Sketching cubic graphs from factored form

~40 min · WMA11 · 1.11

WMA11 · 1.11 · 40 min

A cubic handed to you already factorised isn't asking you to solve anything — it's showing you the finished sketch, one bracket at a time. Every fact the picture needs (where it meets each axis, whether it crosses or turns at each point, which way it opens) is already sitting in the factors. The one verified WMA11 series that examines this exact sketch records a genuinely mark-losing habit — finding the y-intercept correctly and then never putting it on the picture — sitting right alongside a second, similar-sounding habit (drawing a smooth turning point as a sharp cusp) that the real mark scheme explicitly forgives for this question, even though a genuine polynomial graph never actually has one.

Key terms in this lesson

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.

Reading a cubic straight off its factors

A cubic function is $f(x) = ax^3 + bx^2 + cx + d$ with $a \neq 0$ (the degree-3 version of the $a \neq 0$ condition quadratics carry, and for exactly the same reason: without it, the leading term vanishes and the expression isn't a cubic at all). Spec 1.10 covers factorising a polynomial of degree $n \geq 3$ ; spec 1.11 covers sketching the curve that factorisation describes — this lesson is the second half, taking a cubic already given in factored form and turning it into a labelled picture with no further algebra required.

Every root of $f(x)=0$ is an x-intercept of $y=f(x)$ , and a product of factors is zero exactly when one of the factors is — so each linear factor $(x-p)$ in a factorised cubic hands you a root at $x=p$ directly, with no solving. $f(x)=(x+2)(x-1)(x-4)$ has roots at $x=-2$ , $x=1$ and $x=4$ : read straight off, remembering that $(x+2)$ is $(x-(-2))$ , so its root carries the opposite sign to the constant printed inside the bracket.

But a factorised cubic tells you more than WHERE the curve meets the axis — it tells you HOW. Not every root looks the same on the page. A simple root (the factor appears once) is a genuine crossing: the curve is on one side of the axis just before the root and the other side just after. A repeated root — a factor that appears twice, written $(x-p)^2$ — behaves completely differently: the curve touches the axis at $x=p$ and turns back the way it came, without ever crossing to the other side. Distinguishing these two is not a stylistic choice about how to draw the picture; it is a fact about the function's sign, fully determined by the factorisation, and the next block derives exactly why.

The two things a full sketch needs beyond the roots

Two pieces of information turn a set of labelled x-intercepts into a complete sketch, and neither comes from solving anything further — both are read straight off the same factorised expression the roots came from.

The y-intercept. Every curve $y=f(x)$ meets the y-axis exactly once, at $x=0$ , so the y-intercept is $(0, f(0))$ — substitute $x=0$ into every factor and multiply. For $f(x)=(x+2)(x-1)(x-4)$ : $f(0)=(2)(-1)(-4)=8$ , so the curve meets the y-axis at $(0,8)$ . This is a genuinely separate step from finding the roots, not a byproduct of it, and — the point the whole second half of this lesson is built around — computing it correctly is not the same as putting it on the sketch.

Which way the ends point. For very large $|x|$ , the leading term $ax^3$ dominates every other term in the expansion (it grows in proportion to $x^3$ , while $bx^2$ only grows in proportion to $x^2$ ), so the sign of $a$ alone decides the curve's long-run direction. Because $x^3$ is an ODD power, it keeps the sign of $x$ : for a positive leading coefficient (a "positive cubic", the research bank's own description of the one verified WMA11 example on this exact topic, Oct 2021 Q6(a)) the curve comes from below the axis on the far left and rises above it on the far right; for a negative leading coefficient, the reverse. This is the one place a cubic's end-behaviour genuinely differs from a quadratic's: a parabola's two arms always point the SAME way (both up, or both down), because $x^2$ is an even power and keeps no memory of $x$ 's own sign — a cubic's two ends always point OPPOSITE ways, because $x^3$ does.

One more piece of spec 1.11 worth stating explicitly, since it's easy to read past: "use of intersection points of graphs to solve equations." A correctly labelled sketch doesn't just describe the curve — it answers questions about it. The number of times the horizontal line $y=k$ crosses your sketch is the number of real solutions to $f(x)=k$ , exactly the same principle the used for a line meeting a quadratic (spec 1.4), just applied to any horizontal line and a cubic instead of one specific line and a parabola. A rough end-behaviour argument alone — without needing exact turning-point coordinates — is often enough to answer a large- $|k|$ version of this question: for $k$ far above or below everything the curve reaches near its roots, only the branch heading off to $+\infty$ or $-\infty$ can possibly meet it, so there is exactly one solution.

