Deep Dive: Quadratic Functions

*Provenance: compiled 2026-09-29. Read in full: research/veridian/WMA11-verified-facts.md (grepped for "quadratic," "factoris," "turning point," "completing the square," "3TQ" and "sketch," then read at §1's spec table, the General Principles for Pure Mathematics Marking in §3, and every §4.1 entry those greps surfaced — Jan 2023 Q4, Jan 2019 Q9, Jan 2023 Q5, Jan 2025 Q4, Jan 2023 Q7(c)-(d), Jan 2019 Q11(b)-(c), Oct 2021 Q2 and Q6(a), June 2022 Q5(a), June 2023 Q1-Q2, plus §6/§7's coverage and gap notes); the live lesson file content/veridian/curriculum/wma11/quadratic-functions.ts (read in full: header, all three prequestions, both teach blocks, both mechanism blocks, the diagram, the method-comparison, worked-chain, chain-drill, the self-explain prompt, the beyond-spec block, the second teach block, both marked-solutions, all seven trap-taxonomy items, all six MCQs, reference card and closing flag); signal-topic-brief-quadratic-functions.md (read in full); codex-deep-dive-discriminant-with-a-parameter.md (read in full, for scope boundary and to avoid duplicating its §2 formula or its two real anchors); codex-formula-and-method-bank.md §1.1.3 and §2.1.1 (read in full); forge-exam-technique-guide.md §2.2 and forge-vocabulary-precision-list.md (grepped for "quadratic," relevant hits read); signal-frequency-and-boundaries.md (grepped, relevant lines read); content/veridian/question-bank/wma11.ts (checked for overlap — confirms the June 2022 Q5(a) item already seeded there under SPEC_QUADRATIC_REVERSE, not re-seeded here). No new primary-source PDF was fetched for this document; every real-exam fact below is drawn from citations already verified in the sources just listed. Every worked micro-example's arithmetic (all factorisations, completed-square forms, discriminants and roots) was independently checked by hand — expanding each factorisation back out and substituting each claimed root back into the original expression — before being written in here.

#What this document is, and isn't

A single-topic reference for spec 1.3 (quadratic functions and their graphs) and 1.5 (completing the square; solving quadratics by factorisation, formula or completing the square), scoped exactly as signal-topic-brief-quadratic-functions.md already scopes it: core quadratic-function knowledge tested directly, not as a step buried inside a different question type. Four close relatives are deliberately left alone here because each already has its own document or lesson: the discriminant with a parameter (spec 1.4) has its own CODEX deep-dive; quadratic/reciprocal inequalities interpreted graphically (spec 1.7–1.9), "hidden quadratic" substitution (spec 1.1/1.5/1.6), and reconstructing $f(x)$ from a turning point plus a root (the lesson's own second half, spec 1.3/1.5's one confirmed real standalone anchor) all have full worked treatments elsewhere in this course and are not reconstructed below.

What follows instead is the theory the lesson's own prose uses but states only in passing or not at all: the three equivalent algebraic forms of a quadratic and the exact identities converting between them; the factorisation mechanism itself — what the mark scheme's own $|mn|=|a|$, $|pq|=|c|$ condition actually means and how it behaves across every genuinely distinct shape a three-term quadratic can take (repeated root, difference of two squares, a vanishing constant term, no rational factorisation at all); a compact sketching procedure stated as a single reusable method; and a synthesis that separates, honestly and by count, how often this topic is the actual point of a WMA11 question from how often its mechanics are merely a sub-step buried inside a different one.

Two labels are used consistently below:

• [real] — a verified fact from a genuine Pearson question paper, mark scheme or examiner report, with series and question number given.
• [VERIDIAN, illustrative] — a fresh invented micro-example built to demonstrate a mechanism, with no claim to resemble any specific real Pearson question.

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#1. The object itself: three forms, one identity connecting them

A quadratic function is $f(x)=ax^2+bx+c$ with $a,b,c$ constants and $a\neq0$ (spec 1.3) — the graph of $y=f(x)$ is a parabola, and every fact about it in this section follows from being able to move freely between three ways of writing the same expression:

• General (expanded) form: $ax^2+bx+c$. Read off directly: the $y$-intercept is $(0,c)$, since $f(0)=c$.
• Factorised form: $a(x-\alpha)(x-\beta)$, where $\alpha,\beta$ are the roots. Read off directly: the $x$-intercepts are $(\alpha,0)$ and $(\beta,0)$.
• Vertex (completed-square) form: $a(x-h)^2+k$. Read off directly: the turning point is $(h,k)$ — a minimum if $a>0$, a maximum if $a<0$ — and the line of symmetry is $x=h$.

