This handout covers Topic 2, Algebra and graphs. Parts marked (Extended) are only tested on the Extended papers; everything else is for both levels.
Algebra and graphs
IGCSE Mathematics · Topic 2
2.1
Working with algebra
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Know that letters can be used to represent generalised numbers. | |
| 2 Substitute numbers into expressions and formulas. |
Source: Cambridge International syllabus
In algebra 代数 we use letters to stand for numbers. A letter whose value can change is a variable 变量. To substitute 代入 means to put a number in place of a letter.
Worked example. Find the value of $3x^{2} - 2y$ when $x = 4$ and $y = 5$.
Algebra manipulation route
Follow expression work from collecting terms to solving.
| English | Chinese | Pinyin |
|---|---|---|
| algebra | 代数 | dài shù |
| variable | 变量 | biàn liàng |
| substitute | 代入 | dài rù |
2.2
Simplifying and expanding
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Simplify expressions by collecting like terms. | Simplify means give the answer in its simplest form, e.g. $2a + 3b + 5a - 9b = 7a - 6b$. |
| 2 Expand products of algebraic expressions. | e.g. expand $3x(2x - 4y)$. Includes products of two brackets involving one variable, e.g. expand $(2x + 1)(x - 4)$. |
| 3 Factorise by extracting common factors. | Factorise means factorise fully, e.g. $9x^2 + 15xy = 3x(3x + 5y)$. |
| Subject content | Notes and examples |
|---|---|
| 1 Simplify expressions by collecting like terms. | Simplify means give the answer in its simplest form, e.g. $2a^2 + 3ab - 1 + 5a^2 - 9ab + 4 = 7a^2 - 6ab + 3$. |
| 2 Expand products of algebraic expressions. | e.g. expand $3x(2x - 4y)$, $(3x + y)(x - 4y)$. Includes products of more than two brackets, e.g. expand $(x - 2)(x + 3)(2x + 1)$. |
| 3 Factorise by extracting common factors. | Factorise means factorise fully, e.g. $9x^2 + 15xy = 3x(3x + 5y)$. |
| 4 Factorise expressions of the form: • $ax + bx + kay + kby$ • $a^2x^2 - b^2y^2$ • $a^2 + 2ab + b^2$ • $ax^2 + bx + c$ • $ax^3 + bx^2 + cx$. | |
| 5 Complete the square for expressions in the form $ax^2 + bx + c$. |
Source: Cambridge International syllabus
A term 项 is a single part of an expression 表达式, such as $5a$ or $-9b$. Like terms 同类项 have exactly the same letters; you may add or subtract them. The number in front of the letter is the coefficient 系数.
Expanding a bracket with an area model
Collecting like terms: add terms with the same letter
Worked example. Simplify $2a^{2} + 3ab - 1 + 5a^{2} - 9ab + 4$.
Collect like terms: $2a^{2} + 5a^{2} = 7a^{2}$, $3ab - 9ab = -6ab$, $-1 + 4 = 3$. So the answer is
To expand 展开 means to multiply out brackets 括号. Multiply every term inside by the term outside; for two brackets, multiply every term in the first by every term in the second.
Worked examples.