Push the same sign-tracking one step further and it answers a genuinely different question type: not "where does the curve cross" but "for which x is the curve positive (or negative)?" — a full SOLUTION SET read straight off the sketch, not a single value. Start from the sign fixed at the far left by the leading coefficient (negative, for a positive cubic, since $x\to-\infty$ sends $x^3\to-\infty$ too) and step rightward across each root in turn: a SIMPLE root flips the sign (the same fact the mechanism block below derives, now applied at every root along the whole line rather than at one root in isolation); a REPEATED root leaves it unchanged, for the same reason it produces a touch rather than a crossing. A real WMA11 question does exactly this: $f(x)=(3x+20)(x+6)(2x-3)$ , already sketched with three simple roots at $x=-\tfrac{20}{3}$ , $x=-6$ and $x=\tfrac{3}{2}$ — in that left-to-right order, since $-\tfrac{20}{3}\approx-6.67$ sits to the left of $-6$ — and a positive leading coefficient. Starting negative on the far left, the sign flips at each crossing in turn: positive between $-\tfrac{20}{3}$ and $-6$ , negative again between $-6$ and $\tfrac{3}{2}$ , positive again beyond $\tfrac{3}{2}$ — giving $f(x)>0$ exactly on $-\tfrac{20}{3}<x<-6$ or $x>\tfrac{3}{2}$ : two separate regions, not one, because three simple roots flip the sign three separate times. The second marked-solution block below transcribes this exact real question's own mark scheme.

Mechanism

Why a repeated root always produces a smooth turn, never a cusp

Take a cubic with a factor squared at $x=p$ : $f(x) = (x-p)^2 \cdot g(x)$ , where $g$ bundles whatever the other factor is and, crucially, $g(p) \neq 0$ (so $p$ really is a double root and nothing more). Near $x=p$ , $g(x)$ hasn't had room to change much from its value at $p$ — it's , and $p$ isn't one of ITS roots — so close to $p$ , $f(x)$ behaves almost exactly like $g(p) \cdot (x-p)^2$ : a PARABOLA, sitting at $x=p$ , scaled by whatever $g(p)$ happens to be. That is precisely the shape quadratic-functions.ts already derived for the $b^2-4ac=0$ touching case, just relocated onto a cubic's repeated root instead of a whole quadratic's vertex. And a parabola's vertex, established in that same derivation, is always smooth — the curve slows to a stop and reverses direction continuously, the way $(x-p)^2$ itself does at $x=p$ , never with the sudden change of direction a cusp would need. A cusp looks like a sharp 'V': two straightish pieces of curve meeting at a point with an abrupt kink, the way $y=|x|$ behaves at the origin. A polynomial graph is never shaped like that anywhere — not at a repeated root, not anywhere else — because near any point it is locally well-approximated by a smooth power of $(x-p)$ , and every one of those powers curves round continuously rather than folding sharply. This is also why the repeated root always sits at a genuine rather than merely a flat-looking wiggle: the curve is forced to approach zero, touch, and head back the way it came, which is exactly what "turning point" means. Contrast this with a SIMPLE root: $f(x)=(x-p)\cdot g(x)$ with $g(p)\neq 0$ behaves, near $p$ , like $g(p)\cdot(x-p)$ — a straight LINE through zero, not a parabola — and a line through a point genuinely crosses from one side to the other, because $(x-p)$ itself changes sign there while $g(x)$ keeps a steady sign nearby. Same local-approximation idea, two different shapes, because the power of $(x-p)$ involved is different: 1 gives a line (crosses), 2 gives a parabola (touches, smoothly).

Diagram — y = (x+1)(x−2)² — a positive cubic with one simple root and one repeated root

x-axis: x · y-axis: y

y = (x+1)(x−2)²: Leading coefficient +1 (a positive cubic once expanded to x³−3x²+4), root at x=−1 (simple, factor appears once) and a repeated root at x=2 (factor squared). Comes from below the axis on the far left, crosses upward through (−1,0), rises to a local maximum, falls back to touch the axis at (2,0) — a local minimum, sitting exactly on the axis — then rises again to above the axis on the far right.

