Then the output is multiplied by 3: $(6, 30)$ . The two moves act on different coordinates and can be tracked in either order with the same result.
B $(2, 30)$
The horizontal shift has been applied in the wrong direction. $f(x-2)$ shifts the curve RIGHT by 2, not left — $x-2 = 4$ requires $x=6$ , not $x=2$ .
C $(6, 13)$
This adds 3 to the y-coordinate instead of multiplying by it. $y = 3f(x-2)$ scales the output by a factor of 3 ( $10 \to 30$ ), it does not add 3 to it.
D $(4, 30)$
The output has correctly been multiplied by 3, but the horizontal shift has been dropped entirely — the x-coordinate should move from 4 to 6, not stay at 4.

Traps tested: Horizontal shift direction not flipped · Vertical stretch applied as addition · Input change omitted from combined transformation

Question 51 mark

Which of these transformations of $y = f(x)$ will NOT be required on the WMA13 paper?

$y = f(3x + 1)$
Correct. This combines two input changes — a stretch and a translation — inside the same bracket, exactly the excluded form: the spec's own guidance states verbatim that "the graph of y = f(ax + b) will not be required."
B $y = 2f(x)$
A single output change (vertical stretch), one of the four base forms the spec names directly. This is required.
C $y = f(x) - 5$
A single output change (vertical translation), also one of the four base forms. Required.
D $y = f(-x) + 1$
This is the spec's own example of an examinable combination — one input change (reflection, the $a=-1$ case of $f(ax)$ ) and one output change ( $+1$ ), not two input changes sharing a bracket. Required, and explicitly named as such.

Traps tested: Single transformation mistaken for excluded form · Input plus output combination mistaken for excluded form

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jun 2023 · Q6(d) — cited directly in this lesson
Mark scheme: Jan 2022 · Q7(b) — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA13.

The Modulus Function and Combinations of Graph Transformations

The four single transformations, derived rather than listed

Combining two transformations — and the one combination the spec explicitly excludes

Why the input change and the output change can be done in either order

What reflecting the output vs reflecting the input actually does — with real numbers, before any letters

Where y=∣f(x)∣y = |f(x)|y=∣f(x)∣ and y=f(∣x∣)y = f(|x|)y=f(∣x∣) actually come from

Where the exam hides two equations inside |f(x)| = k

Solve |3x − 2| > 2x + 1

Complete the chain — solve |4x + 3| = 2x + 9

Deep dive

Practice this for real

Where $y = |f(x)|$ and $y = f(|x|)$ actually come from