For three brackets (Extended), expand two first, then multiply by the third:
| English | Chinese | Pinyin |
|---|---|---|
| term | 项 | xiàng |
| expression | 表达式 | biǎo dá shì |
| like terms | 同类项 | tóng lèi xiàng |
| coefficient | 系数 | xì shù |
| expand | 展开 | zhǎn kāi |
| brackets | 括号 | kuò hào |
2.2
Factorising
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Simplify expressions by collecting like terms. | Simplify means give the answer in its simplest form, e.g. $2a + 3b + 5a - 9b = 7a - 6b$. |
| 2 Expand products of algebraic expressions. | e.g. expand $3x(2x - 4y)$. Includes products of two brackets involving one variable, e.g. expand $(2x + 1)(x - 4)$. |
| 3 Factorise by extracting common factors. | Factorise means factorise fully, e.g. $9x^2 + 15xy = 3x(3x + 5y)$. |
| Subject content | Notes and examples |
|---|---|
| 1 Simplify expressions by collecting like terms. | Simplify means give the answer in its simplest form, e.g. $2a^2 + 3ab - 1 + 5a^2 - 9ab + 4 = 7a^2 - 6ab + 3$. |
| 2 Expand products of algebraic expressions. | e.g. expand $3x(2x - 4y)$, $(3x + y)(x - 4y)$. Includes products of more than two brackets, e.g. expand $(x - 2)(x + 3)(2x + 1)$. |
| 3 Factorise by extracting common factors. | Factorise means factorise fully, e.g. $9x^2 + 15xy = 3x(3x + 5y)$. |
| 4 Factorise expressions of the form: • $ax + bx + kay + kby$ • $a^2x^2 - b^2y^2$ • $a^2 + 2ab + b^2$ • $ax^2 + bx + c$ • $ax^3 + bx^2 + cx$. | |
| 5 Complete the square for expressions in the form $ax^2 + bx + c$. |
Source: Cambridge International syllabus
To factorise 因式分解 is the opposite of expanding: write the expression as a product of brackets. Always take out the common factor 公因式 first.
Worked example. $9x^{2} + 15xy = 3x(3x + 5y)$, because $3x$ divides both terms.
The following patterns are Extended.
Grouping (four terms): take a common factor from each pair.
Difference of two squares 平方差: $a^{2} - b^{2} = (a + b)(a - b)$.
Perfect square 完全平方: $a^{2} + 2ab + b^{2} = (a + b)^{2}$.
Quadratic 二次 expressions $ax^{2} + bx + c$: find two numbers that multiply to $a \times c$ and add to $b$, then split the middle term.
Worked example. Factorise $2x^{2} + 7x + 3$.
Here $a \times c = 6$ and $b = 7$. The numbers $1$ and $6$ work. Split and group:
For $ax^{3} + bx^{2} + cx$, take out the common $x$ first: $2x^{3} + 7x^{2} + 3x = x(2x^{2} + 7x + 3) = x(x + 3)(2x + 1)$.
| English | Chinese | Pinyin |
|---|---|---|
| factorise | 因式分解 | yīn shì fēn jiě |
| common factor | 公因式 | gōng yīn shì |
| difference of two squares | 平方差 | píng fāng chà |
| perfect square | 完全平方 | wán quán píng fāng |
| quadratic | 二次 | èr cì |
2.2
Completing the square (Extended)
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Simplify expressions by collecting like terms. | Simplify means give the answer in its simplest form, e.g. $2a^2 + 3ab - 1 + 5a^2 - 9ab + 4 = 7a^2 - 6ab + 3$. |
| 2 Expand products of algebraic expressions. | e.g. expand $3x(2x - 4y)$, $(3x + y)(x - 4y)$. Includes products of more than two brackets, e.g. expand $(x - 2)(x + 3)(2x + 1)$. |
| 3 Factorise by extracting common factors. | Factorise means factorise fully, e.g. $9x^2 + 15xy = 3x(3x + 5y)$. |
| 4 Factorise expressions of the form: • $ax + bx + kay + kby$ • $a^2x^2 - b^2y^2$ • $a^2 + 2ab + b^2$ • $ax^2 + bx + c$ • $ax^3 + bx^2 + cx$. | |
| 5 Complete the square for expressions in the form $ax^2 + bx + c$. |
Source: Cambridge International syllabus
A suspension bridge cable hangs in a parabola — the graph of a quadratic.
Completing the square 配方法 rewrites $x^{2} + bx + c$ as $(x + p)^{2} + q$. Take half of the $x$-coefficient, square it, then balance.
Worked example. Write $x^{2} + 6x + 1$ in completed-square form.