(−1, 0) — simple root, genuine crossing: The curve is below the axis just before x=−1 and above it just after: a real sign change, drawn as the curve passing straight through the axis at an angle, not tangentially.
(2, 0) — repeated root, touches and turns smoothly: The curve is above the axis on both sides of x=2 and reaches exactly zero at x=2 itself: no sign change, so the curve does not cross — it dips down to touch the axis and curves smoothly back up, the same shape as a parabola's vertex sitting on the x-axis.
(0, 4) — y-intercept: f(0) = (1)(−2)² = (1)(4) = 4. A separate calculation from either root, and — per the documented error this diagram exists to correct — a separate thing to remember to WRITE on the sketch, not just to have computed on scratch paper.

Common error: Drawing a sharp cusp — a pointed 'V' shape — at the repeated-root turning point, x=2.

Correct: The turning point at a repeated root is always smooth, curving round continuously exactly like a parabola's vertex. A cubic graph is never shaped like |x| anywhere on its domain — it is a polynomial, and every polynomial curve turns smoothly. (The real mark scheme for the question this quote comes from explicitly condones a cusp-like appearance when awarding its shape mark — so this habit did not actually cost marks on that specific paper — but the underlying maths is still what a correct sketch should show.)

examiner-report · Oct 2021 · Q6(a)

Worked, in full

Sketch $y = (x+1)(x-2)^2$ from the factored form alone — every feature, every label

01
Identify every root and its multiplicity from the factors. $(x+1)$ appears once: a simple root at $x=-1$ . $(x-2)$ appears squared: a repeated root at $x=2$ , multiplicity 2.
Earns: Method for locating x-intercepts — each factor set to zero, with its power noted.
02
Decide crossing vs touching for each root using the multiplicity alone, not by testing points yet. Odd multiplicity (here, 1) means a genuine sign change: the curve CROSSES at $x=-1$ . Even multiplicity (here, 2) means the squared factor never goes negative: the curve TOUCHES at $x=2$ and turns back — smoothly, per the mechanism above, never as a cusp.
Earns: Correct crossing/touching classification for both roots, from multiplicity alone.
03
Find the y-intercept. $f(0) = (0+1)(0-2)^2 = (1)(4) = 4$ , so the curve meets the y-axis at $(0, 4)$ .
Earns: Correct y-intercept, calculated — the step the documented trap shows is not, on its own, enough.
04
Determine the end-behaviour. Expanding (or simply multiplying the leading terms of each factor, $x \cdot x^2 = x^3$ ) gives a leading coefficient of $+1$ : a positive cubic. The curve comes from below the axis on the far left and rises to above it on the far right.
Earns: Correct end-behaviour direction from the sign of the leading coefficient.
05
Assemble the sketch and label every feature found above: $(-1,0)$ with the curve crossing through it; $(2,0)$ with the curve touching and smoothly turning back; $(0,4)$ marked on the y-axis; the curve rising from bottom-left to top-right overall. (Not needed for the sketch itself, but as an optional check: expanding gives $f(x)=x^3-3x^2+4$ , and spec 4.2's power rule gives $f'(x)=3x^2-6x=3x(x-2)$ , zero at $x=0$ and $x=2$ — confirming the repeated root at $x=2$ really is a turning point, and that the other turning point sits at $x=0$ , which happens here to coincide with the y-intercept.)
Earns: A complete, correctly-labelled sketch — every coordinate found above actually present on the drawing.

Source — Examiner report, Oct 2021

"failed to label the y intercept"

Complete it yourself

Complete the chain — sketch $y = (x+3)(x-1)^2$ from the factored form

01
Identify the roots and multiplicities. $(x+3)$ appears once: a simple root at $x=-3$ . $(x-1)$ appears squared: a repeated root at $x=1$ , multiplicity 2.
02
Classify each root by its multiplicity. Odd multiplicity at $x=-3$ (power 1): the curve CROSSES. Even multiplicity at $x=1$ (power 2): the curve TOUCHES and turns back smoothly.

Marked, line by line

$f(x) = (x+2)(x-1)(x-3)$ . (a) Find the coordinates of the point where the curve $y=f(x)$ meets the y-axis. (2) (b) Sketch the curve $y=f(x)$ , showing clearly the coordinates of the points where the curve meets the coordinate axes. (3) — VERIDIAN-original question, written in the style real WMA11 factorised-cubic sketch questions use (a short 'find the y-intercept' part followed by a 'sketch, showing the axis-crossing coordinates' part is the natural shape spec 1.10/1.11 combine into). Not a reproduction of Oct 2021 Q6(a) or any other past-paper question — the research bank supplies no question text for that series to reproduce. Per-line mark allocations are VERIDIAN-modelled on the general marking-guidance conventions in WMA11-verified-facts.md §3, not copied from a real mark scheme.

