HookThe formula that helped blow up the global economy
In 2000 a quantitative analyst named David X. Li published a compact mathematical function — a Gaussian copula — that let banks put a single number on how likely it was that a whole basket of mortgages would default together. It was elegant, it was fast, and Wall Street fell in love with it, using it to price trillions of pounds of mortgage-backed securities through the 2000s. The trouble was hidden in the small print of the function: it read the chance of joint default off recent, calm-market data and treated that correlation as if it held in every condition. When the US housing market turned in 2007, the assumption shattered, the model's outputs became fiction, and the securities it had priced helped trigger the worst financial crisis since the 1930s. A 2009 Wired feature called it 'the formula that killed Wall Street'.
The lesson is not that the algebra was wrong — it is that a function is only ever as good as your handling of it and your honesty about where it holds. That is what this entire section trains. You will master the laws of indices and surds, the three techniques that unlock quadratics, simultaneous equations and inequalities, the algebra of polynomials and the factor theorem, how to sketch and transform graphs, how to build composite and inverse functions, how to split a fraction into partial fractions, and finally how to use functions to model the real world — including, exactly as Li failed to, stating the limitations of the model you have built.
ModelIndices and surds — the rules everything else stands on
Two skills open A-level algebra because everything later leans on them. The laws of indices now extend from whole-number powers to all rational exponents: \(a^m\times a^n=a^{m+n}\), \(a^m\div a^n=a^{m-n}\), \((a^m)^n=a^{mn}\), \(a^0=1\), \(a^{-n}=\dfrac{1}{a^n}\), and the two that make A-level different from GCSE — \(a^{1/n}=\sqrt[n]{a}\) and \(a^{m/n}=\left(\sqrt[n]{a}\right)^m\). Reading a fractional power correctly is the whole game: the denominator is a root, the numerator is a power, and a negative sign means 'take the reciprocal'. So \(27^{-2/3}\) is an instruction — 'cube-root it, square the result, then flip'.
Surds are roots that stay irrational, like \(\sqrt2\) or \(\sqrt{12}\), and the spec wants them simplified and their denominators rationalised. Simplify by pulling out square factors: \(\sqrt{12}=\sqrt{4\times 3}=2\sqrt3\). Rationalising the denominator removes a surd from the bottom of a fraction. For a single surd you multiply top and bottom by that surd; for a two-term denominator like \(\sqrt7-\sqrt5\) you multiply by its conjugate \(\sqrt7+\sqrt5\), because \((\sqrt7-\sqrt5)(\sqrt7+\sqrt5)=7-5=2\) clears both roots at once — a difference of two squares doing the heavy lifting.
Evaluate \(27^{-2/3}\), then rationalise \(\dfrac{6}{\sqrt7-\sqrt5}\).
For the index, work through the fraction's parts. The denominator \(3\) is a cube root: \(27^{1/3}=3\). The numerator \(2\) squares it: \(3^2=9\). The minus sign takes the reciprocal: \(27^{-2/3}=\dfrac{1}{9}\).
To rationalise, multiply top and bottom by the conjugate \(\sqrt7+\sqrt5\): \(\dfrac{6}{\sqrt7-\sqrt5}\times\dfrac{\sqrt7+\sqrt5}{\sqrt7+\sqrt5}=\dfrac{6(\sqrt7+\sqrt5)}{(\sqrt7)^2-(\sqrt5)^2}=\dfrac{6(\sqrt7+\sqrt5)}{7-5}=\dfrac{6(\sqrt7+\sqrt5)}{2}=3\sqrt7+3\sqrt5\). The denominator is now a whole number — which is exactly what 'rationalising' means.
ModelQuadratics — completing the square, the discriminant, and solving
A quadratic \(ax^2+bx+c\) has a parabola for its graph, and three techniques unlock it. Completing the square rewrites it as \(a(x+p)^2+q\), which instantly exposes the vertex at \((-p,\ q)\) and hence the maximum or minimum — indispensable, because the turning point is often the whole question. The discriminant \(\Delta=b^2-4ac\) reveals the roots without solving: \(\Delta\gt 0\) gives two distinct real roots, \(\Delta=0\) gives one repeated root (the curve touches the \(x\)-axis), and \(\Delta\lt 0\) gives no real roots (the curve misses the axis). Examiners love handing you an unknown coefficient and asking for the condition that makes the roots real or equal — that is a discriminant inequality in disguise.
