📚 AQA Mathematics: Algebra and Functions Revision Guide | AQA 数学：代数和函数考点精讲
Algebra and functions lie at the heart of AQA A-Level Mathematics, forming the foundation for calculus, trigonometry, and applied units. This revision guide walks you through the essential techniques – from simplifying expressions to modelling with functions – providing clear explanations, key formulas, and examiner tips to help you master each topic and maximise your exam performance.
代数和函数是 AQA A-Level 数学的核心，为微积分、三角学和应用单元奠定了基础。本考点精讲将带你梳理从化简表达式到函数建模的关键技巧，提供清晰的解释、重要的公式和考官提示，帮助你掌握每个主题，在考试中发挥出最佳水平。
1. Simplifying Algebraic Expressions | 化简代数表达式
Simplifying algebraic expressions is the first step to success in algebra. Always collect like terms, apply the distributive law, and handle negative signs with care. For rational expressions, factorise numerators and denominators before cancelling common factors, but remember to state any restrictions (e.g. a denominator cannot be zero).
化简代数表达式是学好代数的第一步。务必合并同类项，正确使用分配律，并小心处理负号。对于分式表达式，应先对分子和分母进行因式分解，再约去公因式，同时记得注明限制条件（例如分母不能为零）。
When working with indices, recall the basic laws:
处理指数运算时，请回顾基本指数定律：
aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ
Be especially careful with negative and fractional powers: a⁻ⁿ = 1/aⁿ and a¹/ⁿ = ⁿ√a. Using these laws consistently will make complex-looking expressions much more manageable.
对负指数和分数指数要格外小心：a⁻ⁿ = 1/aⁿ，a¹/ⁿ = ⁿ√a。熟练运用这些定律可以使看似复杂的表达式变得非常容易处理。
2. Quadratic Equations and the Discriminant | 二次方程与判别式
be solved by factorising, completing the square, or using the quadratic formula. The formula is one of the most important tools in your arsenal:
二次方程 ax² + bx + c = 0 可通过因式分解、配方法或求根公式求解。其中求根公式是你手中最重要的工具之一：
x = (-b ± √b² - 4ac) / (2a)
The expression b² - 4ac is called the discriminant (often denoted by Δ) and determines the nature of the roots:
表达式 b² - 4ac 被称为判别式（常用 Δ 表示），它决定了根的性质：
- Δ > 0: two distinct real roots (两个不等实根)
- Δ = 0: one repeated real root (一个重根)
- Δ < 0: no real roots (没有实根，有一对共轭复数根)
In exam questions, you may be asked to find the range of a parameter for which a quadratic has real roots, or to prove that a quadratic is always positive. Always connect the discriminant to the graph of the quadratic – for a > 0, the graph is U-shaped; for a < 0, it is ∩-shaped.
在考试题中，你可能需要求出使二次方程有实根的参数范围，或者证明某个二次式恒大于零。一定要把判别式与二次函数的图像联系起来：当 a > 0 时，图像开口向上；当 a < 0 时，开口向下。
3. Factorising Polynomials | 多项式因式分解
) to find the first factor, then use polynomial long division or synthetic division to reduce the degree.
对于三次或更高次多项式，因式定理是你最好的帮手。如果多项式 f(x) 满足 f(p) = 0，那么 (x - p) 就是它的一个因式。可以先尝试代入较小的整数值（±1, ±2, ±3）找到第一个因式，然后用多项式长除法或综合除法降低多项式的次数。
Once you have reduced the polynomial to a quadratic, factorise it further if possible. Always write your final answer in fully factorised form, e.g. (x - 2)(x + 1)(x - 3). In AQA exams, you might also need to factorise expressions with a repeated factor, such as (x - 1)²(x + 4).
