Notes · MathematicsUK · A-Levels

Algebra and functions

This is the algebraic backbone of the whole course: indices and surds, quadratics and the discriminant, equations and inequalities, polynomials and the factor theorem, the modulus function, composite and inverse functions, graph transformations and partial fractions. Fluency here underpins calculus, coordinate geometry and every applied topic.

5 sections~13 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 2

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Exam profile

AO1 · Manipulate algebraic expressions and solve equations and inequalities accuratelyAO2 · Reason about functions, their graphs and their transformationsAO3 · Use algebraic models and interpret them in context

Operators:solvesimplifyexpresssketchshow thatfactorisefind

basic level

AS-Level covers indices, surds, quadratics, simultaneous equations, inequalities, polynomials, graphs and transformations.

higher level

The full A-Level adds the modulus function, composite and inverse functions, combined transformations and partial fractions.

Depth

Text

Contents · 5 sections

Algebra and functions

Indices and surds#

FoundationDfE GCE Mathematics B1-B2AQA 7357 B

The square-root function

Fig. 1The principal square root is the non-negative branch; it is defined only for

x \geq 0

The laws of indices are the grammar of algebra and extend the familiar rules to zero, negative and fractional powers. For any positive base

a

and any real powers,

a^m \times a^n = a^{m+n}

\dfrac{a^m}{a^n} = a^{m-n}

and

(a^m)^n = a^{mn}

. From these follow the definitions

a^0 = 1

a^{-n} = \dfrac{1}{a^n}

and

a^{1/n} = \sqrt[n]{a}

, so that a fractional index means 'root then power':

a^{m/n} = \left(\sqrt[n]{a}\right)^m

. Every simplification of a power expression is an application of these rules in some order.

A surd is an irrational root left in exact form, such as

\sqrt{2}

2\sqrt{5}

. Exact answers in surd form are expected wherever a decimal would be only an approximation. The manipulation rules are

\sqrt{ab} = \sqrt{a}\,\sqrt{b}

and

\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}

; these let you simplify

\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}

by extracting the largest square factor.

Rationalising the denominator removes a surd from the bottom of a fraction. For a single surd you multiply top and bottom by that surd:

\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}

. For a two-term denominator you multiply by its conjugate, using the difference of two squares:

\dfrac{1}{2 + \sqrt{3}} = \dfrac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \dfrac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}

, because

(2+\sqrt3)(2-\sqrt3) = 4 - 3 = 1

A common thread is that a fractional or negative index and a surd are two views of the same object, and switching between them unlocks problems. Writing

\dfrac{1}{\sqrt{x}}

x^{-1/2}

is exactly what lets you differentiate or integrate it later, so treat 'convert to index form' as a standard first move whenever roots appear inside calculus.

a^{m/n} = \sqrt[n]{a^{m}} = \left(\sqrt[n]{a}\right)^{m}

Fractional indices

The denominator is the root, the numerator is the power; either order gives the same value for positive $a$ .

\frac{1}{a+\sqrt{b}} = \frac{a-\sqrt{b}}{a^2 - b}

Rationalising with a conjugate

Multiplying by the conjugate uses $(a+\sqrt b)(a-\sqrt b)=a^2-b$ to clear the surd from the denominator.

Worked example

Rationalising a two-term denominator

Express $\dfrac{4}{\sqrt{5} + 1}$ in the form $a + b\sqrt{5}$ where $a$ and $b$ are rational.

Multiply by the conjugate
Multiply top and bottom by $\sqrt{5} - 1$ : $\dfrac{4(\sqrt{5} - 1)}{(\sqrt{5} + 1)(\sqrt{5} - 1)}$ .
Simplify the denominator
$(\sqrt{5} + 1)(\sqrt{5} - 1) = 5 - 1 = 4$ .
$(\sqrt{5}+1)(\sqrt{5}-1) = 5 - 1 = 4$
Divide
$\dfrac{4(\sqrt{5} - 1)}{4} = \sqrt{5} - 1 = -1 + 1\sqrt{5}$ .

Result: $\dfrac{4}{\sqrt{5}+1} = -1 + \sqrt{5}$ , so $a = -1$ , $b = 1$ .

Typical mistakes

Writing $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ , which is false (there is no such law).
Forgetting to simplify a surd fully, e.g. leaving $\sqrt{50}$ instead of $5\sqrt{2}$ .