To solve, factorise when the quadratic is friendly, otherwise use \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\), which is itself just the completed-square form rearranged. The spec also wants equations that are 'quadratic in something else' — a quadratic in \(x^2\), in \(\sqrt x\), or in \(e^x\) — solved by substituting a single letter for that block, cracking the ordinary quadratic, then undoing the substitution.
Express \(2x^2-8x+3\) in completed-square form, state its minimum, and find its roots.
Factor \(2\) from the \(x\)-terms: \(2(x^2-4x)+3\). Complete the square inside: \(x^2-4x=(x-2)^2-4\), so \(2\big((x-2)^2-4\big)+3=2(x-2)^2-8+3=2(x-2)^2-5\). The minimum value is therefore \(-5\), reached at \(x=2\); the vertex is \((2,\ -5)\).
The discriminant is \(\Delta=(-8)^2-4(2)(3)=64-24=40\gt 0\), so there are two distinct real roots. Solving \(2(x-2)^2-5=0\) gives \((x-2)^2=\dfrac{5}{2}\), so \(x=2\pm\sqrt{\dfrac{5}{2}}=2\pm\dfrac{\sqrt{10}}{2}\). Reading the vertex straight off the completed square is far faster than differentiating.
MechanismSimultaneous equations and inequalities
Simultaneous equations ask for the point(s) where two graphs meet. Two straight lines yield to elimination or substitution; the A-level step up is one linear and one quadratic, always cracked by substitution — rearrange the linear equation for one variable and drop it into the quadratic, leaving a single quadratic to solve. Geometrically, the number of solutions is the number of intersections: two solutions means the line cuts the curve twice, one means it is a tangent, none means they miss — and that count is governed by the discriminant of the quadratic you form.
Inequalities need one extra reflex. A linear inequality is solved like an equation, with the single rule that multiplying or dividing by a negative number flips the sign. A quadratic inequality must be solved by sketching: factorise to find where the parabola cuts the axis, then read off the region. For an upward parabola, \(y\gt 0\) is outside the roots and \(y\lt 0\) is between them. Solutions are written with 'and'/'or' or in set notation, for example \(\{x : x\le -2\}\cup\{x : x\ge 3\}\).
Solve simultaneously \(y=x^2-3x+2\) and \(y=2x-2\); then solve \(x^2-x-6\ge 0\).
For the pair, substitute the line into the curve: \(x^2-3x+2=2x-2\Rightarrow x^2-5x+4=0\Rightarrow (x-1)(x-4)=0\), so \(x=1\) or \(x=4\). Back-substitute into \(y=2x-2\): \(x=1\) gives \(y=0\), and \(x=4\) gives \(y=6\). The line meets the curve at \((1,\ 0)\) and \((4,\ 6)\).
For the inequality, factorise: \(x^2-x-6=(x-3)(x+2)\), so the upward parabola cuts the axis at \(x=-2\) and \(x=3\). It is \(\ge 0\) outside the roots, giving \(x\le -2\) or \(x\ge 3\), that is \(\{x : x\le -2\}\cup\{x : x\ge 3\}\). A quick sketch prevents the standard error of writing the 'between the roots' region by mistake.
MechanismPolynomials — the factor theorem and algebraic division
A polynomial is a sum of whole-number powers of \(x\), and this leaf gathers every way of pulling one apart. You expand brackets term by term and collect like terms; you factorise by reversing that process. The new tool is the factor theorem: if \(f(p)=0\) then \((x-p)\) is a factor of \(f(x)\), and conversely. This turns factorising a cubic into a short search — try small values \(p=\pm 1,\ \pm 2,\ \pm 3,\dots\) until one gives zero, and you have prised out a linear factor.
Once you have that factor, algebraic division (or comparing coefficients) reduces the cubic to a quadratic you can factorise the usual way. The same skills let you simplify rational expressions — algebraic fractions — by factorising numerator and denominator fully and cancelling whole common factors, never individual terms. The single most penalised error in the topic is cancelling across a \(+\) sign, for instance striking the \(x\) out of \(\dfrac{x+2}{x+5}\); you may only cancel a bracket that divides the entire top and the entire bottom.
Factorise \(f(x)=2x^3-3x^2-11x+6\) completely.
Search for a root with the factor theorem. Test \(x=3\): \(f(3)=2(27)-3(9)-11(3)+6=54-27-33+6=0\). Since \(f(3)=0\), \((x-3)\) is a factor. Divide \(2x^3-3x^2-11x+6\) by \((x-3)\) to get the quadratic \(2x^2+3x-2\). Factorise that: \(2x^2+3x-2=(2x-1)(x+2)\). Therefore \(f(x)=(x-3)(2x-1)(x+2)\), with roots \(x=3,\ \tfrac12,\ -2\) — three linear factors for a cubic, exactly as expected. Expanding back out is the reliable way to confirm the division was clean.
