IB AQA Mathematics: Algebra and Functions Key Points | IB AQA 数学：代数和函数考点精讲

📚 IB AQA Mathematics: Algebra and Functions Key Points | IB AQA 数学：代数和函数考点精讲

Algebra and functions form the backbone of the IB and AQA Mathematics syllabus, linking numerical manipulation to abstract reasoning and graphical interpretation. Mastering this unit requires fluency in simplifying expressions, solving equations, and understanding how functions behave under transformations. This guide breaks down each key topic with step‑by‑step clarity, pairing concise English explanations with full Chinese translations, so you can revise efficiently and tackle exam questions with confidence.

代数和函数是 IB 与 AQA 数学课程的核心支柱，它将数值运算与抽象推理和图像分析紧密相连。掌握本单元需要熟练化简表达式、解方程，并理解函数在变换下的行为。本文逐一精讲各核心考点，以简洁的英文解释搭配完整的中文译文，帮助你高效复习、自信应对考试。

1. Polynomials and Operations | 多项式及其运算

A polynomial is an expression consisting of variables and coefficients, constructed using only addition, subtraction, multiplication, and non‑negative integer exponents. The degree of a polynomial is the highest power of the variable, and the leading coefficient is the coefficient of that term. You must be able to add, subtract, multiply, and divide polynomials, including long division when simplifying rational expressions.

多项式是由变量和系数通过加法、减法、乘法以及非负整数次幂构成的表达式。多项式的次数是指变量的最高次幂，首项系数是该最高次项的系数。你需要能够进行多项式的加减乘除运算，包括在化简有理式时使用长除法。

When multiplying two polynomials, each term in the first polynomial must be multiplied by each term in the second — a process often called the distributive law or expansion. For division, the algorithm mirrors long division of integers: divide the leading terms, multiply the divisor by the result, subtract, and bring down the next term. The outcome can be expressed as dividend = divisor × quotient + remainder.

两个多项式相乘时，必须将第一个多项式中的每一项与第二个多项式中的每一项相乘——这一过程常被称为分配律或展开。除法算法则与整数的长除法类似：首项相除，将除数乘以结果，相减，并移下下一项。结果可表示为：被除式 = 除式 × 商式 + 余式。

The Remainder Theorem states that when a polynomial f(x) is divided by (x – a), the remainder is f(a). The Factor Theorem follows: if f(a) = 0, then (x – a) is a factor of f(x). These theorems are powerful tools for factorising cubic and higher‑degree polynomials.

余数定理指出，当多项式 f(x) 除以 (x – a) 时，余数为 f(a)。因式定理随之而来：若 f(a) = 0，则 (x – a) 是 f(x) 的一个因式。这些定理是分解三次及更高次多项式的重要工具。

2. Factorisation Techniques | 因式分解技巧

Factorisation reverses expansion, rewriting an expression as a product of simpler factors. Common techniques include extracting a common factor, grouping, recognising difference of two squares (a² – b² = (a – b)(a + b)), and applying the sum or difference of cubes. For quadratics, splitting the middle term or using the ‘ac method’ is often effective.

因式分解是展开的逆过程，即将表达式重写为较简单因式的乘积。常见技巧包括提取公因式、分组分解、识别平方差公式 (a² – b² = (a – b)(a + b))，以及利用立方和与立方差。对于二次式，拆中项法或“ac 法”通常很有效。

When factorising a quadratic of the form ax² + bx + c, find two numbers that multiply to ac and add to b. Rewrite the middle term using these numbers and factor by grouping. This systematic approach works for all factorable quadratics and extends to more complex polynomials when combined with the Factor Theorem.

分解形如 ax² + bx + c 的二次式时，需找到两个数，其乘积为 ac，和为 b。利用这两个数将中项重写，再进行分组分解。这一系统方法适用于所有可分解的二次式，并在结合因式定理后可扩展至更复杂的多项式。

For IB and AQA exams, also practise factorising expressions that appear in rational functions, as simplification often requires cancelling common factors, provided you state the domain restrictions to avoid division by zero.

