EduNinja

E2.1 Introduction to algebra

A letter can stand for a number whose value is not fixed yet or may change. This lets one algebraic expression describe every value that fits the same relationship.

Relationship	Algebra	Meaning
five copies of a number $n$	$5n$	$5\times n$
three more than $n$	$n+3$	add 3 to the value of $n$
£48 shared equally among $x$ people	$48/x$	each share when $x\ne0$
a number $p$ decreased by 7	$p-7$	subtract 7 from the value of $p$

If one notebook costs £ $p$ , then five notebooks cost £ $5p$ . The same expression works whether $p=2$ , $p=3.50$ or another allowed price. The letter represents the number; it is not an abbreviation for the object.

Within one expression, every occurrence of the same letter has the same value. Different letters represent quantities independently unless the context states a relationship between them.

$5n$ means $5\times n$ , not $5+n$ and not the two-digit number “5n”. Keep the operation stated by the relationship: “three more” gives $n+3$ , while “three times” gives $3n$ .

Substitution means replacing each letter with its given numerical value while keeping every operation in the original expression or formula.

Use this order: (1) write the original expression; (2) replace every letter with its value, putting negative values in brackets; (3) evaluate powers and brackets first; (4) multiply or divide; (5) add or subtract; (6) check that every occurrence was replaced.

$F=2a^2+3b,\qquad a=-3,\ b=4$

Substitute before calculating: $F=2(-3)^2+3(4)$ . Then $(-3)^2=9$ , so $F=2(9)+12=30$ . The brackets ensure the square applies to the whole negative value.

A formula may contain several letters. For $s=\frac12(u+v)t$ with $u=20$ , $v=30$ and $t=7$ , substitution gives $s=\frac12(20+30)(7)=175$ . Keep the grouping shown by the formula.

Do not change signs or operations while substituting. In particular, $(-3)^2=9$ but $-3^2=-9$ because the exponent applies before the leading minus when there are no brackets.

E2.2 Algebraic manipulation

Like terms have exactly the same variable part, including the same powers. Their coefficients can be added or subtracted because they count the same kind of algebraic quantity.

Terms	Like?	Reason
$3x$ and $-5x$	yes	both have variable part $x$
$2a^2$ and $7a^2$	yes	both have variable part $a^2$
$4x$ and $4x^2$	no	the powers differ
$3ab$ and $-2ba$	yes	$ab=ba$

Group like terms, then combine only their coefficients: $4a^2-3ab+2+5a^2+7ab-6=(4+5)a^2+(-3+7)ab+(2-6)=9a^2+4ab-4$ .

A reliable check is to identify the variable part of every term before calculating. Constants are like terms with one another, but a constant is not like a term containing a variable.

Do not combine unlike terms: $3x+2x^2$ cannot become $5x^3$ or $5x^2$ . Simplifying changes the form of an expression, not its value.

Expanding removes brackets by multiplying every term in one factor by every term in the other factor. This is the distributive law: $a(b+c)=ab+ac$ .

For one bracket, multiply the outside term by each inside term: $3x(2x-4y)=6x^2-12xy$ . Keep the sign attached to each term.

$(3x+y)(x-4y)$

Make all four products: $3x^2-12xy+xy-4y^2$ . Then collect like terms to obtain $3x^2-11xy-4y^2$ .

With more than two brackets, expand two factors first, simplify, then multiply by the next factor. For example, start $(x-2)(x+3)(2x+1)$ by finding $(x-2)(x+3)=x^2+x-6$ .

A negative term changes the sign of its product. Missing just one cross-product makes a double-bracket expansion incomplete; multiplying only the first terms is not enough.

Factorising is the reverse of expanding. To factorise fully by extraction, place the greatest factor shared by every term outside one pair of brackets.

Find the greatest common numerical factor. For each variable, take the lowest power present in every term. Divide every original term by that common factor to form the bracket.

$18x^2y+24xy^2$

The greatest common numerical factor is $6$ ; both terms also contain $xy$ . Therefore $18x^2y+24xy^2=6xy(3x+4y)$ .

Expand the result to check: $6xy(3x+4y)=18x^2y+24xy^2$ . If the bracket still has a common factor, the expression has not been factorised fully.

A factor must divide every term. Do not extract a higher variable power than the smallest power common to all terms.

Factorise completely by extracting any common factor first, then recognise the structure that remains. Expanding the final factors should reproduce the original expression.

Structure	Factorised form	Recognition cue
$ax+bx+kay+kby$	$(a+b)(x+ky)$	group pairs with a repeated bracket
$a^2x^2-b^2y^2$	$(ax-by)(ax+by)$	difference of two squares
$a^2+2ab+b^2$	$(a+b)^2$	square ends and twice their product
$ax^2+bx+c$	two linear factors	product gives $ac$ and sum gives $b$
$ax^3+bx^2+cx$	$x(ax^2+bx+c)$ first	every term contains $x$

For $6x^2+7x-20$ , use numbers with product $6(-20)=-120$ and sum $7$ : $15$ and $-8$ . Split and group: $6x^2+15x-8x-20=3x(2x+5)-4(2x+5)=(3x-4)(2x+5)$ .

