Algebra and functions — Edexcel A-Level Mathematics

A surd is an irrational root, such as √2 or √3. You simplify surds by removing square factors: √50 = √(25 × 2) = 5√2. You add or subtract like surds, for example 3√2 + 5√2 = 8√2, and multiply using √a × √b = √(ab), so √2 × √8 = √16 = 4. To rationalise a denominator, multiply numerator and denominator by a suitable surd or conjugate. For a single surd, 1/√3 = √3/3. For a binomial denominator, multiply by the conjugate: 1/(2 + √3) = (2 − √3)/((2 + √3)(2 − √3)) = (2 − √3)/(4 − 3) = 2 − √3. You must also simplify expressions such as (3 + √2)(3 − √2) = 9 − 2 = 7. Assessment rewards correct simplification, rationalisation, and exact answers in surd form.

A quadratic f(x) = ax² + bx + c has a parabolic graph; its shape and position follow from a, b and c, with the vertex found by completing the square. Writing ax² + bx + c as a(x + p)² + q gives the minimum or maximum value q at x = −p, and the line of symmetry x = −p. The discriminant Δ = b² − 4ac decides the roots: Δ > 0 gives two distinct real roots, Δ = 0 gives a repeated real root, and Δ < 0 gives no real roots. Solving equations may need factorising, the formula or completing the square. Equations in a function of the unknown, such as x⁴ − 5x² + 4 = 0, are solved by substituting u = x², solving the quadratic in u, then converting back and checking each value.

Simultaneous equations in two variables are solved by finding every pair (x, y) that satisfies both equations at once. Elimination adds or subtracts multiples of the equations to remove one variable; it suits two linear equations, where the result is normally a single pair. Substitution rearranges one equation for one variable and puts that expression into the other; it is the reliable method when one equation is linear and the other is quadratic, because it produces a quadratic in one variable with up to two solutions. Each solution must be a matching pair: substitute each value back to find its partner value, and check both pairs in both original equations. Graphically the solutions are the intersection points of a line and a curve.

An inequality is solved like an equation, with one extra rule: multiplying or dividing by a negative number reverses the inequality sign. Linear inequalities with brackets are expanded first; those with fractions are cleared by multiplying by the positive lowest common denominator. A quadratic inequality is solved by finding the critical values where the expression equals zero, then testing intervals or sketching the parabola. For a > 0, the solution of ax² + bx + c > 0 lies outside the roots, written with ‘or’, while ax² + bx + c < 0 lies between them, written with ‘and’. Set notation may be used instead, for example {x : x < −1} ∪ {x : x > 3}. Graphically, y > x + 1 is the half-plane above the line y = x + 1, and y > ax² + bx + c is the region above the parabola, with a dashed boundary when the inequality is strict.

Polynomial manipulation covers expanding products such as (x + 2)(x² − 3x + 1) and collecting like terms, factorising by common factors and by inspection, and dividing a polynomial by a linear expression. Simple algebraic division may be done by long division or by comparing coefficients. The factor theorem states that (x − a) is a factor of f(x) exactly when f(a) = 0, so substituting a candidate value tests for a factor and helps to factorise fully. Rational expressions are simplified by factorising numerator and denominator completely, then cancelling common factors, remembering that the cancelled factor must not be zero. Division by a linear expression can also reduce an improper rational expression before further simplification.

Sketching shows shape, intercepts, turning points, asymptotes and behaviour as x → ±∞. For a polynomial like y = x³ − 3x, find intercepts, differentiate for stationary points, and note opposite end behaviour. For y = |mx + c|, sketch the V-shape by reflecting the negative part of the line in the x-axis. For y = a/x, the axes are asymptotes (x = 0, y = 0), with branches in opposite quadrants. For y = a/x², branches lie on opposite sides of the vertical asymptote x = 0, but on the same side of the horizontal asymptote y = 0 (both above the x-axis if a > 0). Solving f(x) = g(x) graphically means reading x-coordinates of intersection points. Proportional relationships give straight-line graphs through the origin: y = kx.

A composite function applies one function after another: fg(x) means g first, then f, so fg(x) = f(g(x)). Order matters; fg and gf are generally different. The domain of fg is the set of x in the domain of g for which g(x) lies in the domain of f. An inverse function reverses the mapping: if f(a) = b then f⁻¹(b) = a. To find f⁻¹, write y = f(x), rearrange to make x the subject, then swap x and y. An inverse exists only when f is one-to-one on its domain; restricting the domain can create an inverse, as with f(x) = x² for x ≥ 0. The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x, so points (a, b) and (b, a) correspond, and the domain of f⁻¹ equals the range of f.

Transformations move or reshape the graph of y = f(x). The transformation y = af(x) stretches the graph vertically by scale factor a; if a is negative the graph is also reflected in the x-axis. The transformation y = f(x) + a translates the graph by a units parallel to the y-axis, upwards when a is positive. The transformation y = f(x + a) translates the graph by −a units parallel to the x-axis, so f(x + 3) moves the graph 3 units left. The transformation y = f(ax) stretches the graph horizontally by scale factor 1/a, so f(2x) halves the x-coordinates. Combinations apply in sequence, and the order matters when stretches and translations are mixed; for example y = af(x + b) translates first and then stretches. Key points, asymptotes and intercepts should be tracked through each transformation.

Partial fractions reverse the process of adding algebraic fractions. A rational function with a denominator that factorises into distinct linear factors is split into one fraction per factor. For example, (5x + 1)/((x − 1)(x + 2)) is written as A/(x − 1) + B/(x + 2); multiplying through by the denominator gives 5x + 1 = A(x + 2) + B(x − 1), and substituting x = 1 and x = −2 finds A and B. A repeated linear factor such as (x − 1)² requires two terms, A/(x − 1) + B/(x − 1)². An improper fraction, where the numerator degree is not lower than the denominator degree, is first divided to give a polynomial plus a proper fraction. Denominators may have up to three factors, and numerators are constant or linear as appropriate.

A model turns a real situation into a function, for example height h metres of a thrown ball after t seconds as h(t) = 20t − 5t². You choose variables, state a domain, and interpret outputs in context. Limitations arise because assumptions fail: air resistance, changing gravity, or a domain where h(t) becomes negative. Refinements adjust the function, domain or constants to fit data better, such as adding a drag term or restricting t to 0 ≤ t ≤ 4. In assessment you form a function, use it to predict, then critique it: identify which feature of reality is ignored, explain the effect on accuracy, and propose a specific improvement with justification.