Half of $6$ is $3$, and $3^{2} = 9$:
When the $x^{2}$ has a coefficient, take it out of the first two terms first:
| English | Chinese | Pinyin |
|---|---|---|
| completing the square | 配方法 | pèi fāng fǎ |
2.3
Algebraic fractions (Extended)
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Manipulate algebraic fractions. | Examples include: • $\frac{x}{3} + \frac{x - 4}{2}$ • $\frac{2x}{3} - \frac{3(x - 5)}{2}$ • $\frac{3a}{4} \times \frac{9a}{10}$ • $\frac{3a}{4} \div \frac{9a}{10}$ • $\frac{1}{x - 2} + \frac{x + 1}{x - 3}$. |
| 2 Factorise and simplify rational expressions. | e.g. $\frac{x^2 - 2x}{x^2 - 5x + 6}$. |
Source: Cambridge International syllabus
An algebraic fraction 分式 has algebra on the top or bottom. Add and subtract using a common denominator; multiply and divide as with ordinary fractions.
Worked examples.
To simplify a rational expression 有理式, factorise the top and bottom, then cancel common brackets.
Algebraic fraction route
Simplify algebraic fractions by factorising before cancelling.
| English | Chinese | Pinyin |
|---|---|---|
| algebraic fraction | 分式 | fēn shì |
| rational expression | 有理式 | yǒu lǐ shì |
2.4
Indices in algebra
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Understand and use indices (positive, zero and negative). | e.g. $2^x = 32$. Find the value of $x$. |
| 2 Understand and use the rules of indices. | e.g. simplify: • $(5x^3)^2$ • $12a^5 \div 3a^{-2}$ • $6x^7y^4 \times 5x^{-5}y$. Knowledge of logarithms is not required. |
| Subject content | Notes and examples |
|---|---|
| 1 Understand and use indices (positive, zero, negative and fractional). | e.g. solve: • $32^x = 2$ • $5^{x+1} = 25^x$. |
| 2 Understand and use the rules of indices. | e.g. simplify: • $3x^{-4} \times \frac{2}{3}x^{\frac{1}{2}}$ • $\frac{2}{5}x^{\frac{1}{2}} \div 2x^{-2}$ • $\left(\frac{2x^5}{3}\right)^3$. Knowledge of logarithms is not required. |
Source: Cambridge International syllabus
The laws of indices 指数 work with letters too: $a^{m} \times a^{n} = a^{m+n}$, $a^{m} \div a^{n} = a^{m-n}$, and $(a^{m})^{n} = a^{mn}$.
Worked examples.
You can also solve simple index equations by writing both sides with the same base 底数.
Worked example. Solve $2^{x} = 32$. Since $32 = 2^{5}$, you get $x = 5$.
Algebraic index law lab
Classify index-law examples by the rule being used.
| English | Chinese | Pinyin |
|---|---|---|
| indices | 指数 | zhǐ shù |
| base | 底数 | dǐ shù |
2.5
Equations
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Construct simple expressions, equations and formulas. | e.g. write an expression for a number that is 2 more than $n$. Includes constructing linear simultaneous equations. |
| 2 Solve linear equations in one unknown. | Examples include: • $3x + 4 = 10$ • $5 - 2x = 3(x + 7)$. |
| 3 Solve simultaneous linear equations in two unknowns. | |
| 4 Change the subject of simple formulas. | e.g. change the subject of formulas where: • the subject only appears once • there is not a power or root of the subject. |
| Subject content | Notes and examples |
|---|---|
| 1 Construct expressions, equations and formulas. | e.g. write an expression for the product of two consecutive even numbers. Includes constructing simultaneous equations. |
| 2 Solve linear equations in one unknown. | Examples include: • $3x + 4 = 10$ • $5 - 2x = 3(x + 7)$. |
| 3 Solve fractional equations with numerical and linear algebraic denominators. | Examples include: • $\frac{x}{2x + 1} = 4$ • $\frac{2}{x + 2} + \frac{3}{2x - 1} = 1$ • $\frac{x}{x + 2} = \frac{3}{x - 6}$. |
| 4 Solve simultaneous linear equations in two unknowns. | |
| 5 Solve simultaneous equations, involving one linear and one non-linear. | With powers no higher than two. |
| 6 Solve quadratic equations by factorisation, completing the square and by use of the quadratic formula. | Includes writing a quadratic expression in completed square form. Candidates may be expected to give solutions in surd form. The quadratic formula is given in the List of formulas. |
| 7 Change the subject of formulas. | e.g. change the subject of a formula where: • the subject appears twice • there is a power or root of the subject. |
Source: Cambridge International syllabus
An equation 方程 says two expressions are equal. To solve a linear 一次 equation, do the same operation to both sides until the unknown 未知数 is alone.