For $x^3-25x$ , extract $x$ first, then use a difference of squares: $x(x^2-25)=x(x-5)(x+5)$ .

A single expression may need more than one step. For example, $20x^2-45y^2=5(4x^2-9y^2)=5(2x-3y)(2x+3y)$ .

Pattern conditions matter: $a^2+b^2$ is not a difference of squares, and the middle term of a perfect square must be exactly $2ab$ . This card factorises expressions; it does not solve equations.

Completing the square rewrites $ax^2+bx+c$ as a multiple of one squared binomial plus a constant. It reverses the identity $(x+p)^2=x^2+2px+p^2$ .

$x^2+bx+c=\left(x+ rac b2 ight)^2+c-\left( rac b2 ight)^2$

When $a
e1 $, first factor$ a $from the$ x^2 $and$ x $terms. Equivalently,$ ax^2+bx+c=a\left(x+rac{b}{2a}
ight)^2+c-rac{b^2}{4a} $for$ a
e0$.

$2x^2+12x+5=2(x^2+6x)+5=2[(x+3)^2-9]+5=2(x+3)^2-13$ . Half the coefficient of $x$ inside the bracket to obtain $3$ .

Expand to verify: $2(x+3)^2-13=2(x^2+6x+9)-13=2x^2+12x+5$ .

Factor out only from the terms that contain $x$ before completing the square; do not accidentally divide the outside constant. Rewriting the expression is not the same as solving a quadratic equation.

E2.3 Algebraic fractions

Algebraic fractions follow the same operation rules as numerical fractions. Denominators must be non-zero, and the final expression should be simplified without changing its value.

Operation	Reliable move
add or subtract	use a lowest common denominator, rewrite every numerator, then combine
multiply	factor first, multiply numerators and denominators, then cancel common factors
divide	multiply by the reciprocal of the second fraction, then simplify

$rac{2}{x-1}+ rac{3}{x+2}= rac{2(x+2)+3(x-1)}{(x-1)(x+2)}= rac{5x+1}{(x-1)(x+2)}$

The common denominator must contain every required factor. In the example, $x
e1,-2$ because those values make an original denominator zero.

For multiplication, $rac{4a}{5} imes rac{15}{8a}= rac32$ for $a
e0 $. For division,$ rac{3p}{7}\divrac{9p}{14q}=rac{3p}{7} imesrac{14q}{9p}=rac{2q}{3} $, with$ p
e0 $and$ q
e0$.

Cancel only common factors in a product. Terms joined by $+$ or $-$ cannot be cancelled: in $rac{x+3}{x}$ , the $x$ is not a factor of the whole numerator.

A rational expression simplifies when its numerator and denominator are written as products and a factor common to both is cancelled. The cancelled factor must be non-zero.

Factorise the numerator fully; factorise the denominator fully; identify identical factors; cancel only those factors; state every value excluded by the original denominator; expand the remaining factors only if a different final form is required.

$rac{2x^2-5x-12}{3x^2-12x}= rac{(2x+3)(x-4)}{3x(x-4)}= rac{2x+3}{3x}$

The original denominator is $3x(x-4)$ , so $x
e0,4 $. Although$ (x-4) $disappears from the simplified expression,$ x=4$ is still excluded because it made the original expression undefined.

Check by multiplying the simplified numerator and denominator by the cancelled factor: $rac{2x+3}{3x} imes rac{x-4}{x-4}$ reconstructs the factorised original expression whenever $x
e4$.

Cancellation removes factors, not matching-looking terms. For example, $rac{x+5}{x}$ cannot be reduced, while $rac{x(x+5)}{x}=x+5$ is valid only for $x
e0$.

E2.4 Indices II

An index tells you how a base is used. Positive whole-number indices represent repeated multiplication; zero, negative and fractional indices extend the same pattern consistently.

Index form	Meaning	Condition
$a^n$	multiply $n$ copies of $a$	$n$ is a positive integer
$a^0$	$1$	$a
e0$
$a^{-n}$	$1/a^n$	$a
e0$
$a^{1/n}$	$\sqrt[n]{a}$	for real even roots, $a\ge0$
$a^{m/n}$	$\sqrt[n]{a^m}=(\sqrt[n]{a})^m$	use a real root where defined

$64^{ rac23}=\left(\sqrt[3]{64} ight)^2=4^2=16$

A negative index does not make the value negative: $5^{-2}=1/5^2=1/25$ . It moves a non-zero factor across the fraction line and changes the sign of its index.

For real $x$ , $(64x^4)^{1/2}=8x^2$ because $x^4=(x^2)^2$ and $x^2\ge0$ . Keep root conditions in mind when the variable power is not automatically non-negative.

$a^0=1$ applies only when $a
e0 $;$ 0^0 $is not assigned this value here. Also,$ a^{-n} $means a reciprocal, not$ -a^n$.

Index laws combine powers only when their bases and operation fit the law. They also let an exponential equation be solved by rewriting both sides with one common base.

