Core Knowledge Guide for Year 9 SQA Advanced Mathematics | Year 9 SQA 进阶数学：核心知识点梳理

📚 Core Knowledge Guide for Year 9 SQA Advanced Mathematics | Year 9 SQA 进阶数学：核心知识点梳理

Welcome to the Year 9 SQA Advanced Mathematics core knowledge guide. This article brings together the key topics you will encounter as you build a deeper understanding of algebra, geometry, trigonometry, statistics and beyond. Mastering these foundations will prepare you for the demands of National 5 and Higher Mathematics. Each section is designed to give you clear explanations and useful examples, with English and Chinese versions side by side so you can follow along in either language.

欢迎阅读 Year 9 SQA 进阶数学核心知识点梳理。本文汇总了你将在深入学习中遇到的关键主题，涵盖代数、几何、三角学、统计等。掌握这些基础将为你在 National 5 和 Higher 数学中的挑战做好准备。每个部分都提供了清晰的解释和实用的例子，中英双语对照，帮助你更好地理解。

1. Algebraic Expressions | 代数表达式

Simplifying algebraic expressions means collecting like terms. Only identical letter combinations and powers can be combined. For example, 5a + 3b − 2a + 6b simplifies to 3a + 9b.

化简代数表达式就是合并同类项。只有字母和指数完全相同的项才能合并。例如 5a + 3b − 2a + 6b 化简为 3a + 9b。

apply the distributive law twice, as in (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12.

展开括号时，要把外面的项乘以括号内的每一项。遇到两个括号则两次使用分配律，例如 (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12。

Simple factorising reverses the expansion. For 4x + 8 we find the highest common factor, 4, and write 4(x + 2). For a quadratic like x² + 6x + 8 we look for two numbers that multiply to 8 and add to 6, giving (x + 2)(x + 4).

简单的因式分解是展开的逆运算。对于 4x + 8，我们提取公因数 4，写成 4(x + 2)。对于像 x² + 6x + 8 的二次式，我们寻找两个乘积为 8、和为 6 的数，得到 (x + 2)(x + 4)。

2. Linear Equations and Inequalities | 线性方程与不等式

For 3x − 7 = 2x + 5, subtract 2x from both sides to get x − 7 = 5, then add 7, so x = 12.

解线性方程时，我们通过对方程两边进行相同的运算来分离未知数。对于 3x − 7 = 2x + 5，两边减去 2x 得到 x − 7 = 5，然后加 7，解得 x = 12。

= 2 and x = 1. Always check your solution by substituting back into the original equation.

如果出现括号，先展开。2(x + 4) = 10 化为 2x + 8 = 10，得 2x = 2，x = 1。记得代入原方程检验。

Inequalities are solved similarly, but if you multiply or divide by a negative number you must reverse the inequality sign. The solution set can be shown on a number line with open or closed circles.

不等式的解法类似，但如果乘以或除以负数，必须反转不等号。解集可以在数轴上用空心或实心圆点表示。

3. Sequences and Patterns | 数列与规律

An arithmetic sequence adds a constant difference, d, between terms. The n‑th term formula is u_n 1)d, where a is the first term. For sequence 3, 7, 11, 15, ... we have a = 3, d = 4, so u_n = 3 + 4(n − 1) = 4n − 1.

等差数列相邻两项的差为常数 d。第 n 项公式为 u_n = a + (n − 1)d，其中 a 是首项。数列 3, 7, 11, 15, ... 中 a = 3, d = 4，因此 u_n = 3 + 4(n − 1) = 4n − 1。

You can use the n‑ ) − 1 = 199. Recognising patterns in diagrams also leads to arithmetic or other sequences.

利用第 n 项可求任意项，例如第 50 项为 4(50) − 1 = 199。图形中的模式也常能导出等差数列或其它数列。

Sequences that increase by different amounts each time can be quadratic. Year 9 Advanced Maths may introduce second differences to identify these.

每步增加量不同的数列可能是二次的。Year 9 进阶数学可能会引入二次差分的概念来识别这类数列。

4. Functions and Graphs | 函数与图像

+ 3, we can draw a table of values and plot the straight line. The gradient is 2 and the y‑intercept is 3.

函数把输入值 x 映射到输出值 y。对于一次函数 y = 2x + 3，我们可列出数值表并描点画出直线。斜率为 2，y 轴截距为 3。

y = mx + c

In this form, m is the gradient and c is the y‑pt. Two lines are parallel if they have the same gradient, e.g. y = 2x + 3 and y = 2x − 5.