Solve by doing the same to both sides
Worked example. Solve $5 - 2x = 3(x + 7)$.
Fractional equations (Extended)
A fractional equation 分式方程 has the unknown in a denominator. Multiply both sides by the denominator to clear it.
Worked example. Solve $\dfrac{x}{2x + 1} = 4$.
Simultaneous equations
Simultaneous equations 联立方程 are two equations solved together. For two linear equations, add or subtract to remove one letter.
Worked example. Solve $2x + y = 7$ and $3x - y = 8$.
Adding removes $y$: $5x = 15$, so $x = 3$. Then $y = 7 - 2(3) = 1$.
The solution of simultaneous equations is where their graphs cross
For one linear and one quadratic equation (Extended), substitute the linear into the curve.
Worked example. Solve $y = x + 2$ and $y = x^{2}$.
so $x = 2$ (giving $y = 4$) or $x = -1$ (giving $y = 1$).
Solving quadratic equations (Extended)
There are three methods.
- By factorising: $x^{2} + 5x + 6 = 0 \Rightarrow (x + 2)(x + 3) = 0 \Rightarrow x = -2$ or $x = -3$.
- By completing the square: $x^{2} + 6x + 1 = 0 \Rightarrow (x + 3)^{2} = 8 \Rightarrow x + 3 = \pm 2\sqrt{2} \Rightarrow x = -3 \pm 2\sqrt{2}$.
- By the quadratic formula 求根公式, which is given in the exam:
$$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}.$$
Worked example (formula). Solve $2x^{2} + 3x - 1 = 0$. Here $a = 2$, $b = 3$, $c = -1$:
Changing the subject
To change the subject 公式变形 of a formula means to rearrange it so a chosen letter is alone on one side.
Worked example. Make $r$ the subject of $A = \pi r^{2}$ (Extended, because of the power).
When the letter appears twice (Extended), collect those terms and factorise. To make $x$ the subject of $y = \dfrac{x + 1}{x - 1}$:
Solving an equation
y = ax² + bx + c
Solving means finding the roots — where the curve crosses the x-axis.
| English | Chinese | Pinyin |
|---|---|---|
| equation | 方程 | fāng chéng |
| linear | 一次 | yī cì |
| unknown | 未知数 | wèi zhī shù |
| fractional equation | 分式方程 | fēn shì fāng chéng |
| simultaneous equations | 联立方程 | lián lì fāng chéng |
| quadratic formula | 求根公式 | qiú gēn gōng shì |
| change the subject | 公式变形 | gōng shì biàn xíng |
2.6
Inequalities
Syllabus
| Subject content | Notes and examples |
|---|---|
| Represent and interpret inequalities, including on a number line. | When representing and interpreting inequalities on a number line: • open circles should be used to represent strict inequalities (<, >) • closed circles should be used to represent inclusive inequalities ($\leqslant$, $\geqslant$) e.g. $-3 \leqslant x < 1$ |
| Subject content | Notes and examples |
|---|---|
| 1 Represent and interpret inequalities, including on a number line. | When representing and interpreting inequalities on a number line: • open circles should be used to represent strict inequalities (<, >) • closed circles should be used to represent inclusive inequalities ($\leqslant$, $\geqslant$). e.g. $-3 \leqslant x < 1$ |
| 2 Construct, solve and interpret linear inequalities. | Examples include: • $3x < 2x + 4$ • $-3 \leqslant 3x - 2 < 7$. |
| 3 Represent and interpret linear inequalities in two variables graphically. | The following conventions should be used: • broken lines should be used to represent strict inequalities (<, >) • solid lines should be used to represent inclusive inequalities ($\leqslant$, $\geqslant$) • shading should be used to represent unwanted regions (unless otherwise directed in the question). e.g. graphs of $x < 1$ and $y \geqslant 1$ |
| 4 List inequalities that define a given region. | Linear programming problems are not included. |
Source: Cambridge International syllabus
An inequality 不等式 uses $<$, $>$, $\leqslant$ or $\geqslant$. Solve it like an equation, but reverse the sign if you multiply or divide by a negative number.