Structure	Law
same base multiplied	$a^m a^n=a^{m+n}$
same base divided	$a^m/a^n=a^{m-n}$ , $a
e0$
power raised to a power	$(a^m)^n=a^{mn}$
product raised to a power	$(ab)^n=a^n b^n$
quotient raised to a power	$(a/b)^n=a^n/b^n$ , $b
e0$

Apply the outer index to every factor: $(27x^9)^{2/3}=27^{2/3}x^{9(2/3)}=9x^6$ . The coefficient and variable power are both affected.

$4^{x+1}=8^{x-1}\quad\Longrightarrow\quad2^{2x+2}=2^{3x-3}$

Equal positive bases give equal exponents, so $2x+2=3x-3$ and $x=5$ . This method needs no logarithms; first look for a common base such as $2$ , $3$ , $5$ or a reciprocal power.

Do not add indices when terms are added: $a^m+a^n$ is not generally $a^{m+n}$ . In $(a^m)^n$ , multiply the indices; in $a^m a^n$ , add them.

E2.5 Equations

Algebra models a situation by naming unknown quantities and translating relationships into expressions, equations or formulas. An expression has no equality sign; an equation states that two quantities are equal; a formula links several variables.

Words	Algebraic structure
5 more than $x$	$x+5$
5 less than $x$	$x-5$
5 less than twice $x$	$2x-5$
$y$ is 3 times $x$	$y=3x$
total is 40	add the parts and set the sum equal to 40

Define each unknown with its unit; build each quantity from that definition; use the relationship word to choose the operation; then check that both sides of an equation have the same meaning and units.

$\text{adult tickets}=x,\quad \text{child tickets}=x+6,\quad x+(x+6)=40$

When two unknowns are linked by two independent facts, construct two equations. For example, if $x+y=18$ and adult tickets cost $8 while child tickets cost$ 5, the revenue fact is $8x+5y=111$ .

Keep an equality only when the words assert equality. Do not write $2x+5=\,$ as an expression, and do not reverse phrases such as ‘5 less than $x$ ’.

Solving a linear equation means finding the value that makes both sides equal. Every valid step performs the same reversible operation on both sides, so the balance and the solution are preserved.

Expand brackets; clear numerical fractions if helpful; collect all terms containing the unknown on one side; collect constants on the other; divide by the coefficient; substitute the result into the original equation to check.

$3(2x-5)+4=2(x+7)\;\Longrightarrow\;6x-11=2x+14\;\Longrightarrow\;4x=25\;\Longrightarrow\;x=\frac{25}{4}$

Substitution gives $3(2\cdot25/4-5)+4=53/2$ and $2(25/4+7)=53/2$ , so the value satisfies both sides.

A term changes sign because the same term was added or subtracted on both sides—not because it ‘moves across’. Distribute a negative multiplier to every term inside its bracket.

A fractional equation can be converted into an ordinary linear or quadratic equation by multiplying every term by a common denominator. Values that make an original denominator zero are excluded from the start.

List excluded values; factor denominators if needed; choose a lowest common denominator; multiply every term on both sides by it; solve the resulting equation; reject any excluded or non-satisfying result.

$\frac1{x+1}+\frac9{x+9}=1,\quad x\ne-1,-9$

Multiplying by $(x+1)(x+9)$ gives $(x+9)+9(x+1)=(x+1)(x+9)$ . This simplifies to $x^2-9=0$ , so $x=3$ or $x=-3$ ; both are allowed and satisfy the original equation.

Cancelling a denominator is shorthand for multiplying every term by a non-zero expression. Never multiply only selected terms, and never accept a root that made an original denominator zero.

A solution to two simultaneous linear equations is one ordered pair that satisfies both equations. Elimination removes one unknown by combining aligned equations; substitution replaces one unknown with an equivalent expression.

Structure	Efficient method
matching or easily matched coefficients	elimination
one variable already isolated	substitution
neither is convenient	rearrange or scale first, then choose

$x+2y=13,\quad x+5y=22$

Subtracting the first equation from the second gives $3y=9$ , so $y=3$ . Substitution into $x+2y=13$ gives $x=7$ . The ordered pair is $(7,3)$ .

Check both originals: $7+2(3)=13$ and $7+5(3)=22$ . A pair is not a solution unless it passes both equations.

When subtracting equations, subtract every term, including signs and constants. Do not report separate unpaired values of $x$ and $y$ .

Substitution turns a linear–non-linear simultaneous system into one quadratic equation. Its two roots can produce two intersection points, and each root must be paired with its corresponding value of the other variable.

Rearrange the linear equation for one variable; substitute into the non-linear equation; expand and collect into a quadratic; solve it; substitute each root separately into the linear equation; verify each ordered pair in both originals.

$y=4-x,\quad x^2+2y^2=67$

Substitution gives $x^2+2(4-x)^2=67$ , so $3x^2-16x-35=0=(3x+5)(x-7)$ . Hence $x=7$ gives $y=-3$ , while $x=-5/3$ gives $y=17/3$ .

Do not mix the $y$ value from one root with the other $x$ root. A tangent may give one repeated solution; no real intersection gives no real ordered pair.

A quadratic equation can be solved by factorisation, completing the square or the quadratic formula. The equation must first be written as $ax^2+bx+c=0$ with $a
e0$.