The lesson's own mechanism block derives the general-to-vertex identity in full ($h=-\frac{b}{2a}$, $k=-\frac{b^2-4ac}{4a}$) and is not re-derived here. What is worth stating once, compactly, is that all three forms are the same object — going from any one to any other is always either an expansion (multiply factors, or expand a square: mechanical, never fails) or a solve (find $\alpha,\beta$ by factorising or the formula; find $h,k$ by completing the square). The three "methods" for solving a quadratic that the mark scheme names are, from this angle, just the three routes to whichever form you don't yet have.

A running example, threaded through the rest of this document: $f(x)=6x^2+7x-3$.

• Factorised form: $(3x-1)(2x+3)$ — check: $(3x-1)(2x+3)=6x^2+9x-2x-3=6x^2+7x-3$. ✓ Roots $x=\frac13$ and $x=-\frac32$.
• Vertex form: $6\left(x^2+\frac76x\right)-3 = 6\left[\left(x+\frac7{12}\right)^2-\frac{49}{144}\right]-3 = 6\left(x+\frac7{12}\right)^2-\frac{121}{24}$ — check: $6\times\frac{49}{144}=\frac{49}{24}$, and $-\frac{49}{24}-3=-\frac{121}{24}$. ✓ Turning point $\left(-\frac7{12},-\frac{121}{24}\right)$, a minimum since $a=6>0$.
• Discriminant: $b^2-4ac=49+72=121=11^2$ — a perfect square, which is exactly why an integer/rational factorisation existed to find (§2 below makes this a theorem, not a coincidence). Formula check: $x=\frac{-7\pm11}{12}=\frac{4}{12}\text{ or }\frac{-18}{12}=\frac13\text{ or }-\frac32$. Matches the factorised roots exactly, as it must.

#2. Factorisation, unpacked: what the mark scheme's condition actually means, and where it breaks down

WMA11-verified-facts.md §3 (General Principles for Pure Mathematics Marking) states the factorisation route to the three-term-quadratic method mark with a precise, checkable condition: seek $(mx+p)(nx+q)$ with $mn=a$ and $pq=c$ (given as $|mn|=|a|$, $|pq|=|c|$ — magnitude-matched, because the method mark credits finding a factor pair of the right size, before the signs are necessarily sorted) [real, verified verbatim]. Two consequences worth stating precisely, because neither is obvious just from the condition itself.

First, why the discriminant predicts whether the search can succeed at all. From the formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. If $a,b,c$ are integers, the roots are rational if and only if $\sqrt{b^2-4ac}$ is rational — and a rational square root of an integer exists only when that integer is a perfect square. So: $b^2-4ac$ is a perfect square $\iff$ both roots are rational $\iff$ a factorisation of the form $a(x-\alpha)(x-\beta)$ with rational $\alpha,\beta$ exists (and, after clearing denominators, an integer factor pair $(mx+p)(nx+q)$ exists too). This is not a heuristic to memorise — it is a direct consequence of the formula already derived in the lesson, and it is the actual reason "check the discriminant first" is sound exam advice (already given as a checklist step in codex-formula-and-method-bank.md §2.1.1) rather than a folk rule: a non-square discriminant doesn't just make factorisation harder to find — it makes it provably impossible over the rationals, so no amount of searching for integer factor pairs will ever succeed.

Second, the condition degrades gracefully into every special case below — none of them is a separate rule, each is the same $mn=a$, $pq=c$ condition with one part of the general shape trivial or absent.

#2.1 $a=1$ — the condition simplifies to "two numbers multiplying to $c$, adding to $b$"

[VERIDIAN, illustrative] $x^2+x-12$: need $pq=-12$, $p+q=1$ (since $m=n=1$, the cross-term coefficient is just $p+q$). The pair $4,-3$ works: $4\times(-3)=-12$, $4+(-3)=1$. So $x^2+x-12=(x+4)(x-3)$ — check: $x^2+4x-3x-12=x^2+x-12$. ✓ Roots $x=-4$ and $x=3$.

#2.2 $a\neq1$ — both $m,n$ and $p,q$ have to be found together

The running example already did this: $6x^2+7x-3=(3x-1)(2x+3)$, with $mn=3\times2=6=|a|$ and $pq=(-1)\times3=-3$, so $|pq|=3=|c|$ — both parts of the mark scheme's condition satisfied, and the middle-term check ($3x\times3+(-1)\times2x=9x-2x=7x$) confirms the specific pairing of signs, which the magnitude condition alone does not fix. This is the genuine extra difficulty of $a\neq1$ over $a=1$: satisfying $mn=a,\ pq=c$ is necessary but not sufficient, because several factor pairs of $a$ and of $c$ can satisfy the magnitude condition while only one specific combination of them reproduces the correct middle term.