Worked example. Solve $-3 \leqslant 3x - 2 < 7$.
Add $2$ to all parts, then divide by $3$:
On a number line 数轴, use an open circle for $<$ or $>$ (value not included) and a closed circle for $\leqslant$ or $\geqslant$ (value included).
$-\tfrac{1}{3} \leqslant x < 3$: a closed circle includes the end value, an open circle excludes it.
Regions (Extended)
An inequality in two letters describes a region 区域 of the graph. Draw the boundary line (broken for $<$ or $>$, solid for $\leqslant$ or $\geqslant$) and shade the unwanted side. You may also be asked to list the inequalities that define a given region.
Inequalities
y = ax + b
An inequality asks where the line is above or below a value.
| English | Chinese | Pinyin |
|---|---|---|
| inequality | 不等式 | bù děng shì |
| number line | 数轴 | shù zhóu |
| region | 区域 | qū yù |
2.7
Sequences
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Continue a given number sequence or pattern. | e.g. write the next two terms in this sequence: 1, 3, 6, 10, 15, ... , ... |
| 2 Recognise patterns in sequences, including the term-to-term rule, and relationships between different sequences. | |
| 3 Find and use the $n$th term of the following sequences: (a) linear (b) simple quadratic (c) simple cubic. | e.g. find the $n$th term of 2, 5, 10, 17 |
| Subject content | Notes and examples |
|---|---|
| 1 Continue a given number sequence or pattern. | Subscript notation may be used, e.g. $T_n$ is the $n$th term of sequence $T$. |
| 2 Recognise patterns in sequences, including the term-to-term rule, and relationships between different sequences. | Includes linear, quadratic, cubic and exponential sequences and simple combinations of these. |
| 3 Find and use the $n$th term of sequences. |
Source: Cambridge International syllabus
Romanesco broccoli: self-similar spirals form a natural number pattern.
A sequence 数列 is a list of numbers that follow a rule. The term-to-term rule 递推规则 tells you how to get the next term from the one before.
To find the rule for the term in position $n$ (the $n$th term), look at how the terms change.
Linear sequence (the terms go up by the same amount each time). The difference is the multiple of $n$.
Worked example. Find the $n$th term of $2, 5, 8, 11, \dots$
The terms go up by $3$, so start with $3n$. Since $3 \times 1 = 3$ but the first term is $2$, subtract $1$: the $n$th term is $3n - 1$.
Quadratic sequence (the differences themselves change by the same amount). The second difference equals $2 \times$ the coefficient of $n^{2}$.
Worked example. Find the $n$th term of $2, 5, 10, 17, \dots$
First differences are $3, 5, 7$; the second difference is $2$, so the $n^{2}$ part is $1n^{2}$. Subtracting $n^{2}$ ($1, 4, 9, 16$) from the sequence leaves $1, 1, 1, 1$. So the $n$th term is $n^{2} + 1$.
A cubic 三次 sequence such as $1, 8, 27, 64, \dots$ has $n$th term $n^{3}$. An exponential sequence 指数数列 such as $2, 6, 18, 54, \dots$ multiplies by a fixed number each time; here the $n$th term is $2 \times 3^{\,n-1}$.
Number sequences
Build an arithmetic (add d) or geometric (times r) sequence term by term.
| English | Chinese | Pinyin |
|---|---|---|
| sequence | 数列 | shù liè |
| term-to-term rule | 递推规则 | dì tuī guī zé |
| cubic | 三次 | sān cì |
| exponential sequence | 指数数列 | zhǐ shù shù liè |
2.8
Direct and inverse proportion (Extended)
Syllabus
| Subject content | Notes and examples |
|---|---|
| Express direct and inverse proportion in algebraic terms and use this form of expression to find unknown quantities. | Includes linear, square, square root, cube and cube root proportion. Knowledge of proportional symbol ($\propto$) is required. |
Source: Cambridge International syllabus
Two quantities are in direct proportion 正比例 if one is always a fixed multiple of the other: $y \propto x$ means $y = kx$, where $k$ is a constant 常数. They are in inverse proportion 反比例 if one rises as the other falls: $y \propto \dfrac{1}{x}$ means $y = \dfrac{k}{x}$. The symbol $\propto$ is read "is proportional to". You can also have proportion to a square, square root, cube or cube root.