Method	Best use	Result
factorisation	factors are visible or easy to find	exact roots
completing the square	vertex form or structural insight is useful	exact or surd roots
quadratic formula	works for every quadratic	exact or rounded roots

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

For $2x^2-x-6=0$ , $(2x+3)(x-2)=0$ , so $x=-3/2$ or $x=2$ . The zero-product rule applies only after one side equals zero.

For $x^2-6x+1=0$ , $(x-3)^2-8=0$ , so $x=3\pm\sqrt8=3\pm2\sqrt2$ . This is completed-square form and an exact surd solution.

The discriminant $b^2-4ac$ predicts two distinct real roots when positive, one repeated real root when zero, and no real roots when negative.

Keep the $\pm$ when taking a square root, place the whole numerator over $2a$ , and round only at the final step when a decimal accuracy is requested.

Changing the subject isolates a chosen variable using inverse operations while preserving equality. When the subject appears more than once, collect all its terms and factor it out before dividing.

Where the subject appears	Key move
once in a chain of operations	reverse the operations in a valid order
in a denominator	clear denominators first
in two or more terms	collect those terms, then factor
raised to a power	isolate the power, then take the appropriate root

$2mh=g(1-h)\;\Longrightarrow\;2mh=g-gh\;\Longrightarrow\;h(2m+g)=g\;\Longrightarrow\;h=\frac{g}{2m+g}$

$m=2p+\sqrt{\frac{x}{y}}\;\Longrightarrow\;m-2p=\sqrt{\frac{x}{y}}\;\Longrightarrow\;x=y(m-2p)^2$

From $A=\pi r^2$ , a radius is non-negative, so $r=\sqrt{A/\pi}$ . Without a contextual sign restriction, solving $x^2=k$ gives $x=\pm\sqrt{k}$ for $k\ge0$ .

Do not divide by a factor that could be zero without recording the restriction. Squaring can hide a sign condition, so check the rearranged formula against the original context.

E2.6 Inequalities

An inequality describes a set of possible values rather than one value. Its symbol controls both the endpoint and the direction shown on a number line.

Inequality	Endpoint	Values shown
$x<a$	open circle at $a$	to the left
$x\le a$	closed circle at $a$	to the left
$x>a$	open circle at $a$	to the right
$x\ge a$	closed circle at $a$	to the right

$-2<x\le4$

The compound inequality means values greater than $-2$ and at most $4$ . Draw an open circle at $-2$ , a closed circle at $4$ , and one continuous segment between them.

An open circle excludes its endpoint; a closed circle includes it. The arrow or shaded segment shows the allowed values, not merely the direction in which the symbol points.

A linear inequality is solved with balance-preserving operations like a linear equation, except that multiplying or dividing both sides by a negative number reverses the inequality sign.

Define the unknown, translate phrases precisely—‘more than’ gives $>$ , ‘at least’ gives $\ge$ , ‘fewer than’ gives $<$ and ‘at most’ gives $\le$ —then check the units and context.

$4-3x\ge\frac{6-x}{5}\;\Longrightarrow\;20-15x\ge6-x\;\Longrightarrow\;-14x\ge-14\;\Longrightarrow\;x\le1$

The final sign reverses because both sides are divided by $-14$ . Adding or subtracting a negative number does not by itself reverse the sign.

If $1<x\le5$ and integer values are requested, list $2,3,4,5$ . Respect both endpoints and the stated number set.

Test one value inside the solution and one outside in the original inequality. This catches a reversed sign or an incorrectly included endpoint.

Do not replace $<$ by $\le$ when solving. An answer may need interpretation—for example, a whole-number count can require the least or greatest admissible integer.

A linear inequality in $x$ and $y$ describes one side of a boundary line. Several inequalities overlap to form a feasible region containing every point that satisfies all of them.

Inequality type	Boundary line
strict: $<$ or $>$	broken line; boundary excluded
inclusive: $\le$ or $\ge$	solid line; boundary included

Replace the inequality by an equality and draw its boundary; choose a test point not on the line; substitute it to decide which side satisfies the inequality; following the Cambridge convention, shade the unwanted side unless the question directs otherwise; repeat for every inequality and label the unshaded overlap $R$ .

$x\ge2,\qquad y\ge x,\qquad 2x+y\le8$

The boundaries $x=2$ , $y=x$ and $2x+y=8$ are all solid. A point in $R$ must lie right of $x=2$ , on or above $y=x$ , and on or below $2x+y=8$ .

Shading the unwanted region means the solution is the part left unshaded. Do not infer the correct side from the line’s gradient; use a test point.

To recover inequalities from a drawn region, identify each boundary equation, read whether the line is included, then determine which side contains the region.

Write the equation of each boundary; use $<$ or $>$ for a broken line and $\le$ or $\ge$ for a solid line; select a point clearly inside the region; substitute it to choose the correct sign; verify that every listed inequality contains the whole region.

Boundary	Equation form
vertical line through $a$	$x=a$
horizontal line through $b$	$y=b$
sloping line	find $y=mx+c$ or an equivalent form such as $ax+by=c$

If $R$ lies below a broken line $x+y=4$ , above a solid line $y=1.5$ , and below a solid line $y=2x+1$ , then $x+y<4$ , $y\ge1.5$ , and $y\le2x+1$ .