ModelSketching graphs and transforming them
You are expected to sketch a family of curves from their equations — not plot points, but capture shape, intercepts and end behaviour. Polynomials: a cubic \(y=(x-a)(x-b)(x-c)\) crosses the axis at its roots and runs off to \(\pm\infty\) at the ends. Reciprocals: \(y=\dfrac{a}{x}\) is a hyperbola with the axes as asymptotes — lines the curve approaches but never touches — while \(y=\dfrac{a}{x^2}\) sits entirely above the \(x\)-axis. The modulus of a linear function, \(y=|x-2|\), is a V-shape: the ordinary line with everything below the axis reflected upward, its vertex where the bracket is zero. Direct and inverse proportion give \(y=kx\) (a line through the origin) and \(y=\dfrac{k}{x}\) (a hyperbola).
On top of these sit the transformations, and the four to know cold are: \(y=f(x)+a\) shifts up by \(a\); \(y=f(x+a)\) shifts left by \(a\); \(y=af(x)\) is a vertical stretch of scale factor \(a\); and \(y=f(ax)\) is a horizontal stretch of scale factor \(\dfrac{1}{a}\). Two are counter-intuitive and cost marks every year: \(f(x+a)\) moves left, not right, and \(f(ax)\) squashes by the reciprocal of \(a\). The rule of thumb: changes inside the bracket act on \(x\) and behave 'backwards'; changes outside act on \(y\) and behave as written.
The point \((4,\ 3)\) lies on \(y=f(x)\). Find its image on \(y=f(x-2)+5\), on \(y=f(2x)\), and on \(y=3f(x)\).
For \(y=f(x-2)+5\): the \(-2\) inside shifts right \(2\) (inside changes run backwards) and the \(+5\) outside shifts up \(5\), so \((4,\ 3)\to(6,\ 8)\). For \(y=f(2x)\): a horizontal stretch of scale factor \(\dfrac{1}{2}\) halves the \(x\)-coordinate while \(y\) is untouched, giving \((2,\ 3)\). For \(y=3f(x)\): a vertical stretch of scale factor \(3\) triples the \(y\)-coordinate, giving \((4,\ 9)\). Tracking one known point through each rule is the safest way to keep the direction right.
MechanismComposite and inverse functions
A function is a rule assigning exactly one output to each input in its domain, and two ways of building new functions are examined. A composite function \(fg(x)\) means 'do \(g\) first, then \(f\)' — feed \(x\) into \(g\), then that result into \(f\). The order is the trap: read composites right-to-left, and expect \(fg(x)\neq gf(x)\) in general. The inverse function \(f^{-1}\) runs \(f\) backwards, so that \(f^{-1}(f(x))=x\). To find it, write \(y=f(x)\), rearrange to make \(x\) the subject, then swap the letters.
Two facts about inverses earn marks. First, the graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\), so a sketch of one hands you the other. Second, the domain and range swap: the domain of \(f^{-1}\) is the range of \(f\), and its range is the domain of \(f\). An inverse exists only where the function is one-to-one — each output coming from a single input — which is why a domain is often restricted before an inverse is taken.
Let \(f(x)=3x-1\) and \(g(x)=x^2+2\). Find \(fg(x)\), \(gf(x)\) and \(f^{-1}(x)\).
Compose \(fg\) by doing \(g\) first: \(fg(x)=f(x^2+2)=3(x^2+2)-1=3x^2+5\). Now the other order: \(gf(x)=g(3x-1)=(3x-1)^2+2=9x^2-6x+1+2=9x^2-6x+3\). The two disagree, confirming \(fg\neq gf\). For the inverse of \(f\): set \(y=3x-1\), rearrange to \(x=\dfrac{y+1}{3}\), then relabel to \(f^{-1}(x)=\dfrac{x+1}{3}\). Check it undoes \(f\): \(f(2)=5\) and \(f^{-1}(5)=\dfrac{5+1}{3}=2\), back where we started.