在此形式中，m 表示斜率，c 表示 y 轴截距。若两条直线斜率相同则它们平行，例如 y = 2x + 3 与 y = 2x − 5。

ent is 3 and the line passes through (1, 7), then 7 = 3(1) + c gives c = 4, so y = 3x + 4.

已知斜率和一个点求直线方程时，可将坐标代入 y = mx + c 求出 c。若斜率为 3 且过点 (1, 7)，则 7 = 3(1) + c 得 c = 4，所以直线方程为 y = 3x + 4。

5. Pythagoras' Theorem | 勾股定理

In a right‑ s of the other two sides: a² + b² = c². The hypotenuse is always opposite the right angle.

在直角三角形中，斜边的平方等于两直角边的平方和：a² + b² = c²。斜边总是直角的对边。

To find a missing hypotenuse, say legs 5 cm and 12 cm, calculate c = √5² + 12² = √169 = 13 cm. To find a shorter side, e.g. a leg when c = 10 and b = 6, a = √10² − 6² = √64 = 8.

求斜边时，如两直角边为 5 cm 和 12 cm，计算 c = √5² + 12² = √169 = 13 cm。求直角边时，比如已知 c = 10，b = 6，则 a = √10² − 6² = √64 = 8。

Pythagoras also applies in 3D problems, such as finding the diagonal of a rectangular box. You first apply the theorem in one face, then use that result with the height.

勾股定理也用于三维问题，例如求长方体对角线。先在平面上使用定理，再结合高度计算。

6. Trigonometry | 三角学初步

Right‑angled triangle trigonometry uses sine, cosine and tangent ratios. Remember the mnemonic SOH CAH TOA:

直角三角形边的三角比使用正弦、余弦和正切。记住口诀 SOH CAH TOA：

sin θ = opposite / hypotenuse

cos θ = adjacent / hypotenuse

tan θ = opposite / adjacent

he hypotenuse is 12 and the angle is 30°, the opposite side = 12 × sin 30° = 12 × 0.5 = 6.

求未知边时，先根据已知角标注对边、邻边和斜边，选择合适的三角比求解。例如，若斜边为 12，角为 30°，则对边 = 12 × sin 30° = 12 × 0.5 = 6。

tions. If tan θ = 1.5, then θ = tan⁻¹(1.5) ≈ 56.3°. The calculator must be in degree mode.

求角时使用反三角函数。若 tan θ = 1.5，则 θ = tan⁻¹(1.5) ≈ 56.3°。计算器必须设置为度数模式。

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7. Perimeter, Area and Volume | 周长、面积与体积

². For a semicircle, use half the circumference plus the diameter. Always give answers with appropriate units.

圆的周长即圆周，C = 2πr，面积 A = πr²。半圆的周长为半圆周长加直径。答案务必带上适当的单位。

s. Find the total area by adding or subtracting. For a triangle, area = ½ × base × height.

复合图形可分解为矩形、三角形和圆形。通过加减求出总面积。三角形面积 = ½ × 底 × 高。

cylinder is a prism with circular base, so V = πr²h. Surface area sums all the face areas.

直棱柱的体积 = 底面积 × 高。圆柱是以圆为底的棱柱，故 V = πr²h。表面积等于所有面的面积之和。

ave volume ⁴⁄₃πr³ and surface area 4πr², introduced in advanced Year 9 or National 5 work.

棱锥或圆锥的体积为 ⅓ × 底面积 × 高。球的体积为 ⁴⁄₃πr³，表面积为 4πr²，这些内容在 Year 9 进阶或 National 5 中会介绍。

8. Statistics and Probability | 统计与概率

The three common averages are mean, median and mode. The mean is sum of data divided by count. The median is the middle value when data are ordered. The mode is the most frequent value.

三种常用的平均数是均值、中位数和众数。均值是数据总和除以数据个数。中位数是数据排序后的中间值。众数是出现次数最多的值。

ximum − minimum. A box plot can display the median, quartiles and range, giving a clear picture of distribution.

极差衡量离散程度：极差 = 最大值 − 最小值。箱形图能显示中位数、四分位数和极差，清晰地呈现分布情况。

vourable outcomes / total outcomes. Probabilities always lie between 0 and 1. For combined events, tree diagrams help multiply probabilities along branches.