在 IB 和 AQA 考试中，也需练习分解有理函数中出现的表达式，因为化简常常需要约去公因式，此时必须注明定义域限制以避免除以零。

3. Quadratic Functions and Their Graphs | 二次函数及其图象

The general form of a quadratic function is f(x) = ax² + bx + c, where a ≠ 0. Its graph is a parabola that opens upward if a > 0 and downward if a < 0. The y‑intercept is c, and the axis of symmetry is the vertical line x = –b/(2a). The vertex, or turning point, lies on this axis.

二次函数的一般形式为 f(x) = ax² + bx + c，其中 a ≠ 0。其图象是一条抛物线，若 a > 0 则开口向上，若 a < 0 则开口向下。y 轴截距为 c，对称轴为直线 x = –b/(2a)。顶点（即转折点）位于该对称轴上。

To sketch a quadratic efficiently, find the intercepts (setting x = 0 and y = 0), the axis of symmetry, and the vertex. The vertex form, obtained by completing the square, reveals the coordinates of the vertex directly: f(x) = a(x – h)² + k gives vertex (h, k).

要高效绘制二次函数草图，需找出截距（令 x = 0 和 y = 0）、对称轴以及顶点。通过配方法得到的顶点式可直接显示顶点坐标：f(x) = a(x – h)² + k 的顶点为 (h, k)。

The discriminant Δ = b² – 4ac determines the nature of the roots. If Δ > 0, the quadratic has two distinct real roots; if Δ = 0, it has one repeated real root (the graph touches the x‑axis); if Δ < 0, it has no real roots (the graph does not cross the x‑axis).

判别式 Δ = b² – 4ac 决定根的性质。若 Δ > 0，二次式有两个相异实根；若 Δ = 0，有一个重根（图象与 x 轴相切）；若 Δ < 0，无实根（图象不与 x 轴相交）。

4. Completing the Square and the Discriminant | 配方法与判别式

Completing the square transforms a quadratic from standard form to vertex form, making it easier to solve equations, find the range, and graph the function. The process involves halving the coefficient of x, squaring it, and adjusting the constant term: x² + bx = (x + b/2)² – (b/2)².

配方法将二次式从标准形式转化为顶点式，使解方程、求值域和绘图更加容易。其步骤包括将 x 的系数减半、平方，再调整常数项：x² + bx = (x + b/2)² – (b/2)²。

When the coefficient of x² is not 1, factor it out of the x² and x terms first. For example, 2x² + 8x + 5 becomes 2(x² + 4x) + 5, then complete the square inside the bracket to obtain 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 3. This form immediately shows the minimum value is –3 at x = –2.

当 x² 系数不为 1 时，需先从 x² 和 x 项中将其提出。例如，2x² + 8x + 5 化为 2(x² + 4x) + 5，然后对括号内的部分配方，得到 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 3。该形式直接表明函数在 x = –2 处取得最小值 –3。

The discriminant not only classifies roots but also helps solve inequalities and analyse intersections. When asked to ‘find the set of values of k for which the equation has real roots’, express the condition Δ ≥ 0 and solve the resulting inequality.

判别式不仅用于对根进行分类，也可用于解不等式和分析图象的交点。当问题要求“求使方程有实根的 k 的取值范围”时，应将条件表达为 Δ ≥ 0，再解所得不等式。

5. Functions: Domain, Range and Notation | 函数：定义域、值域与符号

A function f from set X to set Y assigns exactly one output in Y for each input in X. The domain is the complete set of possible input values (x‑values), while the range is the set of all outputs f(x) produced. Understanding how to express these using inequality and interval notation is essential.

从集合 X 到集合 Y 的函数 f，为 X 中的每个输入恰好指定 Y 中的一个输出。定义域是可能输入值（x 值）的完整集合，值域是所有输出 f(x) 的集合。理解如何用不等式和区间记号表示定义域和值域至关重要。

When a function involves a denominator, square root, or logarithm, restrictions apply: the denominator cannot be zero, the radicand must be non‑negative, and the argument of a logarithm must be positive. Always state these restrictions explicitly, as they define the natural domain.