Active revision

Simplify $\dfrac{\sqrt{75} + \sqrt{12}}{\sqrt{3}}$ and express $\dfrac{5}{3 - \sqrt{2}}$ with a rational denominator.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Quadratic functions and the discriminant#

StandardDfE GCE Mathematics B3AQA 7357 B

The three discriminant cases

Fig. 2From left to right:

\Delta > 0

(two roots),

\Delta = 0

(repeated root),

\Delta < 0

(no real roots).

A quadratic

y = ax^2 + bx + c

(with

a \neq 0

) has a parabolic graph, opening upwards when

a > 0

and downwards when

a < 0

. Completing the square rewrites it as

y = a\left(x + \dfrac{b}{2a}\right)^2 + \left(c - \dfrac{b^2}{4a}\right)

, which reads off the vertex (turning point) at

x = -\dfrac{b}{2a}

and the minimum or maximum value directly. This 'completed-square' form is the most informative way to write a quadratic.

Completing the square also derives the quadratic formula, and you should be able to reproduce the derivation. Starting from

ax^2 + bx + c = 0

, divide by

a

, complete the square to

\left(x + \dfrac{b}{2a}\right)^2 = \dfrac{b^2 - 4ac}{4a^2}

, take square roots and rearrange to obtain

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

. The formula is not a magic recipe but the algebraic consequence of completing the square in general.

The quantity under the root,

\Delta = b^2 - 4ac

, is the discriminant, and its sign classifies the roots. If

\Delta > 0

there are two distinct real roots (the parabola crosses the

x

-axis twice); if

\Delta = 0

there is one repeated root (the parabola touches the axis); if

\Delta < 0

there are no real roots (the parabola does not meet the axis). Questions frequently give a condition — 'the equation has equal roots', 'the line is a tangent' — that translates into a statement about

\Delta

Because the discriminant condition converts a geometric statement about intersection into an algebraic inequality in a parameter, it is a favourite route into problem solving. 'Find the values of

k

for which

x^2 + kx + 9 = 0

has real roots' becomes

k^2 - 36 \geq 0

, i.e.

k \leq -6

k \geq 6

— a discriminant inequality solved like any other.

x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}

The quadratic formula

Obtained by completing the square on $ax^2+bx+c=0$ ; the term under the root is the discriminant.

\Delta = b^{2} - 4ac \;\begin{cases}> 0 & \text{two real roots}\\ = 0 & \text{one repeated root}\\ < 0 & \text{no real roots}\end{cases}

The discriminant classifies the roots

The sign of $\Delta$ tells you how many times the parabola meets the x-axis.

Worked example

Completed square and vertex

Write $y = 2x^2 - 12x + 23$ in completed-square form and state the minimum point.

Take the coefficient out
$y = 2(x^2 - 6x) + 23$ .
Complete the square inside
$x^2 - 6x = (x - 3)^2 - 9$ , so $y = 2\big((x-3)^2 - 9\big) + 23$ .
$x^2 - 6x = (x-3)^2 - 9$
Expand the constant
$y = 2(x - 3)^2 - 18 + 23 = 2(x - 3)^2 + 5$ .
Read the vertex
The square is least (zero) at $x = 3$ , giving $y = 5$ .

Result: $y = 2(x-3)^2 + 5$ ; the minimum point is $(3, 5)$ .

Typical mistakes

Sign slips in $-\dfrac{b}{2a}$ when locating the vertex from the completed square.
Using $\Delta \geq 0$ where the question demands two distinct roots (which needs $\Delta > 0$ ).

Active revision

The equation $2x^2 + (k+1)x + 8 = 0$ has a repeated root. Find the possible values of $k$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Equations, inequalities and polynomials#

StandardDfE GCE Mathematics B4-B6AQA 7357 B

A fully factorised cubic

Fig. 3

y = (x-1)(x+2)(x-3)

crosses the axis at each of its three simple roots.

Simultaneous equations may be solved by elimination or substitution; when one equation is linear and one quadratic, substitute the linear into the quadratic to obtain a single quadratic. The number of intersection points of a line and a curve is then read from the discriminant of that quadratic: two solutions (the line is a secant), one (a tangent) or none. This links algebra directly to coordinate geometry.