5 marks available

(a) — 2 marks

01
$f(0) = (0+2)(0-1)(0-3)$
Method mark for substituting x=0 into every factor.
M1
02
$f(0) = (2)(-1)(-3) = 6$ , so the curve meets the y-axis at $(0, 6)$ .
Accuracy mark for the correct coordinate — not just the value 6, but stated as the point (0,6).
A1

(b) — 3 marks

101
x-intercepts at $(-2, 0)$ , $(1, 0)$ and $(3, 0)$ , from setting each factor to zero.
Independent mark for all three x-intercepts correctly found and shown — no method beyond reading the factors is required, which is exactly what a B mark is for: an independent accuracy mark where there is no method, and the general marking guidance's own example of when a B mark applies is "often given for a comment or for a graph."
B1
102
Curve drawn coming from below the axis on the far left, crossing at each of the three x-intercepts in turn (each root here is simple, so every crossing is a genuine sign change, none a touch), and rising above the axis on the far right.
Independent mark for correct overall shape: three genuine crossings (since every factor here appears once — no repeated root to touch at), and correct end-behaviour (positive cubic, leading coefficient +1, bottom-left to top-right).
B1
103
$(0, 6)$ — found in part (a) — marked explicitly on the sketch itself.
Independent mark for the y-intercept appearing ON THE DRAWING, not merely calculated in part (a). This is a separate mark from part (a)'s A1: a candidate can find the correct value and still lose this one by never transferring it to the picture — see the wrong path below.
B1

Marked, line by line

Figure 4 shows a sketch of part of the curve $C$ with equation $y=f(x)$ , where $f(x) = (3x+20)(x+6)(2x-3)$ . Use the given information to state the values of $x$ for which $f(x) > 0$ . (2) — REAL WMA11 exam question, Jan 2023 Q10(a) (Publications Code WMA11_01_MS_2301, QP Log Number P72066A), re-extracted from the real mark scheme and examiner report PDFs for this lesson's mark-scheme-bullet coverage audit rather than transcribed from any prior summary — this is the second of this lesson's two named research-bank anchors (WMA11-verified-facts.md §4.1) getting a genuinely real transcription. The real question's part (b) (expand the three brackets) and part (c) (find where the tangent at the y-intercept cuts $C$ again, using differentiation) test different spec points and aren't reproduced here — only part (a), the graph-reading skill this lesson is actually about.

2 marks available

(a) — 2 marks

01
One correct region stated, e.g. $x > \tfrac{3}{2}$ (condoned with $\leq$ / $\geq$ for this mark, and equivalent notation such as $x \in (\tfrac{3}{2}, \infty)$ also accepted).
Real mark-scheme wording: 'One of $-\tfrac{20}{3} < x < -6$ , $x > \tfrac{3}{2}$ ' — either region alone, with either inequality style, earns this mark on its own.
M1
02
Both correct regions stated together: $-\tfrac{20}{3} < x < -6$ or $x > \tfrac{3}{2}$ .
Real mark-scheme wording: 'Both $-\tfrac{20}{3} < x < -6$ , $x > \tfrac{3}{2}$ ' — the second region, correctly bounded on both sides and correctly ordered (the more negative value, $-\tfrac{20}{3}\approx-6.67$ , written first), is what this mark actually checks; the mark scheme's own notation leniency only covers the M1 mark above, not this one.
A1