Worked example. $y$ is in direct proportion to $x$, and $y = 12$ when $x = 3$. Find $y$ when $x = 7$.
First find $k$: $12 = k \times 3$, so $k = 4$ and $y = 4x$. Then $y = 4 \times 7 = 28$.
Inverse proportion
y = a/x
Inverse proportion: as x doubles, y halves — a reciprocal curve with two asymptotes.
| English | Chinese | Pinyin |
|---|---|---|
| direct proportion | 正比例 | zhèng bǐ lì |
| constant | 常数 | cháng shù |
| inverse proportion | 反比例 | fǎn bǐ lì |
2.9
Graphs in practical situations
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Use and interpret graphs in practical situations including travel graphs and conversion graphs. | e.g. interpret the gradient of a straight-line graph as a rate of change. |
| 2 Draw graphs from given data. | e.g. draw a distance–time graph to represent a journey. |
| Subject content | Notes and examples |
|---|---|
| 1 Use and interpret graphs in practical situations including travel graphs and conversion graphs. | Includes estimation and interpretation of the gradient of a tangent at a point. |
| 2 Draw graphs from given data. | |
| 3 Apply the idea of rate of change to simple kinematics involving distance–time and speed–time graphs, acceleration and deceleration. | |
| 4 Calculate distance travelled as area under a speed–time graph. | Areas will involve linear sections of the graph only. |
Source: Cambridge International syllabus
The gradient 斜率 (steepness) of a graph shows a rate of change 变化率.
- A travel graph 行程图 (distance–time graph) has gradient equal to speed; a flat part means the object is not moving.
- A conversion graph 换算图 is a straight line used to change between two units (for example, miles and kilometres).
On a distance–time graph the gradient is the speed; a flat section means the object has stopped.
Speed–time graphs (Extended)
On a speed–time graph the gradient is the acceleration 加速度 (or deceleration 减速度 if the speed falls), and the area 面积 under the graph is the distance 距离 travelled.
Worked example. A car speeds up from rest to $20\text{ m/s}$ in $8\text{ s}$, then stays at $20\text{ m/s}$ for $12\text{ s}$. Find the acceleration and the total distance.
Acceleration $= \dfrac{20}{8} = 2.5\text{ m/s}^{2}$. The distance is the area: a triangle plus a rectangle,
On a speed–time graph the gradient is the acceleration and the area underneath is the distance travelled.
Real-life graphs
y = ax + b
A distance–time or cost graph is read from its gradient and its intercept.
| English | Chinese | Pinyin |
|---|---|---|
| gradient | 斜率 | xié lǜ |
| rate of change | 变化率 | biàn huà lǜ |
| travel graph | 行程图 | xíng chéng tú |
| conversion graph | 换算图 | huàn suàn tú |
| acceleration | 加速度 | jiā sù dù |
| deceleration | 减速度 | jiǎn sù dù |
| area | 面积 | miàn jī |
| distance | 距离 | jù lí |
2.10
Graphs of functions
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Construct tables of values, and draw, recognise and interpret graphs for functions of the following forms: • $ax + b$ • $\pm x^2 + ax + b$ • $\frac{a}{x} \ (x \neq 0)$ where $a$ and $b$ are integer constants. | |
| 2 Solve associated equations graphically, including finding and interpreting roots by graphical methods. | e.g. find the intersection of a line and a curve. |
| Subject content | Notes and examples |
|---|---|
| 1 Construct tables of values, and draw, recognise and interpret graphs for functions of the following forms: • $a x^n$ (includes sums of no more than three of these) • $a b^x + c$ where $n = -2, -1, -\frac{1}{2}, 0, \frac{1}{2}, 1, 2, 3$; $a$ and $c$ are rational numbers; and $b$ is a positive integer. | Examples include: • $y = x^3 + x - 4$ • $y = 2x + \frac{3}{x^2}$ • $y = \frac{1}{4} \times 2^x$. |
| 2 Solve associated equations graphically, including finding and interpreting roots by graphical methods. | e.g. finding the intersection of a line and a curve. |
| 3 Draw and interpret graphs representing exponential growth and decay problems. |
Source: Cambridge International syllabus
To draw a graph, make a table of values 数值表: choose values of $x$, work out $y$, then plot the coordinates 坐标 and join them with a smooth curve.