Linear inequalities are solved like equations, with the crucial rule that multiplying or dividing by a negative number reverses the inequality sign. Quadratic inequalities are solved by first finding the roots of the corresponding equation, then deciding which region satisfies the inequality — most reliably by sketching the parabola. For

x^2 - x - 6 > 0

the roots are

-2

and

3

, and since the parabola opens upwards the expression is positive outside the roots, giving

x < -2

x > 3

The factor theorem states that

(x - a)

is a factor of a polynomial

p(x)

if and only if

p(a) = 0

. Together with algebraic (long) division this lets you fully factorise cubics and quartics: find one root by trial from the factors of the constant term, divide out the corresponding linear factor, and factorise the remaining quotient. The related remainder theorem gives the remainder on dividing

p(x)

(x - a)

as simply

p(a)

Fully factorised form is the key to sketching a polynomial and to solving polynomial inequalities, because the graph crosses the

x

-axis at each simple root and touches at each repeated root, and the sign of the polynomial alternates between consecutive simple roots. Reading a solution set such as '

p(x) \geq 0

' from a sketch is far safer than trying to reason about signs algebraically.

(x - a)\ \text{is a factor of}\ p(x) \iff p(a) = 0

The factor theorem

A value $a$ is a root exactly when $(x-a)$ divides the polynomial exactly.

Solution set of a quadratic inequality

Fig. 4

x^2 - x - 6 > 0

holds outside the roots:

x < -2

x > 3

Worked example

Factor theorem and full factorisation

Factorise $p(x) = x^3 + 2x^2 - 5x - 6$ completely.

Find a root
Try $x = -1$ : $p(-1) = -1 + 2 + 5 - 6 = 0$ , so $(x + 1)$ is a factor.
Divide
Dividing gives $p(x) = (x + 1)(x^2 + x - 6)$ .
Factorise the quotient
$x^2 + x - 6 = (x + 3)(x - 2)$ .

Result: $p(x) = (x + 1)(x + 3)(x - 2)$ , with roots $-1$ , $-3$ and $2$ .

Typical mistakes

Giving the solution of $x^2 - x - 6 > 0$ as $-2 < x < 3$ (that is the region where it is negative).
Dividing a polynomial incorrectly and not checking by multiplying back or using $p(a) = 0$ .

Active revision

Given that $x = 2$ is a root of $p(x) = 2x^3 - 3x^2 - 11x + 6$ , factorise $p(x)$ completely and hence solve $p(x) < 0$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Functions: composite, inverse and the modulus#

AdvancedDfE GCE Mathematics B7AQA 7357 B

A function and its inverse

Fig. 5

y = x^2

(for

x \geq 0

) and its inverse

y = \sqrt{x}

are reflections in

y = x

A function is a rule assigning to each input in its domain exactly one output; the set of outputs is the range. Function notation

f(x)

, together with a stated domain, defines the function completely, and much of the A-Level treatment is about combining and inverting functions carefully. Being explicit about domain and range is essential, because it decides whether an inverse exists at all.

A composite function applies one function to the output of another:

fg(x) = f\big(g(x)\big)

, read 'do

g

first, then

f

'. The order matters — in general

fg \neq gf

— and the domain of

fg

is restricted to those

x

for which

g(x)

lies in the domain of

f

. Careful bracketing and working from the inside out avoid the commonest slips.

The inverse function

f^{-1}

reverses

f

, so

f^{-1}f(x) = x

and

ff^{-1}(x) = x

. An inverse exists only when

f

is one-to-one (each output comes from exactly one input), which is why domains are often restricted to make a function invertible. To find

f^{-1}

, set

y = f(x)

, rearrange to make

x

the subject, and swap the letters; graphically,

y = f^{-1}(x)

is the reflection of

y = f(x)

in the line

y = x

, and the domain and range swap over.

The modulus function

|x|

returns the non-negative size of

x

|x| = x

for

x \geq 0

and

|x| = -x

for

x < 0

. The graph of

y = |f(x)|

reflects any part of

y = f(x)

that lies below the

x

-axis up above it. Solving modulus equations and inequalities such as

|2x - 4| = 6

is done either by squaring both sides or by considering the two cases

2x - 4 = 6

and

2x - 4 = -6

, and you must check the solutions.

fg(x) = f\big(g(x)\big), \qquad f^{-1}f(x) = x

Composite and inverse

Composition applies g then f; the inverse undoes f, and exists only when f is one-to-one.

|x| = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases}

The modulus function

The modulus gives the non-negative magnitude; the graph of $|f(x)|$ reflects negative parts above the axis.

The modulus graph

Fig. 6

y = |2x - 4|

has a corner where

2x - 4 = 0

, at

x = 2

Worked example

Finding an inverse

The function $f$ is defined by $f(x) = \dfrac{2x + 1}{x - 3}$ for $x > 3$ . Find $f^{-1}(x)$ .