#2.3 $b=0$ — a difference of two squares, whenever $a$ and $c$ have opposite signs

[VERIDIAN, illustrative] $9x^2-16=0$. Here $p+q$ (the coefficient of $x$) must be $0$, forcing $q=-p$, so the condition becomes $mn=9$, $-p^2=-16\Rightarrow p^2=16\Rightarrow p=\pm4$. Taking $m=n=3,\ p=-4,\ q=4$: $(3x-4)(3x+4)=9x^2+12x-12x-16=9x^2-16$. ✓ Roots $x=\pm\frac43$. This is the familiar identity $a^2-b^2=(a-b)(a+b)$ applied to $9x^2-16=(3x)^2-4^2$ — a genuine, common sub-case of the same $mn=a,pq=c$ condition, not a different technique, and it only produces two real roots because $a$ and $c$ have opposite signs here; if they had the same sign ($9x^2+16=0$, say), $p^2=-16$ has no real solution and the discriminant test confirms it directly ($0-4(9)(16)<0$).

#2.4 $c=0$ — factor out $x$, and don't lose the root doing it

[VERIDIAN, illustrative] $x^2+5x=0$: the general condition still applies with $q=0$ forced (since $pq=c=0$ needs one of $p,q$ to vanish), which is exactly what "factor $x$ out" means: $x(x+5)=0$, so $x=0$ or $x=-5$. The one live trap here is mechanical and structurally identical to a real, cited WMA11 error family: dividing both sides by $x$ to simplify — turning $x^2+5x=0$ into $x+5=0$ — silently discards the root $x=0$, in exactly the way that dividing $36k^2-20k<0$ through by $k$ was confirmed, on a real script, to discard $k=0$ ("many candidates ignored the solution, $k=0$" [real, Jan 2023 Q4]). No real WMA11 anchor testing this exact bare $c=0$ shape has been found in this course's research — the parallel is a structural one (dividing an equation by an unknown-sign or possibly-zero quantity always risks losing a root), not a second citation for this specific case.

#2.5 A perfect square trinomial — the repeated-root case, factorisation and discriminant agreeing

[VERIDIAN, illustrative] $4x^2-20x+25=0$. Discriminant: $400-400=0$ — one repeated root expected. Factorising: $mn=4,\ pq=25$, and since the discriminant is zero the two roots coincide, so the factor pair must be $(2x-5)(2x-5)=(2x-5)^2$ — check: $4x^2-10x-10x+25=4x^2-20x+25$. ✓ Root $x=\frac52$, repeated. This is the factorised-form mirror of the vertex-form fact the lesson's diagram already states (a zero discriminant means the vertex sits exactly on the axis): here it shows up as the two brackets becoming identical rather than as a vertex height of zero — the same fact, read off a different form.

#2.6 No rational factorisation at all — recognise it and switch routes immediately

[VERIDIAN, illustrative] $x^2+3x-5=0$: discriminant $9-4(1)(-5)=9+20=29$, not a perfect square. By §2's own theorem, no factorisation with rational $p,q$ exists — searching for integer factors of $-5$ that sum to $3$ (the pairs are only $\pm1,\mp5$, giving sums $\pm4$, neither equal to $3$) confirms this directly and would never succeed no matter how long the search continued. The formula gives the roots directly, in exact surd form: $x=\frac{-3\pm\sqrt{29}}{2}$ — and, per the mark scheme's own exact-answer convention already verified in this course (WMA11-verified-facts.md §3), left in this form rather than rounded, since no calculator instruction or decimal request was given.

#3. Sketching a quadratic: one procedure, three possible starting forms

Spec 1.3 pairs "quadratic functions" with "their graphs" in the same clause — sketching is not adjacent content, it is the spec point's own second half. No real WMA11 question asking a candidate to sketch a bare quadratic from scratch has been found in this course's research (consistent with signal-topic-brief-quadratic-functions.md's own finding that this topic essentially never stands alone as a full question) — so what follows is method, grounded in the algebra already established above and in the two real sketch-labelling conventions this course has confirmed on a structurally identical curve family (a cubic, §4 below), not a reproduction of a confirmed quadratic-specific mark scheme.