当函数包含分母、平方根或对数时，存在限制条件：分母不能为零，被开方数必须非负，对数的真数必须为正。务必明确写出这些限制，因为它们决定了自然定义域。

The range of a function is often found by considering the graph or by using transformations. For quadratic functions, the range is y ≥ k if the parabola opens upward, or y ≤ k if downward, where k is the y‑coordinate of the vertex.

函数的值域通常可通过图象或变换求得。对于二次函数，若抛物线开口向上，则值域为 y ≥ k；若开口向下，则为 y ≤ k，其中 k 是顶点的 y 坐标。

6. Composite and Inverse Functions | 复合函数与反函数

A composite function fg(x) means apply g first, then apply f to the result: fg(x) = f(g(x)). For the composition to be defined, the range of the inner function g must be a subset of the domain of f. The order matters — fg(x) is generally not equal to gf(x).

复合函数 fg(x) 表示先应用 g，再将其结果应用到 f：fg(x) = f(g(x))。要使复合有意义，内部函数 g 的值域必须是 f 的定义域的子集。顺序很重要——通常 fg(x) 不等于 gf(x)。

The inverse function f⁻¹(x) reverses the effect of f. To find it, write y = f(x), swap x and y, then solve for y. The inverse exists only if f is one‑to‑one (injective). The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f.

反函数 f⁻¹(x) 的作用是逆转 f 的效果。求反函数时，写出 y = f(x)，交换 x 和 y，再解出 y。反函数仅当 f 为一一映射（单射）时才存在。f⁻¹ 的定义域是 f 的值域，f⁻¹ 的值域是 f 的定义域。

Graphically, the inverse function is the reflection of y = f(x) in the line y = x. You can use this symmetry to sketch inverses and check that f(f⁻¹(x)) = x and f⁻¹(f(x)) = x for all x in the appropriate domains.

从图形上看，反函数是 y = f(x) 关于直线 y = x 的反射。你可以利用这种对称性来绘制反函数草图，并验证在相应定义域内满足 f(f⁻¹(x)) = x 和 f⁻¹(f(x)) = x。

7. Transformations of Graphs | 图象变换

Transformations move, stretch, or reflect the graph of a function. The common types are translation, stretch, and reflection. Knowing the precise mapping notation — such as f(x) + a, f(x + a), af(x), f(ax) — is crucial for both sketching and describing transformations.

变换可移动、拉伸或反射函数图象。常见类型包括平移、伸缩和对称。准确掌握映射记号——例如 f(x) + a、f(x + a)、af(x)、f(ax)——对于画图和描述变换至关重要。

Transformation	Effect on y = f(x)	中文说明
Vertical translation +k	y = f(x) + k	竖直上移 k 个单位
Horizontal translation –h	y = f(x + h)	水平左移 h 个单位
Vertical stretch factor a	y = af(x)	竖直方向伸缩 a 倍
Horizontal stretch factor 1/b	y = f(bx)	水平方向伸缩 1/b 倍
Reflection in x‑axis	y = –f(x)	关于 x 轴对称反射
Reflection in y‑axis	y = f(–x)	关于 y 轴对称反射

When combining transformations, apply them step by step, respecting the order: horizontal shifts and stretches inside the function argument, then vertical shifts and stretches outside. Exam questions often require you to reconstruct the equation of a transformed function from a given sequence.

组合变换时，应逐步应用，注意次序：首先处理函数括号内的水平平移与伸缩，然后是外部的竖直平移与伸缩。考试题常要求根据给定的变换序列重新列出变换后的函数方程。

8. Exponential and Logarithmic Functions | 指数函数与对数函数

Exponential functions have the form f(x) = aˣ with a > 0, a ≠ 1. The most important base is e, approximately 2.718, giving the natural exponential function eˣ. Its graph passes through (0, 1), approaches the x‑axis as x → –∞, and grows rapidly for positive x.