Set y equal to f(x)
$y = \dfrac{2x + 1}{x - 3}$ .
Make x the subject
$y(x - 3) = 2x + 1 \Rightarrow yx - 3y = 2x + 1 \Rightarrow x(y - 2) = 3y + 1$ .
$x(y-2) = 3y + 1$
Solve for x and swap
$x = \dfrac{3y + 1}{y - 2}$ , so $f^{-1}(x) = \dfrac{3x + 1}{x - 2}$ .

Result: $f^{-1}(x) = \dfrac{3x + 1}{x - 2}$ .

Typical mistakes

Assuming $fg = gf$ ; composition is generally non-commutative.
Squaring a modulus equation and forgetting to reject solutions that do not satisfy the original.

Active revision

For $f(x) = 3x - 1$ and $g(x) = x^2$ with $x \geq 0$ , find $fg(x)$ , $gf(x)$ and $f^{-1}(x)$ , and solve $|3x - 1| = 5$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Graphs, transformations and partial fractions#

AdvancedDfE GCE Mathematics B8-B9AQA 7357 B

A translation of a parabola

Fig. 7

y = (x - 2)^2 + 1

y = x^2

translated 2 right and 1 up.

Knowing the shapes of standard graphs lets you sketch quickly and check answers. You should recognise straight lines, quadratics and cubics, the reciprocal

y = \dfrac{1}{x}

(with its two asymptotes) and

y = \dfrac{1}{x^2}

, roots and exponentials. A sketch should show the essential features — intercepts, turning points, asymptotes and behaviour for large

|x|

— rather than being an accurate plot.

Transformations of

y = f(x)

follow fixed rules.

y = f(x) + a

is a translation by

a

up;

y = f(x + a)

is a translation by

a

to the left (note the sign);

y = kf(x)

is a vertical stretch by factor

k

;

y = f(kx)

is a horizontal stretch by factor

\dfrac{1}{k}

; and

y = -f(x)

and

y = f(-x)

are reflections in the

x

- and

y

-axes respectively. The counter-intuitive cases are the horizontal ones, where the effect on

x

is the opposite of what the sign suggests.

Combined transformations must be applied in a valid order, and the order matters when a stretch and a translation act in the same direction. For

y = f(2x - 4)

, factorise the inside as

f\big(2(x - 2)\big)

to see it as a horizontal stretch of factor

\tfrac{1}{2}

followed by a translation of 2 to the right, rather than misreading it as a translation of 4.

Partial fractions split a single algebraic fraction into a sum of simpler ones, which is indispensable for integration and for binomial expansions. A proper fraction with distinct linear factors in the denominator, such as

\dfrac{5x + 1}{(x - 1)(x + 2)}

, is written as

\dfrac{A}{x - 1} + \dfrac{B}{x + 2}

; the constants are found by multiplying up and substituting convenient values of

x

. Repeated factors

(x - a)^2

require both

\dfrac{B}{x - a}

and

\dfrac{C}{(x - a)^2}

terms.

y = f(x + a) + b \ \text{translates } y=f(x) \text{ by } \begin{pmatrix} -a \\ b \end{pmatrix}

Translation rule

Inside the bracket shifts horizontally by $-a$ (opposite sign); outside shifts vertically by $+b$ .

\frac{5x + 1}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}

Partial fractions (distinct linear factors)

Multiply up and substitute $x = 1$ and $x = -2$ to find $A$ and $B$ quickly.

Worked example

Partial fractions with distinct linear factors

Express $\dfrac{5x + 1}{(x - 1)(x + 2)}$ in partial fractions.

Set up the form
Write $\dfrac{5x + 1}{(x - 1)(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$ , so $5x + 1 = A(x + 2) + B(x - 1)$ .
Substitute x = 1
$5(1) + 1 = A(3) \Rightarrow 6 = 3A \Rightarrow A = 2$ .
Substitute x = -2
$5(-2) + 1 = B(-3) \Rightarrow -9 = -3B \Rightarrow B = 3$ .

Result: $\dfrac{5x + 1}{(x - 1)(x + 2)} = \dfrac{2}{x - 1} + \dfrac{3}{x + 2}$ .

Typical mistakes

Translating $y = f(x + a)$ to the right instead of the left.
Omitting the $\dfrac{C}{(x-a)^2}$ term for a repeated linear factor in partial fractions.

Active revision

Express $\dfrac{11 - 3x}{(x - 3)(x + 1)}$ in partial fractions, and describe the single transformation mapping $y = x^2$ onto $y = (x + 4)^2$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Algebra and functions

Key points

Rationalising a two-term denominator

Key points

Completed square and vertex

Key points

Factor theorem and full factorisation

Key points

Finding an inverse

Key points

Partial fractions with distinct linear factors

Sources