Named traps

cusp-at-a-repeated-root-turning-point: A real WMA11 examiner report on a factored-cubic sketch question states directly: "a few candidates were inclined to feature a cusp at the minimum point" (Oct 2021, Q6(a)) rather than a smooth turning point. The mechanism above explains why this is never mathematically correct: near a repeated root, a cubic behaves exactly like a scaled parabola, and a parabola's vertex is always smooth — a polynomial graph is never shaped like a sharp 'V' anywhere on its domain. Worth stating precisely, though, since it is easy to overstate: the REAL mark scheme for this exact question explicitly condones it — "Condone with no axes and condone cusp like appearance for the turning points" is the mark scheme's own wording for the shape-method mark — so this specific habit did not, in fact, cost the candidates who did it any marks on this specific paper. That is a fact about this one scheme's leniency, not a licence to draw one: the actual mathematics of the curve still calls it wrong, and the y-intercept-omission trap below shows what a mark scheme looks like when a comparable-sounding habit genuinely does cost the mark.
y-intercept-left-unlabelled: A second verified, quoted trap from the same question: "[a surprising number of otherwise-correct sketches] failed to label the y intercept" (Oct 2021, Q6(a)). The specific failure mode worth naming is that this is NOT usually a calculation error — the value is often found correctly elsewhere in the working — it is an omission at the very last step, transferring a number that already exists onto the picture the question actually asked for. Treat "find it" and "label it on the sketch" as two separate items on a checklist, not one. Unlike the cusp trap above, this one is not condoned: the real mark scheme ties an independent B1 mark specifically to the y-intercept appearing ON the sketch, and that mark is genuinely lost when it doesn't.
critical-values-stated-instead-of-inequalities: A real, examiner-confirmed trap on the real WMA11 question the second marked-solution above transcribes: "Both marks were lost by a large proportion of candidates who wrote down one or more critical values only, not realising that inequalities were required" (Jan 2023, Q10(a)). Finding the roots (the critical values) is necessary but not sufficient — the question asks for every x where $f(x)>0$ , which describes REGIONS between and beyond those values, not a list of the values themselves. The roots are the boundary of the answer, not the answer.
compound-inequality-region-not-bounded-both-sides: A second real, examiner-confirmed trap from the same question, distinct from writing critical values alone: "Of those who scored just one mark it was usually for writing $x>\tfrac{3}{2}$ together with an error such as $x>-6$ " (Jan 2023, Q10(a)) — one region built correctly, the other built as a one-sided inequality when the sign pattern demands it be bounded on BOTH sides. This happens whenever a cubic has three simple roots: crossing the first root turns the sign positive, but crossing the SECOND root turns it negative again, so the positive stretch that started at the first root has already ended by the second — an inequality stated from only one boundary silently keeps going past the point where the sign changed back.
compound-inequality-boundary-order-reversed: A third real, examiner-confirmed error on the same question, different again from the two above: "Some responses had inequality signs used incorrectly e.g. $-6 < x < -\tfrac{20}{3}$ " (Jan 2023, Q10(a)) — the two correct boundary VALUES, written in the wrong order. Since $-\tfrac{20}{3}\approx-6.67$ is more negative than $-6$ , the compound inequality $-6<x<-\tfrac{20}{3}$ describes no real values of x at all — nothing is simultaneously greater than $-6$ and less than a smaller number — a fluent-looking answer that is actually empty. The smaller (more negative) value always goes on the left of a compound inequality.
repeated-root-treated-as-a-crossing: Not a phrase the research bank's examiner-report extracts quote by name for this topic — named here because it is the natural counterpart failure to the cusp trap above, and a real risk this lesson would understate by only covering the smoothness of the turning point and not whether a turning point is drawn at all. Drawing the curve crossing straight through a repeated root, the way it would at a simple root, ignores that a squared factor can never change sign: there is no sign change to cross through, so the curve must touch and turn back, not pass through.
end-behaviour-direction-reversed: Also not a phrase the research bank quotes for this specific topic — named here as a mechanically-motivated risk, not a documented one. A cubic's two ends point in OPPOSITE directions (unlike a quadratic's, which always point the same way), and it is a genuinely easy slip to import quadratic intuition and draw both ends of a positive cubic curving upward, or to read the sign of the leading coefficient backwards. Check with one value far from every root: for a positive cubic, $f(-100)$ should come out strongly negative and $f(100)$ strongly positive.

In your own words

In one sentence: why must a repeated (squared) factor's root always produce a smooth turning point on a cubic's graph, never a sharp cusp?

Beyond the spec

Spec 1.10/1.11 covers factorisation and sketching for "simple cubics," and every verified WMA11 example in this research pass uses roots of multiplicity 1 or 2 only. Seeing what a multiplicity-3 root does removes a natural but wrong generalisation this lesson could otherwise leave behind: that ANY repeated factor means "touches." It doesn't — it depends on whether the repetition count is even or odd, and multiplicity 2 is only the smallest even case.