Line style decides strict versus inclusive; location decides the direction. A correct boundary equation with the wrong inequality sign does not define the same region.

For Cambridge IGCSE Mathematics E2.6, you must represent, solve and interpret inequalities and identify regions. Linear programming problems are explicitly not included.

Included	Not included
draw boundary lines with correct solid or broken style	formulate a business optimisation model
shade unwanted regions and identify the overlap	optimise an objective function systematically
read or list inequalities defining a region	use vertex testing as a general linear-programming procedure

A question may still ask you to read a largest or smallest value directly from a supplied region. Use the graph as directed, but do not add an unrequested linear-programming method or extend the syllabus into optimisation theory.

This card controls the assessable boundary; it does not create an additional calculation method or a new card type.

E2.7 Sequences

To continue a sequence, identify a rule that works between every displayed pair of consecutive terms. The position of a term may be written with subscript notation such as $u_1,u_2,u_3$ .

Pattern check	What to calculate	Typical continuation
additive	first differences	add or subtract the same amount
multiplicative	ratios of consecutive non-zero terms	multiply or divide by the same factor
alternating or cyclic	separate odd/even positions or repeating operations	repeat the full cycle
changing differences	differences, then second or third differences	extend the difference pattern first

For $6,13,32,69,130,\ldots$ , the terms match $n^3+5$ : $1^3+5=6$ , $2^3+5=13$ , and so on. The next term is $6^3+5=221$ .

For $100,50,25,12.5,6.25,\ldots$ , each term is half the previous one, so the next term is $3.125$ .

A rule must fit all shown transitions. Do not assume a constant difference after checking only the first pair, and do not confuse the term value $u_n$ with its position $n$ .

The way differences or ratios behave reveals a sequence family. Recognising the family narrows the possible term-to-term and position-to-term rules.

Sequence family	Diagnostic pattern	Common nth-term shape
linear	constant first difference	$an+b$
quadratic	constant second difference	$an^2+bn+c$
cubic	constant third difference	$an^3+bn^2+cn+d$
exponential	constant non-zero ratio	$ar^{n-1}$

$4,9,14,19,\ldots$ is linear because first differences are $5$ . $3,10,29,66,\ldots$ is cubic because it matches $n^3+2$ . $1,4,16,64,\ldots$ is exponential with ratio $4$ .

A simple combination such as $n^3+2^n$ may not show an immediately constant difference or ratio. Compare the terms with familiar powers, subtract the identifiable component, and test the remaining pattern.

A curved growth pattern is not automatically exponential. Use constant differences or ratios as evidence, and verify the proposed relationship against every given term.

An nth-term rule gives the value at position $n$ directly. Its algebraic form should match the recognised sequence family and reproduce every supplied term.

Family	Starting move
linear	constant difference $a$ gives $an+b$ ; use one term to find $b$
quadratic	constant second difference is $2a$ ; subtract $an^2$ and find the remaining linear rule
cubic	constant third difference is $6a$ ; subtract $an^3$ and analyse the remainder
exponential	ratio $r$ gives $ar^{n-1}$ , where $a$ is the first term

$4,9,14,19,\ldots:\quad u_n=5n-1$

Check positions $n=1,2,3$ before accepting a rule. For $u_n=5n-1$ , these give $4,9,14$ , matching the sequence.

To find a term, substitute its positive integer position. To decide whether $331$ belongs to $u_n=5n-1$ , solve $5n-1=331$ : $n=66.4$ , not a positive integer, so $331$ is not a term.

For $24,12,6,3,\ldots$ , $u_n=24(1/2)^{n-1}$ . The exponent is $n-1$ so that $u_1=24$ .

Matching only the next term does not prove an nth-term rule. Verify all displayed terms, and when testing membership require $n$ to be a permitted positive integer.

E2.8 Proportion

Proportion states how one quantity scales with another. Replace the symbol $\propto$ by an equation containing a constant of proportionality $k$ , use known values to find $k$ , then use the equation for the unknown quantity.

Relationship	Algebraic model
$y$ directly proportional to $x^p$	$y=kx^p$
$y$ inversely proportional to $x^p$	$y=\dfrac{k}{x^p}$
linear	$p=1$
square / square root	$p=2$ / $p= frac12$
cube / cube root	$p=3$ / $p= frac13$

Translate the words into a model; substitute one complete known pair to calculate $k$ ; write the fully determined formula; substitute the new value; solve and check whether the direction and scale are sensible.

$y\propto\frac1{x^2},\quad 7.5=\frac{k}{4^2}\Rightarrow k=120,\quad y=\frac{120}{5^2}=4.8$

If $p$ is directly proportional to $(q+2)^2$ , the whole bracket is squared: $p=k(q+2)^2$ . Do not replace it by $kq^2+2$ .

For $y=kx^p$ , multiplying $x$ by a factor $a$ multiplies $y$ by $a^p$ . For $y=k/x^p$ , it multiplies $y$ by $1/a^p$ . Thus halving the distance in an inverse-square relationship multiplies the result by $4$ .

The symbol $\propto$ is not an equality until $k$ is included. ‘Inverse’ places the full stated expression in the denominator, and roots must apply to exactly the quantity named.