事件的概率 = 有利结果数 / 全部可能结果数。概率值总在 0 到 1 之间。对于复合事件，树状图有助于分支相乘计算概率。

P(A or B) = P(A) + P(B). Year 9 Advanced Math may include conditional probability basics.

若 A 与 B 相互独立，则 P(A 且 B) = P(A) × P(B)。对于互斥事件，P(A 或 B) = P(A) + P(B)。Year 9 进阶数学可能初步涉及条件概率。

9. Transformations and Symmetry | 变换与对称

Four key transformations are reflection, rotation, translation and enlargement. Reflection requires a mirror line; rotation needs a centre, angle and direction (clockwise or anticlockwise).

四种基本变换是反射、旋转、平移和放大。反射需指定镜面线；旋转需指出中心、角度和方向（顺时针或逆时针）。

Translation moves a shape by a vector, e.g. (3, −2) means 3 right and 2 down. Enlargement changes size from a centre with a scale factor. If the scale factor is k, distances from the centre are multiplied by k.

平移通过向量移动图形，如 (3, −2) 表示右移 3、下移 2。放大从一个中心点按比例因子缩放。比例因子为 k 时，各点到中心的距离都乘以 k。

Symmetry can be reflective (lines of symmetry) or rotational (order of rotation). A regular polygon of n sides has n lines of symmetry and rotational order n.

对称分为镜面对称（对称轴）和旋转对称（旋转阶数）。正 n 边形有 n 条对称轴和 n 阶旋转对称。

Combining transformations leads to single equivalent transformations, an idea explored in more advanced geometry tasks.

多个变换可以合成一个等效变换，这一概念会在更高级的几何题目中探讨。

10. Ratio and Proportion | 比与比例

A ratio compares two or more quantities in the same units. Simplify ratios by dividing by the highest common factor, e.g. 18 : 24 simplifies to 3 : 4. A ratio in the form 1 : n is often useful.

比用来比较两个或多个同类量。通过除以最大公因数化简比，如 18 : 24 化简为 3 : 4。写成 1 : n 的形式常常很方便。

y in a given ratio, add the parts to find the total number of shares, then multiply. Share £50 in the ratio 2 : 3 means 2 + 3 = 5 shares, each £10, so £20 and £30.

按比例分配时，将份额总数相加得到总份数，然后相乘。按 2 : 3 分 £50，共有 2 + 3 = 5 份，每份 £10，所以分得 £20 和 £30。

y is directly proportional to x, then y = kx, where k is the constant of proportionality.

正比例表示两个量以相同速率增减。若 y 与 x 成正比，则 y = kx，k 为比例常数。

g values using the unitary method or by setting up equations. For instance, if 5 pens cost £2, find the cost of 8 pens: cost per pen = £0.40, so 8 × 0.40 = £3.20.

可以通过单位法或建立等式求未知值。例如，5 支笔 £2，求 8 支笔的价钱：每支 £0.40，8 × 0.40 = £3.20。

11. Quadratic Expressions and Equations | 二次表达式与方程

ly each term: x² − 2x + 5x − 10 = x² + 3x − 10. Always carefully combine the middle terms.

两个一次式相乘得到一个二次式。对于 (x + 5)(x − 2)，各项相乘：x² − 2x + 5x − 10 = x² + 3x − 10。中间项合并要细心。

Factorising a quadratic of the form x² + bx + c involves finding two numbers that multiply to c and add to b. For x² + 5x + 6, the numbers 2 and 3 work, giving (x + 2)(x + 3).

对形如 x² + bx + c 的二次式因式分解，要找到两个数其积为 c、和为 b。例如 x² + 5x + 6，取 2 和 3，得 (x + 2)(x + 3)。

each bracket to zero. For x² + 5x + 6 = 0 we have (x + 2)(x + 3) = 0, so x = −2 or x = −3.

通过因式分解解二次方程：令表达式等于零，分解因式，再令每个括号为零。对于 x² + 5x + 6 = 0，有 (x + 2)(x + 3) = 0，所以 x = −2 或 x = −3。

If the quadratic has a coefficient of x² other than 1, a more careful factorising method is needed, often splitting the middle term. This builds the foundation for National 5 techniques.

如果 x² 的系数不为 1，就需要更系统的因式分解法，常涉及拆分中项。这为 National 5 的解题技巧打下基础。

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