Take $f(x) = (x-p)^3 \cdot h(x)$ with $h(p) \neq 0$ — a TRIPLE root. Near $x=p$ , this behaves like $h(p)\cdot(x-p)^3$ , the shape of $y=x^3$ itself relocated to sit at $p$ : the gradient is zero there (it flattens, the way $y=x^3$ flattens at the origin), but $(x-p)^3$ is an ODD power, so it still changes sign as $x$ passes through $p$ — negative just before, zero at $p$ , positive just after (or the reverse, depending on the sign of $h(p)$ ). The curve does not turn back the way it does at a double root; it flattens out horizontally for an instant and then continues through to the other side. This is called a : stationary, because the gradient is momentarily zero, but an inflection rather than a turning point, because the curve keeps going in the same overall direction rather than reversing. The general pattern, of which multiplicity 1 and 2 are the two cases this lesson actually needs: ODD multiplicity (1, 3, 5, ...) always crosses (multiplicity 1 at full speed, multiplicity 3 flattening through, multiplicity 5 flattening even more) and EVEN multiplicity (2, 4, ...) always touches and turns back. It is the parity of the exponent, not "repeated vs. not," that is the real determining fact — and P1 only ever asks about the smallest instance of each: 1 and 2.

Retrieval — with feedback on every choice

Question 1

3 marks

Which is a correct, fully-labelled description of the graph of $y = (x+1)(x-2)^2$ ?

Question 2

3 marks

$g(x) = -(x+3)(x-2)^2$ . What is the y-intercept, and which way does the curve open?

Question 3

2 marks

$(x+4)(x-1)^2 = 0$ . How many DISTINCT real values of $x$ satisfy this equation?

Question 4

2 marks

A real WMA11 question sketches $f(x) = (3x+20)(x+6)(2x-3)$ — three simple roots at $x=-\tfrac{20}{3}$ , $x=-6$ and $x=\tfrac{3}{2}$ , positive leading coefficient. For which values of $x$ is $f(x) > 0$ ?

Reference — not a study method, a lookup

Roots straight from the factors: (x−p) → root at p. Set each factor to zero, watching for sign flips.
Simple factor (power 1) → curve CROSSES. Squared factor (power 2) → curve TOUCHES and turns back smoothly — never a cusp.
y-intercept: substitute x=0 into every factor, multiply. Finding it and labelling it ON the sketch are two separate marks.
Leading coefficient positive → bottom-left to top-right ('positive cubic'). Negative → top-left to bottom-right — opposite ends, unlike a quadratic's matching arms.
Reading f(x)>0 (or <0) off the sketch: start from the far-left sign fixed by the leading coefficient, then flip the sign at every SIMPLE root crossed (a REPEATED root leaves it unchanged) — three simple roots can produce two separate positive regions, not one.
A cubic has at most 2 turning points, never 1 — and a repeated (squared) root is always exactly one of them.