把多项式降次为二次式后，如果可能就继续分解。最终答案一定要写成完全因式分解的形式，例如 (x - 2)(x + 1)(x - 3)。在 AQA 考试中，你可能还需要分解含有重因式的表达式，比如 (x - 1)²(x + 4)。
4. Inequalities and Set Notation | 不等式与集合表示法
Solving linear and quadratic inequalities requires similar steps to solving equations, but with one crucial difference: multiplying or dividing both sides by a negative number reverses the inequality sign. For quadratic inequalities like x² - 5x + 6 > 0, sketch the graph or use a sign table to identify intervals where the expression is positive.
解线性不等式和二次不等式的步骤与解方程类似，但有一个关键区别：两边同乘或同除以一个负数时，不等号要改变方向。对于 x² - 5x + 6 > 0 这类二次不等式，可以画出草图或使用符号表来确定表达式为正的区间。
et notation and interval notation. AQA often expects answers in forms such as {x : x < -2} ∪ {x : x > 3} or using interval brackets: (-∞, -2) ∪ rackets for strict inequalities (<, >), square brackets for inclusive inequalities (≤, ≥).
你必须熟练掌握集合表示法和区间表示法。AQA 通常会要求以 {x : x < -2} ∪ {x : x > 3} 或者区间形式 (-∞, -2) ∪ (3, ∞) 作答。注意开区间与闭区间的区别：严格不等号用圆括号，包含等号时用方括号。
5. Functions: Domain and Range | 函数：定义域与值域
A function maps each input (x) to exactly one output (f(x)). The domain is the set of all possible input values, while the range is the set of all possible output values. For a function given by an algebraic formula, the natural domain is all real numbers except those that cause division by zero or negative values under an even root.
函数将每个输入 (x) 唯一地映射到一个输出 (f(x))。定义域是所有可能输入值的集合，值域则是所有可能输出值的集合。对于用代数表达式给出的函数，其自然定义域为所有不会导致分母为零或偶次根号下为负数的实数。
eting the square is a powerful technique here: f(x) = a(x + p)² + q, so the vertex is at (–p, q) and the range is easily read off.
要寻求值域，需考虑图像的形状和转折点。对于二次函数 f(x) = ax² + bx + c，若 a > 0，值域为 [f(xₘᵢₙ), ∞)；若 a < 0，值域为 (-∞, f(xₘₐₓ)]。配方法是一个很有力的技巧：f(x) = a(x + p)² + q，其顶点为 (–p, q)，由此可直接读出值域。
6. Composite and Inverse Functions | 复合函数与反函数
Composite functions combine two functions: (f ∘ g)(x) = f(g(x)). The order matters – apply the inner function first, then the outer. The domain of f ∘ g(x) is in the domain of f. Always check these conditions carefully to avoid losing marks.
复合函数是将两个函数组合在一起：(f ∘ g)(x) = f(g(x))。顺序至关重要——先代入内层函数，再代入外层。f ∘ g 的定义域由那些既在 g 的定义域内，又使 g(x) 落在 f 的定义域内的 x 值构成。务必仔细检查这些条件，以防丢分。
he inverse exists only if f is one-to-one on the given domain. You can restrict a function' n to make it one-to-one; for example, f(x) = x² with domain x ≥ 0 has the inverse f⁻¹(x) = √x.
反函数 f⁻¹(x) 可以撤销 f 的作用。求反函数的方法是：令 y = f(x)，交换 x 和 y，然后解出 y。反函数仅在原函数在给定义域上是一一映射时才存在。你可以通过限制定义域来使它成为一一映射；例如，f(x) = x² 在定义域 x ≥ 0 上的反函数为 f⁻¹(x) = √x。
7. Transformations of Graphs | 图像变换
u to sketch new functions quickly. The four basic transformations applied to y = f(x) are:
理解图像变换可以让你快速画出新函数的草图。应用于 y = f(x) 的四种基本变换是：
- y = f(x) + a: translation upwards by a (向上平移 a)
- y = f(x + a): translation to the left by a (向左平移 a，注意 a > 0 时向左)
- y = a f(x): vertical stretch by factor a (垂直方向拉伸 a 倍)
- y = f(ax): horizontal stretch by factor 1/a (水平方向拉伸 1/a 倍)
Reflections are also common: y = –f(x) is a reflection in the x-axis, and y = f(– n the y-axis. When performing multiple transformations, always follow the correct sequence – hen translating. Being systematic with transformation order will prevent confusing errors.