指数函数的形式为 f(x) = aˣ，其中 a > 0 且 a ≠ 1。最重要的底数是 e，约等于 2.718，它给定了自然指数函数 eˣ。其图象经过点 (0, 1)，当 x → –∞ 时趋近于 x 轴，而在正方向快速增长。

Logarithms are the inverses of exponentials. If y = aˣ, then x = logₐy. The natural logarithm, ln x, uses base e. Key laws include: logₐ(xy) = logₐx + logₐy, logₐ(x/y) = logₐx – logₐy, and logₐ(xⁿ) = n logₐx. These are essential for solving exponential equations.

对数是指数的逆运算。若 y = aˣ，则 x = logₐy。自然对数 ln x 的底数为 e。关键法则包括：logₐ(xy) = logₐx + logₐy，logₐ(x/y) = logₐx – logₐy，logₐ(xⁿ) = n logₐx。这些法则对于解指数方程至关重要。

To solve an equation like 2e³ˣ = 10, divide by 2, take natural logs on both sides: ln(e³ˣ) = ln 5, which simplifies by the inverse property to 3x = ln 5, so x = (ln 5)/3. Similarly, use the definition of logarithms to switch between exponential and logarithmic forms.

解方程 2e³ˣ = 10 时，先除以 2，再两边取自然对数：ln(e³ˣ) = ln 5，利用反函数性质简化为 3x = ln 5，因此 x = (ln 5)/3。类似地，可使用对数定义在指数形式与对数形式之间切换。

9. Solving Equations and Inequalities | 方程与不等式的求解

Solving equations requires isolating the variable using inverse operations, maintaining balance. For quadratic equations, after setting to zero, use factorisation, completing the square, or the quadratic formula: x = [–b ± √(b² – 4ac)] / (2a). Always verify solutions in the original equation, especially when squaring or taking logarithms.

解方程需要利用逆运算将变量分离，并保持等号两边平衡。对于二次方程，在移项为零后，可使用因式分解、配方法或求根公式：x = [–b ± √(b² – 4ac)] / (2a)。务必在原方程中验证解，尤其在涉及平方或取对数时。

Inequalities are similar to equations but require special handling when multiplying or dividing by a negative number, which reverses the inequality sign. Quadratic inequalities are best solved by sketching the graph, identifying the intervals where the function lies above or below the x‑axis.

不等式与方程类似，但在乘以或除以负数时需特殊处理，此时不等号方向反转。二次不等式最好通过绘制草图来解决，确定函数位于 x 轴上方或下方的区间。

For example, to solve x² – 5x + 6 > 0, factorise to (x – 2)(x – 3) > 0. The critical values are 2 and 3. Testing intervals gives the solution x < 2 or x > 3. Use open or closed circles on a number line to distinguish strict from non‑strict inequalities.

例如，解 x² – 5x + 6 > 0 时，因式分解为 (x – 2)(x – 3) > 0。临界值为 2 和 3。测试各区间可得解为 x < 2 或 x > 3。在数轴上使用空心或实心圆圈以区分严格与不严格的不等号。

10. Partial Fractions | 部分分式

Partial fraction decomposition rewrites a rational expression as a sum of simpler fractions, making integration, differentiation, and series expansion more manageable. The process depends on factoring the denominator completely and then determining the constants for each linear or irreducible quadratic factor.

部分分式分解将有理式重写为几个较简单的分式之和，使得积分、求导和级数展开更易处理。其步骤取决于完全分解分母，再针对每一个一次因式或不可约二次因式确定常数。

For a rational expression with distinct linear factors, such as (px + q)/[(x – a)(x – b)], write it as A/(x – a) + B/(x – b). Multiply through by the denominator and substitute convenient x‑values (like the roots) to solve for A and B. For repeated factors include terms with powers of the denominator.