E2.9 Graphs in practical situations

A practical graph connects two measured quantities. Read the axes, units and scale first; then interpret a coordinate, interval or gradient in the context rather than as an isolated number.

Distance–time feature	Meaning
rising straight segment	moving away at constant speed
falling straight segment	moving back at constant speed
horizontal segment	stationary
steeper segment	greater speed
intersection of two journeys	same place at the same time

To read a value, start at the known coordinate, move parallel to an axis until reaching the graph, then move parallel to the other axis and read the scale. Interpolate carefully between labelled marks.

A conversion graph maps one unit or currency to another. Read in either direction using the same line; a straight line through the origin represents a constant conversion factor.

A downward distance–time segment means returning toward the reference point, not travelling at negative speed. A horizontal segment means stopped, not zero distance from the start.

A graph should preserve every supplied value and make the relationship readable. For a journey, each segment must begin where the preceding event ends.

Label both axes with quantity and unit; choose a uniform scale covering all data; plot each coordinate accurately; join points with straight segments when the rate is constant or as directed; check endpoints, stops and continuity against the context.

Journey statement	Graph action
starts later	first point has the stated later time
travels at constant speed	draw a straight sloping segment
rests for a time interval	draw a horizontal segment of that duration
returns to the start	finish on distance $0$

If a cyclist travels $12$ km home at $24$ km/h, the return takes $12/24=0.5$ h, or $30$ minutes. Use that duration to place the final endpoint.

Do not join points before checking the event order and units. A visually plausible line is wrong if its endpoint time, distance or constant-rate gradient does not match the data.

A gradient is a rate of change: vertical change divided by horizontal change. Its meaning and units come from the graph axes.

Graph	Gradient means	Units example
distance–time	speed	km/h or m/s
speed–time	acceleration	m/s²
horizontal distance–time segment	zero speed	distance unit per time unit
horizontal speed–time segment	zero acceleration	speed unit per time unit

$a=\frac{\Delta v}{\Delta t}=\frac{12-0}{40-0}=0.3\text{ m/s}^2$

A negative speed–time gradient represents deceleration. When the question asks for the deceleration, report its positive magnitude unless a signed acceleration is requested.

For a curve, draw a tangent that touches at the required point and follows the local direction. Choose two well-separated points on the tangent—not necessarily on the curve—and calculate rise divided by run to estimate the instantaneous rate.

Do not use the height of a speed–time graph as acceleration: height is speed, while gradient is acceleration. Keep time and speed units consistent before dividing.

On a speed–time graph, area equals speed multiplied by time, so the area between the graph and the time axis is the distance travelled.

Linear section	Area
rectangle	base $imes$ height
triangle	$frac12 imes$ base $imes$ height
trapezium	$frac12 imes$ (sum of parallel sides) $imes$ separation

Split the region at every change of gradient; label the time width and speed height of each rectangle, triangle or trapezium; convert units before multiplying; calculate each area; add all non-overlapping parts.

$d=\frac12(6+12)(30)+12(60)=990\text{ m}$

Speed in m/s multiplied by time in seconds gives metres. Speed in km/h multiplied by minutes requires converting minutes to hours first.

After finding total distance, average speed is $\text{total distance}/\text{total time}$ . It is not generally the arithmetic mean of the displayed speeds.

Cambridge limits these area calculations to linear graph sections. Avoid double-counting when subdividing, and do not use area under a distance–time graph as distance.

E2.10 Graphs of functions

A function graph shows every plotted pair $(x,f(x))$ . A reliable graph begins with an accurate table of values, then connects points according to the function’s continuous shape and domain.

Choose the stated $x$ values; substitute each one carefully; keep enough decimal accuracy for plotting; label axes and use a uniform scale; plot coordinates; draw a smooth curve or straight line as appropriate; check intercepts, separate branches and overall shape.

Form	Recognition feature
$ax+c$	straight line
$ax^2+bx+c$	parabola with one turning point
cubic power combination	S-like or turning cubic shape
$a/x+c$ or $a/x^2+c$	separated reciprocal branches; excluded denominator values
$ab^x+c$	exponential curve approaching a horizontal level
specified $ax^n$ combinations	shape depends on the allowed power $n$ and coefficients

For the specified power forms, $n$ may be $-2,-1,-\tfrac12,0,\tfrac12,1,2,3$ , and no more than three $ax^n$ terms are combined. Respect real-domain restrictions for roots and negative powers.

Do not join reciprocal branches across an undefined value or force a curve through an uncalculated point. A sketch shows key structure; a drawn graph from a table must also preserve scale and plotted accuracy.

A graphical solution is an $x$ -coordinate where two required expressions have equal $y$ values. Roots are intersections with the $x$ -axis; simultaneous solutions are intersections of two graphs.

Equation	Graphical target
$f(x)=0$	where $y=f(x)$ crosses or touches the $x$ -axis
$f(x)=k$	intersections of $y=f(x)$ with horizontal line $y=k$
$f(x)=g(x)$	intersections of $y=f(x)$ and $y=g(x)$

To solve $x^3+4x^2-x-6=0$ using an existing graph of $y=x^3+4x^2-4$ , rearrange to $x^3+4x^2-4=x+2$ . Draw $y=x+2$ and read the intersection $x$ -coordinates.