对称变换也很常见：y = –f(x) 是关于 x 轴的对称，y = f(–x) 是关于 y 轴的对称。当进行多个变换时，应遵循正确的顺序——通常是先拉伸再平移。有系统地按顺序处理变换可以避免混乱和错误。
8. Exponential and Logarithmic Functions | 指数函数与对数函数
The exponential function f(x) = eˣ m function f(x) = ln x is its inverse, defined only for x > 0, with ln 1 = 0 and ln e = 1.
指数函数 f(x) = eˣ 具有导数等于自身的独特性质。它的图像经过点 (0, 1)，处处为正，并且当 x 很大时迅速增长。自然对数函数 f(x) = ln x 是它的反函数，仅当 x > 0 时有定义，且满足 ln 1 = 0，ln e = 1。
Be comfortable switching between exponential and logarithmic forms: aˣ = b ⟺ x = logₐ b. For the natural base, eˣ = y ⟺ h and decay contexts, such as population growth P = P₀eᵏᵗ or radioactive decay N = N₀e⁻ᵏᵗ.
要熟练掌握指数式与对数式之间的转换：aˣ = b ⟺ x = logₐ b。对于自然底数，eˣ = y ⟺ x = ln y。在建模中，指数函数常出现在增长和衰减的情境中，例如人口增长 P = P₀eᵏᵗ 或放射性衰变 N = N₀e⁻ᵏᵗ。
9. Laws of Logarithms and Solving Equations | 对数定律与解方程
The laws of logarithms are essential for simplifying expressions and solving exponential equations. For any base, including e and 10, the key laws are:
对数定律对于化简表达式和解指数方程至关重要。对于任何底数（包括 e 和 10），核心定律有：
log(xy) = log x + log y, log(x/y) = log x - log y, log(xⁿ) = n log x
When solving equations like 2e³ˣ Always check that your solutions fall within the valid domain of the logarithmic function.
解像 2e³ˣ⁻¹ = 10 这样的方程时，先将指数项分离出来，然后两边同时取自然对数。你也可以通过取指数来解对数方程：ln(2x + 1) = 3 等价于 2x + 1 = e³。务必要检查所求得的解是否在对数函数的有效定义域内。
AQA exam questions often combine log laws with hidden quadratics. For instance, e²ˣ - 5eˣ + 6 = 0 can be turned into a standard quadratic by substituting u = eˣ g u² - 5u + 6 = 0 and converting back. Such substitution techniques are highly examinable.
AQA 考试题经常将对数定律与“隐藏的二次方程”结合起来。例如，e²ˣ - 5eˣ + 6 = 0 可以通过设 u = eˣ 转化为标准的二次方程 u² - 5u + 6 = 0，求解后再回代。这类代换技巧在考试中非常常见。
10. Modelling with Functions | 函数建模应用
e at which a quantity reaches half its initial value, or the long-term behaviour as t → ∞.
现实情境的建模题要求你解读给定的函数，通常会附带说明物理情境的图像。你可能需要求出最大值或最小值、某个量达到其初始值一半的时刻，或者当 t → ∞ 时的长期行为。
Transformations are often applied in modelling – ems can be solved by setting A = ½A₀ and solving for t. Similar principles apply to Newton's law of cooling or logistic growth models you may encounter in AQA papers.
变换也常常出现在建模中——例如，平移或拉伸基本增长曲线来拟合数据。在用指数衰减建模时，A = A₀e⁻ᵏᵗ 是常见公式；半衰期问题可以通过令 A = ½A₀ 并求解 t 来解决。类似的原则也适用于你在 AQA 试卷中可能遇到的牛顿冷却定律或逻辑斯谛增长模型。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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