对于具有相异一次因式的有理式，如 (px + q)/[(x – a)(x – b)]，可将其写为 A/(x – a) + B/(x – b)。等式两边同乘分母，再代入特定的 x 值（如各跟）求解 A 和 B。若因式重复，则需包含分母幂次逐增的各项。

When an irreducible quadratic factor like (x² + c) appears, the corresponding numerator must be linear: (Cx + D)/(x² + c). This form appears frequently in IB HL and AQA Further Pure modules. Practice identifying the correct setup to save time in exams.

若出现不可约二次因式，如 (x² + c)，对应的分子必须为一次式：(Cx + D)/(x² + c)。此形式在 IB HL 和 AQA 进阶纯数模块中经常出现。通过练习准确地列出设定式，可在考试中节省时间。

11. Modulus Functions and Absolute Value | 绝对值函数

The modulus function |x| returns the non‑negative magnitude of x, defined as x if x ≥ 0 and –x if x < 0. Its graph is a V‑shape with vertex at the origin. Transformations apply as usual, so y = |x – 2| shifts the vertex to (2, 0).

绝对值函数 |x| 返回 x 的非负大小，定义为：当 x ≥ 0 时取 x，当 x < 0 时取 –x。其图象为 V 形，顶点在原点。变换照常适用，因此 y = |x – 2| 将顶点移至 (2, 0)。

Equations involving modulus often produce two cases. For |f(x)| = k, set f(x) = k or f(x) = –k. For inequalities like |f(x)| < k, rewrite as –k < f(x) < k; for greater‑than, use f(x) < –k or f(x) > k. Sketching the graph helps visualise the solution sets.

含绝对值的方程通常产生两种情况。对于 |f(x)| = k，令 f(x) = k 或 f(x) = –k。对于形如 |f(x)| < k 的不等式，可改写为 –k < f(x) < k；若是大于号，则用 f(x) < –k 或 f(x) > k。绘制图象有助于直观理解解集。

Modulus functions also arise in defining composition and inverses. The absolute value can split the domain into regions where the function behaves differently, so always consider piecewise definitions when differentiating or integrating.

绝对值函数也出现在复合函数和反函数的定义中。绝对值会将定义域划分为函数行为不同的区域，因此在求导或积分时务必考虑分段定义。

12. Systems of Equations and Simultaneous Equations | 方程组与联立方程

A system of equations consists of two or more equations that share variables. Solutions are the set of values satisfying all equations simultaneously. Linear systems can be solved by substitution, elimination, or graphically by finding intersection points of the corresponding lines.

方程组由两个或更多含有共同变量的方程组成。解是同时满足所有方程的数值集。线性方程组可通过代入法、消元法求解，也可通过图像法找出对应直线的交点。

For a system with a quadratic and a linear equation, substitution is the most reliable method. Replace one variable in the quadratic using the linear equation, solve the resulting single‑variable equation, and back‑substitute. The number of intersections can be 0, 1, or 2, which connects to the discriminant concept.

对于一次与二次方程组成的方程组，代入法是最可靠的方法。利用一次方程将二次方程中的某个变量替换，解出所得的单变量方程，再回代求解。交点的个数可以是 0、1 或 2，这与判别式的概念相关联。

Three equations in three unknowns extend the elimination method, requiring systematic reduction. Always check solutions by substituting back into the original equations. Graphical interpretation in 3D involves planes intersecting at a point, a line, or having no common intersection.

三元一次方程组需将消元法系统地进行，逐步消减变量。务必代回原方程检验解。其三维几何含义为平面交于一点、一直线或无共同交点。

Word problems often lead to simultaneous equations. Translating a real‑world scenario into algebraic language and then solving is a key skill tested in both IB and AQA examinations.

应用题常常导向联立方程。将现实情境转化为代数语言并求解，是 IB 和 AQA 考试中都重点考查的关键技能。

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IB AQA Mathematics: Algebra and Functions Key Points | IB AQA 数学：代数和函数 考点精讲