Draw any required line accurately with a ruler, identify every intersection within the stated domain, project vertically to the $x$ -axis and report values to the precision supported by the grid.

Zero, one, two or more visible intersections mean the equation has that many graphical solutions in the shown interval. A tangent contact counts as one repeated root.

Do not read the $y$ -coordinate when the question asks for $x$ , and do not invent accuracy beyond the graph scale. Every algebraic rearrangement must preserve the same equality.

Exponential change multiplies by the same factor over equal time intervals. Growth curves rise increasingly quickly; decay curves fall quickly at first and then level towards zero.

Situation	Model	Multiplier	Graph behaviour
growth at $r\%$ per interval	$P=P_0(1+r/100)^t$	greater than $1$	increasing
decay at $r\%$ per interval	$P=P_0(1-r/100)^t$	between $0$ and $1$	decreasing

Choose sensible time values including $t=0$ ; calculate the corresponding quantities; plot $(t,P)$ with labelled units; join with a smooth curve; check that the vertical intercept is the initial value $P_0$ .

$P=40000(1.15)^t$

The bacteria model starts at $40000$ and multiplies by $1.15$ each hour. At $t=3$ , $P=40000(1.15)^3=60835$ . Equal vertical additions would indicate linear, not exponential, growth.

For $M=20(0.9)^t$ , the mass stays positive and approaches $0$ as time increases. Read a threshold time from the first whole $t$ for which the curve is below the stated level.

A percentage decrease uses $1-r/100$ , not $r/100$ . Do not draw exponential decay crossing below zero when the model has a positive initial amount and positive multiplier.

E2.11 Sketching curves

A sketch records a function’s essential structure rather than plotting a dense table: family shape, intercepts, turning points, symmetry, asymptotes and end behaviour must agree with its equation.

Family and syllabus form	Essential sketch features
linear: $ax+by=c$	straight line; $x$ - and $y$ -intercepts
quadratic: $y=ax^2+bx+c$	parabola; axis of symmetry; one maximum or minimum; roots
cubic: $y=ax^3+b$ or $y=ax^3+bx^2+cx$	cubic end direction; roots; up to two turning points
reciprocal: $y=a/x+b$	two branches; vertical asymptote $x=0$ ; horizontal asymptote $y=b$
exponential: $y=ar^x+b$	$y$ -intercept $a+b$ ; horizontal asymptote $y=b$ ; growth or decay direction

Identify the family and leading sign; find exact intercepts where possible; find required symmetry, turning points and asymptotes; place and label these features; draw a smooth shape with correct end behaviour that approaches but does not cross a vertical asymptote.

$y=x^2+10x+14=(x+5)^2-11$

The completed-square form gives the minimum $(-5,-11)$ and axis of symmetry $x=-5$ . The positive squared coefficient makes the parabola open upwards; roots, if required, are symmetric about $x=-5$ .

Factorisation reveals intercept behaviour. In $y=(x+1)(x-3)^2$ , the graph crosses at $x=-1$ but touches and turns at the repeated root $x=3$ ; $y=9$ when $x=0$ .

For $y=2/x-1$ , the graph is undefined at $x=0$ , so $x=0$ is vertical; as $|x|$ grows, $2/x$ approaches $0$ , so $y=-1$ is horizontal.

Do not use differentiation to find cubic turning points in this objective. A sketch is still constrained: labelled roots, repeated-root behaviour, symmetry, turning points and asymptotes must match the equation.

E2.12 Differentiation

The gradient of a curve at one point is the gradient of its tangent there. A tangent follows the curve’s local direction but is treated as a straight line for the gradient calculation.

Mark the required point; place a ruler so the line just touches and matches the curve’s local direction; draw a long tangent; choose two well-separated, readable points on the tangent; calculate vertical change divided by horizontal change; include appropriate units.

$\text{gradient}=\frac{y_2-y_1}{x_2-x_1}$

The calculation points need not lie on the original curve. A longer triangle reduces the effect of reading error; retain the sign of the rise and run.

A rising tangent has positive gradient, a falling tangent negative gradient, and a horizontal tangent gradient $0$ . The estimate depends on the accuracy of both tangent and coordinate readings.

A chord through two curve points estimates an average gradient over an interval, not the instantaneous gradient at the named point.

The derivative $\dfrac{dy}{dx}$ gives the gradient function. For each allowed power term, multiply by the old power and reduce that power by one.

$\frac{d}{dx}(ax^n)=anx^{n-1}$

Original term	Derivative
$ax^n$	$anx^{n-1}$
$bx$	$b$
constant $c$	$0$

$y=5+8x-\frac43x^3\quad\Longrightarrow\quad\frac{dy}{dx}=8-4x^2$

Here $a$ is rational, $n$ is a non-negative integer, and the syllabus uses simple sums of no more than three such terms. Write the derivative in $dy/dx$ notation when required.

If $\frac{d}{dx}(3x^q)=15x^4$ , compare coefficient and power: $q-1=4$ and $3q=15$ , so $q=5$ .