1. Plot the $y$-intercept: $(0,c)$, read directly off any general-form expression with no working needed.
2. Plot the $x$-intercept(s), from whichever form is fastest: factorise if the discriminant is a perfect square, otherwise use the formula. If the discriminant is negative, there are none — the curve sits entirely above the axis ($a>0$) or entirely below it ($a<0$), and no intercept should be drawn.
3. Plot the turning point, from the completed-square form: $\left(-\frac{b}{2a},\,-\frac{b^2-4ac}{4a}\right)$. If two $x$-intercepts are already known, its $x$-coordinate can be found faster, without completing the square at all, as their midpoint — a direct consequence of the parabola's symmetry about the line $x=-\frac{b}{2a}$ already established in the lesson.
4. Orient the curve by the sign of $a$: arms up (minimum) if $a>0$, arms down (maximum) if $a<0$.
5. Draw the turning point smooth, never as a sharp point, and label every plotted point. Both conventions are real and confirmed — on a cubic, not a quadratic, but for a reason that transfers directly: a repeated root produces the same "touch, don't cross" geometry on both curve families, and an examiner report on a cubic-sketching question records both faults directly: some candidates drew "a cusp at the minimum point" instead of a smooth turn, and others left the $y$-intercept unlabelled on an otherwise correct sketch [real, Oct 2021 Q6(a) — already cited for this purpose in the lesson's own diagram]. The geometric fact underneath (squaring never produces a corner) is curve-family-independent; the specific citation is not being claimed as a quadratic-sketching finding, only as the nearest real evidence this course has for a fault that would look identical on a parabola.

Worked through, on the running example $f(x)=6x^2+7x-3$: $y$-intercept $(0,-3)$; $x$-intercepts $\left(\frac13,0\right)$ and $\left(-\frac32,0\right)$ (from §1's factorisation); turning point $\left(-\frac7{12},-\frac{121}{24}\right)$, which is also the midpoint of $\frac13$ and $-\frac32$: $\frac{\frac13+\left(-\frac32\right)}{2}=\frac{-\frac76}{2}=-\frac7{12}$ ✓, confirming step 3's shortcut against the completed-square answer already found in §1; minimum, arms up, since $a=6>0$.

The mirror case, $a<0$: [VERIDIAN, illustrative] $g(x)=-2x^2+8x-3$. Completing the square: $-2(x^2-4x)-3=-2\left[(x-2)^2-4\right]-3=-2(x-2)^2+5$ — turning point $(2,5)$, a maximum, since $a=-2<0$. Discriminant: $64-24=40$, positive but not a perfect square, so two irrational roots exist: $x=\frac{-8\pm\sqrt{40}}{-4}=2\mp\frac{\sqrt{10}}2$, i.e. $x=2-\frac{\sqrt{10}}2\approx0.42$ and $x=2+\frac{\sqrt{10}}2\approx3.58$. The sketching procedure above runs identically for this curve; only the orientation in step 4 and the sign of the turning-point test in step 3 flip, exactly as the lesson's own diagram key point states in prose ("the $a<0$ mirror") — this is that statement's first fully worked numeric instance.

#4. The three solving routes, and the marking conventions that actually govern them

The lesson's method-comparison block already derives and compares all three routes on one shared example ($2x^2-7x-15=0$) in full M/A/B detail and is not reconstructed here. Two real marking conventions are worth stating with their own citations, because both apply specifically to this topic and both cost marks in a way that has nothing to do with getting the algebra wrong.

A numerically correct pair of roots is not itself the method mark. Confirmed verbatim on a real hidden- quadratic question whose final step is an ordinary three-term-quadratic solve: "It must be clear they are solving the given quadratic not their incorrect one. Answers just written down scores M0" [real, Jan 2023 Q5(b) mark scheme, via forge-exam-technique-guide.md §2.2]. The three named routes (§2 above, or the formula, or completing the square) are not merely the fastest ways to solve a three-term quadratic — per this convention, a correct pair of roots with none of the three visible in the working scores the same as a wrong pair: zero, for that mark. This is the same discipline the lesson's own trap-taxonomy already documents from the calculator-technology angle (Jan 2025 Q4(b): using a calculator with no shown method "could have been penalised") — the convention above is the general form of it, independent of whether a calculator was involved at all.