Not affiliated with or endorsed by Pearson Edexcel. This lesson has been through a mark-scheme-bullet coverage audit (course-ultra §1.2+§1.4) since PR #352's four-persona council: both real anchor questions the research bank names for this exact topic (WMA11-verified-facts.md §4.1, "Jan 2023 Q10, Oct 2021 Q6") were re-extracted directly from the actual Pearson Edexcel IAL WMA11/01 mark scheme and examiner report PDFs for this pass (cross-checked with pdftotext -layout AND -raw, every numeric claim independently recomputed), not trusted from the research bank's own prior summary or from this file's own earlier text. Two real findings resulted. First: Jan 2023 Q10(a) (Publications Code WMA11_01_MS_2301, QP Log Number P72066A) — $f(x)=(3x+20)(x+6)(2x-3)$ , state where $f(x)>0$ from a given sketch — was a genuinely untaught spec-1.11 subskill (reading a solution set, not just a shape, off a sign pattern); it now has real coverage above (a teach paragraph, a second marked-solution block transcribing the real M1/A1 mark scheme, three named traps, and a fourth MCQ), all built from the real question, real mark scheme and real examiner report rather than invented. Second, and a correction to this file's own earlier framing: Oct 2021 Q6(a) (Publications Code WMA11_01_2110_MS, QP Log Number P66645A) is real — $f(x)=2(x+1)(x-3)^2$ — and this lesson's two quotes from its examiner report ('a few candidates were inclined to feature a cusp at the minimum point [on the x-axis]' and '[a surprising number of otherwise-correct sketches] failed to label the y intercept [at (0, 18)]') both remain accurate quotes, but this lesson previously treated them as two comparably mark-costly habits. They are not: the real mark scheme's own wording for the shape-method mark reads 'Condone with no axes and condone cusp like appearance for the turning points' — the cusp habit was noted but explicitly forgiven under this exact scheme, while the y-intercept omission genuinely costs its own independent B1 mark. The lede, worked content, diagram and trap-taxonomy above are corrected to hold that distinction accurately: a cusp is still mathematically wrong on every polynomial graph, full stop, but it was not, in fact, equally mark-costly on the one real series this lesson cites it from. Every OTHER specific cubic used throughout this lesson — f(x)=(x+1)(x−2)², h(x)=(x+3)(x−1)², g(x)=−(x+3)(x−2)², and the first marked-solution's three-distinct-root example f(x)=(x+2)(x−1)(x−3) — remains VERIDIAN-original, independently hand-verified, and not a reproduction of either real question. The first marked-solution's per-line mark codes remain VERIDIAN-modelled on the general marking-guidance conventions verbatim-verified in the research bank's §3 (including the B-mark definition quoted directly in that block: 'often given for a comment or for a graph'), because that question is itself VERIDIAN-original and has no real scheme to transcribe from; the SECOND marked-solution block transcribes a real scheme instead, as detailed above. Two of the trap-taxonomy items (repeated-root-treated-as-a-crossing, end-behaviour-direction-reversed) remain explicitly flagged in their own text as mechanically-motivated risks this lesson names for completeness, not phrases either real examiner report quotes for this specific topic — the cusp, y-intercept, and all three new Jan 2023 Q10(a) traps carry direct citation. Full bullet-by-bullet accounting of this audit pass is logged in research/veridian/WMA11-verified-facts.md's "Lesson-audit log" section.

Question 13 marks

Which is a correct, fully-labelled description of the graph of $y = (x+1)(x-2)^2$ ?

Crosses the x-axis at $(-1,0)$ ; touches and turns back smoothly at $(2,0)$ ; crosses the y-axis at $(0,4)$ ; rises from below the axis on the far left to above it on the far right.
Correct — every feature present and correctly classified: simple root crosses, repeated root touches (smoothly, never a cusp), y-intercept found AND stated, end-behaviour matches the positive leading coefficient.
BSame as above, but drawn with a sharp cusp at $(2,0)$ instead of a smooth turn
This is the documented, quoted error on this exact question type: "a few candidates were inclined to feature a cusp at the minimum point" (Oct 2021, Q6(a)). A repeated root always produces a smooth turning point — the curve locally behaves like a scaled parabola there, and a parabola's vertex never has a sharp corner.
CSame as A, but the curve crosses straight through the axis at $(2,0)$ exactly as it does at $(-1,0)$
A repeated (squared) factor can never change sign, so there is no sign change to cross through at x=2 — the curve must touch and turn back. Treating a double root the same way as a simple one ignores exactly the information the squared factor was there to convey.
DSame shape as A, but with no y-intercept marked on the sketch, since the question only asked for where the curve meets the AXES and the y-intercept was already found separately
The y-intercept IS one of the points where the curve meets the coordinate axes — it meets the y-axis there. A real examiner report records this exact omission on otherwise-correct sketches: "[a surprising number of otherwise-correct sketches] failed to label the y intercept" (Oct 2021, Q6(a)). Finding a value and putting it on the picture are two separate steps.

Traps tested: Cusp at a repeated root turning point · Repeated root treated as a crossing · Y intercept left unlabelled

Question 23 marks

$g(x) = -(x+3)(x-2)^2$ . What is the y-intercept, and which way does the curve open?

$(0, -12)$ ; comes from ABOVE the axis on the far left and falls below it on the far right
Correct. $g(0)=-(3)(-2)^2=-(3)(4)=-12$ . Expanding gives leading coefficient $-1$ (a negative cubic), so the two ends point opposite ways from a positive cubic: down on the right, up on the left.
B $(0, -12)$ ; comes from below the axis on the far left and rises above it on the far right
The y-intercept value is right; the end-behaviour direction is backwards. This is the negative-leading-coefficient case: the two ends point opposite ways to a POSITIVE cubic, not the same way. Checking one far value settles it: $g(-100)$ is strongly POSITIVE here (the overall minus sign combined with a large negative $(x+3)$ makes the product positive), not negative.
C $(0, 6)$ ; falls from above to below the axis
The repeated factor $(x-2)^2$ has been evaluated as $(0-2)=-2$ instead of $(0-2)^2=4$ — the square was dropped. Correctly, $g(0)=-(3)(4)=-12$ , not $-(3)(-2)=6$ .
D $(0, 12)$ ; falls from above to below the axis
The end-behaviour direction here is right, but the overall minus sign in front of the whole expression has been dropped: $g(0) = -(3)(4) = -12$ , not $+12$ . The minus sign outside the brackets applies to the WHOLE product, not just to one factor inside it.