The points where a graph crosses the $x$-axis (the horizontal axis 坐标轴) are the roots 根 — the solutions of $y = 0$.
You can solve an equation by reading a graph. The intersection point 交点 of a line and a curve gives the solution of the two equations together.
For exponential growth 指数增长 and exponential decay 指数衰减, the graph of $y = a\,b^{x} + c$ rises (or falls) faster and faster and flattens towards a horizontal line.
Graphing a quadratic
y = ax² + bx + c
Drag a, b and c and watch the parabola move — its turning point and where it crosses the axes.
| English | Chinese | Pinyin |
|---|---|---|
| table of values | 数值表 | shù zhí biǎo |
| coordinates | 坐标 | zuò biāo |
| axis | 坐标轴 | zuò biāo zhóu |
| roots | 根 | gēn |
| intersection point | 交点 | jiāo diǎn |
| exponential growth | 指数增长 | zhǐ shù zēng zhǎng |
| exponential decay | 指数衰减 | zhǐ shù shuāi jiǎn |
2.11
Sketching curves
Syllabus
| Subject content | Notes and examples |
|---|---|
| Recognise, sketch and interpret graphs of the following functions: (a) linear (b) quadratic. | Knowledge of symmetry and roots is required. Knowledge of turning points is not required. |
| Subject content | Notes and examples |
|---|---|
| Recognise, sketch and interpret graphs of the following functions: (a) linear (b) quadratic (c) cubic (d) reciprocal (e) exponential. | Functions will be equivalent to: • $ax + by = c$ • $y = ax^2 + bx + c$ • $y = ax^3 + b$ • $y = ax^3 + bx^2 + cx$ • $y = \frac{a}{x} + b$ • $y = ar^x + b$ where $a$, $b$ and $c$ are rational numbers and $r$ is a rational, positive number. Knowledge of turning points, roots and symmetry is required. Knowledge of vertical and horizontal asymptotes is required. Finding turning points of quadratics by completing the square is required. |
Source: Cambridge International syllabus
A quick sketch should show the right shape and the key features: where it crosses the axes, any symmetry 对称, and any line the curve gets close to.
| Function | Shape |
|---|---|
| linear, $y = mx + c$ | straight line; gradient $m$, $y$-intercept 截距 $c$ |
| quadratic, $y = ax^{2} + bx + c$ | a parabola 抛物线 (U-shape if $a>0$, $\cap$-shape if $a<0$) |
| cubic, $y = ax^{3} + bx + c$ | an S-shaped curve |
| reciprocal, $y = \dfrac{a}{x} + b$ | two separate curves |
| exponential, $y = a\,r^{x} + b$ | fast growth or decay |
The basic graph shapes; knowing each shape lets you sketch quickly from the equation.
For a parabola, completing the square gives the turning point 转折点 (the lowest or highest point). For example $y = (x + 3)^{2} - 8$ has its turning point at $(-3, -8)$.
Completing the square, $y=(x+3)^2-8$, shows the turning point $(-3,-8)$ and the line of symmetry $x=-3$.
An asymptote 渐近线 is a line that the curve gets closer and closer to but never touches — for example, the $x$-axis for $y = \dfrac{a}{x}$, or the line $y = b$ for $y = a\,r^{x} + b$.