Quoting the formula before substituting protects the method mark against an arithmetic slip. This is a general marking-guidance rule, not specific to the quadratic formula, but the quadratic formula is its paper-defining example: where a memorised formula is quoted in letters first, the method mark can still be scored even if a later substitution goes wrong; where it is not quoted, the method mark has to be inferred from the working, and "may be lost if there is any mistake in the working" [real, WMA11-verified-facts.md §3, General Principles for Pure Mathematics Marking, verbatim]. Since the quadratic formula is not in the exam formula booklet (confirmed directly against the booklet's own P1 section, which contains only mensuration and the cosine rule — already established in the lesson and in paper-anatomy.ts), writing it out in full, in letters, is not a presentational nicety here: it is the one line standing between an arithmetic slip and a lost method mark, on the one formula in this topic that has to be produced from memory every time.

#5. Synthesis: how often is this topic actually the point of a question, versus a step inside one

The honest answer has two very different numbers in it, and conflating them would misrepresent the evidence.

As the stated target of a question — rare, and confirmed exactly once. Across the five WMA11 series with a fully-read examiner report (Jan 2019, Oct 2020, Oct 2021, Jan 2023, Jan 2025), no question exists whose entire point is core spec 1.3/1.5 content in isolation — sketching a bare quadratic, or solving one with no further context. The one confirmed real exception found anywhere in this course's research sits outside that five-series base: June 2022 Q5(a), the reverse-reconstruction question already fully taught in the lesson's second half and seeded in content/veridian/question-bank/wma11.ts. That is 1 series of 7 touched (the 5 fully-read plus the 2 partially-read June series) confirmed to test this topic as a standalone question target — not a likelihood, a count of what has actually been checked.

As a mechanical sub-step inside a differently-labelled question — common, and traceable question by question. The act of solving a three-term quadratic (or reading a discriminant, or completing a square) turns up as an unlabelled step inside at least one question in 4 of the 5 fully-read series (all but Oct 2020), and in both partially-read June series:

Every row except June 2022's is already fully documented, with its own real marks and traps, in that question's own topic's lesson file or CODEX deep-dive, and is not re-taught here. What this table adds that no single one of those documents states on its own is the aggregate picture: the mechanics of this topic are close to ubiquitous on this paper — appearing in some form in the large majority of series this course has actually read — while the topic as the named subject of a question is close to absent. A student who can solve a three-term quadratic fluently by all three routes, but has never seen this fact stated, would reasonably expect a dedicated "solve this quadratic" question worth its own several marks; the real record says that question essentially doesn't exist on this paper — the fluency is instead the toll paid, over and over, to reach the question that is actually being asked.

#6. Honest limits of the evidence

Every count in §5 is a count of series this course's research has actually read, not an estimate over the full population of WMA11 papers ever sat. Two series (Oct 2022, Jan 2024) are confirmed as genuine Pearson mark schemes but were never mined question-by-question, so this topic's presence or absence in them is unknown, not counted as an absence. The two June series were read for one or two targeted questions each, not the full paper, so a sub-step appearance elsewhere in either paper would not have been found by this pass. No likelihood percentage or grade boundary is stated or implied anywhere in this document — the evidence available does not support either, and none is claimed. §2.4's structural parallel between dividing by $x$ and dividing by $k$ is explicitly not a second citation for the $c=0$ case; it is one citation (Jan 2023 Q4) applied by analogy, and is labelled as such at the point it is made.

#7. What this document deliberately does not repeat

The lesson file's three prequestions, both teach blocks, its full prose derivation of the discriminant and the quadratic formula from completing the square, its diagram of the three discriminant cases, its complete method-comparison on $2x^2-7x-15=0$, its self-explain prompt, its beyond-spec proof of the sum-and-product-of- roots identity (and its own explicit note that this is not examinable at P1), its second teach block and marked-solution on reversing the direction (June 2022 Q5(a), already fully taught and fully seeded in the question bank), all seven trap-taxonomy items, all six MCQs, its reference card and its closing flag all already exist in quadratic-functions.ts and are not reconstructed above. codex-deep-dive-discriminant-with- a-parameter.ts's general theory of $\Delta(p)$ as a function of a parameter, its own two real anchors worked on their own numbers, and its own sub-case taxonomy are a different document's territory and are only cross-referenced here, not repeated. This document's own contribution was the three-form identity stated as a single running example (§1); the factorisation mechanism's exact behaviour across every distinct algebraic shape it can take, including the theorem connecting the discriminant to whether a rational factorisation can exist at all (§2); a single reusable sketching procedure with its first full numeric worked instance in both orientations (§3); two marking-convention citations specific to the basic solving act rather than to any parameter or substitution variant of it (§4); and the standalone-versus-embedded count that sits alongside — not beneath — all of that existing content (§5).