Traps tested: End behaviour direction reversed · Repeated factor not squared when evaluating · Overall negative coefficient dropped

Question 32 marks

$(x+4)(x-1)^2 = 0$ . How many DISTINCT real values of $x$ satisfy this equation?

$2$ — $x=-4$ and $x=1$ , even though $x=1$ is a repeated root counted twice inside the factorisation
Correct. The equation has three linear factors in total, but two of them are identical — $(x-1)$ appearing twice contributes only ONE distinct x-value, $x=1$ , however many times it's counted in the multiplicity.
B $3$ — there are three factors in total, so three roots
This counts $x=1$ twice because the factor $(x-1)$ appears twice — correct for the polynomial's total DEGREE and its multiplicity count, but not for the number of genuinely different x-values that solve the equation, which is what "distinct" asks for.
C $1$ — only $x=-4$ , since a squared factor doesn't produce a real solution
A squared factor being zero is still the factor being zero: $(x-1)^2=0$ genuinely requires $x-1=0$ , i.e. $x=1$ , which is a perfectly real value. Squaring does not remove a root from existing — it only removes the SIGN CHANGE that would otherwise happen there.
DIt cannot be determined without expanding the equation first
The factored form already states every root and its multiplicity directly — expanding would only make this information harder to see, not easier.

Traps tested: Multiplicity confused with distinct intercept count · Squared factor wrongly excluded from real roots · Factored form treated as insufficient for root count

Question 42 marks

$-\tfrac{20}{3} < x < -6$ or $x > \tfrac{3}{2}$
Correct — the real mark scheme's own answer. Starting negative on the far left, the sign flips at every simple root crossed: positive between $-\tfrac{20}{3}$ and $-6$ , negative again between $-6$ and $\tfrac{3}{2}$ , positive again beyond $\tfrac{3}{2}$ — two separate positive regions, because the sign changes back and forth three times across three simple roots.
B $x > -6$
This is the real, documented one-mark error on this exact question: correctly finds the right-hand boundary of the FIRST positive region but then treats everything above $-6$ as positive, silently swallowing the genuinely negative stretch between $-6$ and $\tfrac{3}{2}$ where the sign has already flipped back.
C $-6 < x < -\tfrac{20}{3}$ or $x > \tfrac{3}{2}$
The two boundary values for the first region are the right numbers, in the wrong order. $-\tfrac{20}{3}\approx-6.67$ is MORE negative than $-6$ , so it belongs on the left of the inequality; written the other way round, $-6<x<-\tfrac{20}{3}$ describes no real values of x at all.
D $x = -\tfrac{20}{3}, -6, \tfrac{3}{2}$
These are the critical values (the roots) — necessary for finding the answer, but not the answer itself. The question asks where $f(x)$ is POSITIVE, which describes the regions between and beyond these values, not a list of the values where $f(x)=0$ .

Traps tested: Compound inequality region not bounded both sides · Compound inequality boundary order reversed · Critical values stated instead of inequalities

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: Oct 2021 · Q6(a) — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA11.

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Up next

The Sine Rule, the Cosine Rule, and the Ambiguous Case

Two rules solve every non-right-angled triangle between them, and only one of the two can ever hand you two correct answers from the same working. The sine rule and the cosine rule both find missing sides and missing angles — but only the sine rule, used to find an angle, can leave you with a genuine fork: \sin\theta and \sin(180° - \theta) are the same number, so the same piece of working can end in either an acute triangle or an obtuse one, and the algebra alone never tells you which. The one verified WMA11 examiner finding on this exact spec point — spec 3.1's own words name it, "including the ambiguous case of the sine rule" — describes this landing on real scripts, not staying a textbook warning: most students reached the correct acute angle and stopped, when the question needed the other one.

60 min