| English | Chinese | Pinyin |
|---|---|---|
| symmetry | 对称 | duì chèn |
| intercept | 截距 | jié jù |
| parabola | 抛物线 | pāo wù xiàn |
| turning point | 转折点 | zhuǎn zhé diǎn |
| asymptote | 渐近线 | jiàn jìn xiàn |
2.12
Differentiation (Extended)
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Estimate gradients of curves by drawing tangents. | |
| 2 Use the derivatives of functions of the form $ax^n$, where $a$ is a rational constant and $n$ is a positive integer or zero, and simple sums of not more than three of these. | $\frac{\mathrm{d}y}{\mathrm{d}x}$ notation will be expected. |
| 3 Apply differentiation to gradients and stationary points (turning points). | |
| 4 Discriminate between maxima and minima by any method. | Maximum and minimum points may be identified by: • an accurate sketch • use of the second differential • inspecting the gradient either side of a turning point. Candidates are not expected to identify points of inflection. |
Source: Cambridge International syllabus
Differentiation 微分 finds the gradient of a curve at any point. You can estimate it by drawing a tangent 切线 (a line that just touches the curve) and measuring its gradient.
The gradient of a curve at a point equals the gradient of the tangent there — what differentiation finds.
The exact rule: if $y = ax^{n}$, then the derivative 导数 is
Differentiate a sum term by term.
Worked example. If $y = x^{3} + 2x^{2} - 5x$, then $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2} + 4x - 5$.
A stationary point 驻点 (turning point) is where the gradient is zero, so set $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$.
Worked example. Find the turning point of $y = x^{2} - 6x + 5$.
The turning point is $(3, -4)$. To decide whether a turning point is a maximum 最大值 or a minimum 最小值, check the sign of the gradient on each side, or use the second derivative (positive means a minimum).
Gradient of a curve
y = ax³ + bx² + cx + d
Move the point: the tangent shows the gradient there, which is what differentiation finds.
| English | Chinese | Pinyin |
|---|---|---|
| differentiation | 微分 | wēi fēn |
| tangent | 切线 | qiè xiàn |
| derivative | 导数 | dǎo shù |
| stationary point | 驻点 | zhù diǎn |
| maximum | 最大值 | zuì dà zhí |
| minimum | 最小值 | zuì xiǎo zhí |
2.13
Functions (Extended)
Syllabus
| Subject content | Notes and examples |
|---|---|
| 1 Understand functions, domain and range and use function notation. | Examples include: • $f(x) = 3x - 5$ • $g(x) = \frac{3(x + 4)}{5}$ • $h(x) = 2x^2 + 3$. |
| 2 Understand and find inverse functions $f^{-1}(x)$. | |
| 3 Form composite functions as defined by $gf(x) = g(f(x))$. | e.g. $f(x) = \frac{3}{x + 2}$ and $g(x) = (3x + 5)^2$. Find $fg(x)$. Give your answer as a fraction in its simplest form. Candidates are not expected to find the domains and ranges of composite functions. This topic may include mapping diagrams. |
Source: Cambridge International syllabus
A function 函数 turns each input into one output. We write $f(x)$, for example $f(x) = 3x - 5$, so $f(2) = 1$. The set of allowed inputs is the domain 定义域; the set of possible outputs is the range 值域.
The inverse function 反函数 $f^{-1}(x)$ undoes the function. To find it, write $y = f(x)$, swap the roles, and make $x$ the subject.
Worked example. Find the inverse of $f(x) = 3x - 5$.
A function as a machine; the inverse runs it backwards - reverse the order, undo each step
A composite function 复合函数 applies one function after another: $gf(x)$ means "do $f$ first, then $g$".
Worked example. If $f(x) = 2x$ and $g(x) = x + 3$, then
Functions
y = f(x)
A function turns each input into exactly one output — watch its shape.
| English | Chinese | Pinyin |
|---|---|---|
| function | 函数 | hán shù |
| domain | 定义域 | dìng yì yù |
| range | 值域 | zhí yù |
| inverse function | 反函数 | fǎn hán shù |
| composite function | 复合函数 | fù hé hán shù |
2.13
Exam tips
- When you expand brackets, multiply every term and watch the signs, especially with a minus in front: $-(x - 3) = -x + 3$.
- To solve an equation, do the same thing to both sides. When you multiply or divide an inequality by a negative number, flip the sign.
- Factorise fully: take out the highest common factor first, then look for a difference of two squares or a quadratic pattern.
- A quadratic usually has two solutions — give both. Check whether the question wants factorising, the formula, or completing the square.
- When substituting into a formula, put each value in brackets first, so signs and powers come out right.