MechanismPartial fractions — running addition in reverse
Partial fractions run algebraic fractions backwards: instead of adding \(\dfrac{2}{x-1}+\dfrac{3}{x+2}\) into a single fraction, you split one fraction back into that sum. It is the standard preparation for integrating and for binomial expansions later in the course, so it is examined both for itself and as the opening move in longer questions. The layout of the split depends on the denominator: each distinct linear factor \((x-a)\) contributes one term \(\dfrac{A}{x-a}\), while a repeated (squared) linear factor \((x-a)^2\) contributes two, \(\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}\) — the spec goes up to squared linear terms.
The method: write the target as that sum with unknown constants on top, multiply through by the whole denominator, then find the constants. The fastest route is substitution, choosing values of \(x\) that switch factors off. For a repeated factor, substitution pins down the constant over the squared term and comparing coefficients of \(x\) finishes the job. Always check the numerator is of lower degree than the denominator first; if it is not, divide out the whole part before splitting.
Express \(\dfrac{5x+1}{(x-1)(x+2)}\) in partial fractions, then split \(\dfrac{3x-1}{(x-1)^2}\).
Write \(\dfrac{5x+1}{(x-1)(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{x+2}\) and multiply up: \(5x+1=A(x+2)+B(x-1)\). Substitute \(x=1\): \(6=3A\Rightarrow A=2\). Substitute \(x=-2\): \(-9=-3B\Rightarrow B=3\). So it splits as \(\dfrac{2}{x-1}+\dfrac{3}{x+2}\).
For the repeated factor, write \(\dfrac{3x-1}{(x-1)^2}=\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}\), so \(3x-1=A(x-1)+B\). Substitute \(x=1\): \(B=2\). Compare the \(x\)-coefficients: \(A=3\). Hence \(\dfrac{3}{x-1}+\dfrac{2}{(x-1)^2}\). Each answer can be checked by recombining over the common denominator.
CaseFunctions as models — and the limits Li ignored
The final leaf steps back and asks what a function is for: modelling the real world. Any model is a function with a job — cost as a function of quantity, height as a function of time, risk as a function of correlation — and its most important property is not its formula but its domain of validity, the range of inputs over which it can be trusted. Leaf 2.11 explicitly wants you to state a model's limitations and suggest refinements, because a function that fits today's data can be nonsense outside the region it was built for.
That is exactly what sank the Gaussian copula in the opening story. Li's formula was a legitimate function; the failure was using it far outside where its assumptions held, treating a correlation measured in calm markets as if it were fixed for all conditions. The A-level version is tamer, but the discipline is identical: give a sensible domain, check the outputs stay physically meaningful, and say plainly where the model would break and how you would improve it — a wider data range, an extra variable, or a curve where you first assumed a straight line.
A tray is folded from a \(24\text{ cm}\) square of card by cutting a square of side \(x\) from each corner and turning up the sides. Its volume is modelled by \(V(x)=x(24-2x)^2\). Find \(V(4)\) and state the model's valid domain.
Substitute \(x=4\): \(V(4)=4\,(24-8)^2=4\times 16^2=4\times 256=1024\text{ cm}^3\). Now the limitation: \(x\) is a length cut from the card, so \(x\gt 0\); and the two corners along one side use up \(2x\) of the \(24\text{ cm}\), so \(2x\lt 24\), that is \(x\lt 12\). The model is valid only for \(0\lt x\lt 12\). Outside that interval the formula still returns numbers — \(V(15)=15\times(-6)^2=540\) is a positive value — but they describe no real tray at all. Stating the domain is the point: a function is only a model where its inputs actually mean something.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
This section is the spine of the Pure papers, so accuracy of method is everything and full working is rewarded line by line. On a completed-square question, write \(a(x+p)^2+q\) and then read the vertex straight off rather than differentiating — it is faster and less error-prone. On any 'real roots / equal roots / no real roots' question, translate the words into a discriminant condition immediately: 'equal roots' means \(b^2-4ac=0\), and an unknown coefficient turns that into an equation or inequality to solve.
For the factor theorem, show the substitution that gives zero ('\(f(3)=0\), so \((x-3)\) is a factor') before you divide, because that line carries a mark of its own. When simplifying algebraic fractions or forming partial fractions, factorise fully first and never cancel across a \(+\). On transformations, state the transformation in words and its direction — the marks are lost on \(f(x+a)\) moving left and \(f(ax)\) stretching by \(\tfrac{1}{a}\), so track a single point through to be sure. With composite functions, write the inner evaluation first and say which function you applied, since muddling \(fg\) with \(gf\) is the classic dropped mark. Finally, on any modelling part, always give a domain and name one honest limitation — the examiner is explicitly looking for the sentence that says where the model stops being trustworthy.