Do not leave a constant in the derivative or reduce the coefficient instead of the power. The rule applies term by term to the stated polynomial scope.

Once a derivative is known, substituting an $x$ -value gives a gradient. A stationary point occurs where that gradient is zero.

Required result	Derivative move
gradient at $x=a$	calculate $dy/dx$ at $a$
points with gradient $m$	solve $dy/dx=m$ , then find each $y$
stationary points	solve $dy/dx=0$ , then substitute each $x$ into the original function
tangent equation	use derivative gradient and point in $y-y_1=m(x-x_1)$

$y=x^3-3x+4,\quad\frac{dy}{dx}=3x^2-3=0\quad\Longrightarrow\quad x=\pm1$

Substitute into the original function, not the derivative: $y(1)=2$ and $y(-1)=6$ , so the stationary points are $(1,2)$ and $(-1,6)$ .

If the curve passes through $(2,6)$ and the derivative gives gradient $7$ , then $y-6=7(x-2)$ , so the tangent is $y=7x-8$ .

Solving $dy/dx=0$ gives only the $x$ -coordinates. A complete stationary point requires the corresponding $y$ from the original function.

A stationary point is classified by how the function behaves around it. Cambridge accepts an accurate sketch, the gradient signs on either side, or the second derivative.

Evidence	Maximum	Minimum
gradient before and after	$+ o-$	$- o+$
second derivative at the point	$d^2y/dx^2<0$	$d^2y/dx^2>0$
accurate sketch	curve turns from rising to falling	curve turns from falling to rising

$\frac{dy}{dx}=3x^2-3,\qquad\frac{d^2y}{dx^2}=6x$

At $x=1$ , $d^2y/dx^2=6>0$ , so $(1,2)$ is a minimum. At $x=-1$ , $d^2y/dx^2=-6<0$ , so $(-1,6)$ is a maximum.

State both the classification and the evidence. Merely writing ‘maximum’ or ‘minimum’ does not show how the decision follows from the derivative or sketch.

Points of inflection are not required. If the second derivative is zero, this test alone is inconclusive; use an allowed alternative rather than assigning a maximum or minimum automatically.

E2.13 Functions

A function assigns exactly one output to each allowed input. In $f(x)$ , $x$ is the input and $f(x)$ is its output.

Term	Meaning
domain	all allowed input values
range	all output values produced from that domain
$f(a)$	substitute $a$ for every $x$ in the rule

$f(x)=3x-5,\qquad \text{domain}=\{-3,0,2\}$

Evaluate the rule for each domain value: $f(-3)=-14$ , $f(0)=-5$ and $f(2)=1$ . Therefore the range is $\{-14,-5,1\}$ .

An input can be an expression. For example, $f(x-3)=3(x-3)-5=3x-14$ . Substitute the whole expression in place of $x$ before simplifying.

To find an input from an output, form an equation. If $f(x)=7$ , solve $3x-5=7$ , giving $x=4$ . This reverses one evaluation; it is not yet a formula for the inverse function.

The range comes from the stated domain, not from every value the formula could accept. If two inputs give the same output, list that output only once in the range set.

An inverse function undoes the original function. If $f$ takes an input to an output, $f^{-1}$ takes that output back to the input.

Write $y=f(x)$ ; interchange $x$ and $y$ ; rearrange to make $y$ the subject; then write the result as $f^{-1}(x)$ .

$f(x)=4x-1:\quad x=4y-1\quad\Longrightarrow\quad y=\frac{x+1}{4}\quad\Longrightarrow\quad f^{-1}(x)=\frac{x+1}{4}$

Check by composing the functions: $f(f^{-1}(x))=4\left( rac{x+1}{4} ight)-1=x$ . Applying them in the opposite order also returns the starting value wherever both are defined.

If $h^{-1}(x)=5$ , apply $h$ to both sides: $x=h(5)$ . This is often quicker than first finding an inverse formula.

The inverse reverses the mapping, so the original range becomes the inverse domain and the original domain becomes the inverse range.

$f^{-1}(x)$ does not mean $1/f(x)$ . The superscript $-1$ names the inverse operation; a reciprocal divides $1$ by the function value.

A composite function uses the output of one function as the input of another. The function nearest $x$ is applied first.

$gf(x)=g(f(x)),\qquad fg(x)=f(g(x))$

Expression	First operation	Second operation
$gf(x)$	find $f(x)$	substitute into $g$
$fg(x)$	find $g(x)$	substitute into $f$
$ff(x)$	find $f(x)$	apply $f$ again

$f(x)=2x+1,\quad g(x)=x^2+4:\qquad gf(x)=(2x+1)^2+4=4x^2+4x+5$

For $f(x)=\dfrac{3}{x+2}$ and $g(x)=(3x+5)^2$ , $fg(x)=\dfrac{3}{(3x+5)^2+2}=\dfrac{1}{3x^2+10x+9}$ . Substitute the entire inner expression, expand carefully, then simplify the fraction.

For a numerical composite such as $fg(1)$ , calculate $g(1)$ first and then use that result as the input to $f$ . Writing the intermediate value makes the order visible.

In general, $gf(x)$ and $fg(x)$ are different because the order changes. This syllabus does not require finding the domains or ranges of composite functions.

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