A-Level Maths: Quadratic Functions Exam Guide | A-Level 数学：二次函数考点精讲

📚 A-Level Maths: Quadratic Functions Exam Guide | A-Level 数学：二次函数考点精讲

Quadratic functions are one of the most fundamental topics in A-Level Mathematics, underpinning everything from algebraic manipulation and graph sketching to calculus and applied modelling. A deep understanding of their forms, roots, discriminant and transformations is essential for high marks on both pure and applied papers. This revision guide walks you through every crucial concept with clear English-Chinese explanations, worked examples and exam-focused tips.

二次函数是 A-Level 数学中最基础的课题之一，支撑着从代数运算、图像绘制到微积分和应用建模的方方面面。深入理解其形式、根、判别式和变换，对于在纯数和应用卷中取得高分至关重要。本篇复习指南通过清晰的中英对照解释、例题和聚焦考点的提示，带你掌握每一个关键概念。

1. Standard Form and Key Terms | 标准形式与关键术语

A quadratic function is any function that can be written in the standard form y = ax² + bx + c, where a, b and c are constants and a ≠ 0. The condition a ≠ 0 ensures the expression is genuinely quadratic, not linear. The term ax² is called the quadratic term, bx the linear term and c the constant term.

二次函数是任何可以写成标准形式 y = ax² + bx + c 的函数，其中 a、b、c 为常数且 a ≠ 0。条件 a ≠ 0 确保了表达式确实是二次的，而不是线性的。ax² 项称为二次项，bx 为一次项，c 为常数项。

The coefficient a determines the width and the direction of the parabola; the larger |a|, the steeper the graph. The coefficient b affects the axis of symmetry, while c gives the y-intercept directly.

系数 a 决定抛物线的宽度和开口方向；|a| 越大，图像越陡峭。系数 b 影响对称轴的位置，而 c 直接给出 y 轴截距。

2. The Parabola and Its Vertex | 抛物线及其顶点

The graph of a quadratic function is a smooth curve called a parabola. If a > 0, the parabola opens upwards and has a minimum point; if a < 0, it opens downwards and has a maximum point. The turning point of the parabola is called the vertex.

二次函数的图像是一条光滑的曲线，称为 抛物线。若 a > 0，抛物线开口向上，有最低点；若 a < 0，开口向下，有最高点。该转折点称为顶点。

For the standard form y = ax² + bx + c, the x-coordinate of the vertex is given by x = −b / (2a). Substituting this back into the function yields the y-coordinate. The vertex is therefore (−b/(2a), f(−b/(2a))), which can also be written as (−b/(2a), (4ac − b²)/(4a)).

对于标准形式 y = ax² + bx + c，顶点的 x 坐标由 x = −b / (2a) 给出。将其代回函数可得 y 坐标。因此顶点为 (−b/(2a), f(−b/(2a)))，也可以写成 (−b/(2a), (4ac − b²)/(4a))。

Vertex = ( −b/(2a) , (4ac − b²)/(4a) )

The vertical line x = −b/(2a) is the axis of symmetry, which divides the parabola into two mirror-image halves.

直线 x = −b/(2a) 是 对称轴，它将抛物线分成两个镜像对称的部分。

3. Completing the Square | 配方法

Completing the square converts the standard form into vertex form: y = a(x − h)² + k, where (h, k) is the vertex. This method is essential for finding maxima/minima and sketching graphs quickly.

配方法将标准形式转化为 顶点式：y = a(x − h)² + k，其中 (h, k) 为顶点。此方法对于快速求最值和画图至关重要。

Example: Express y = 2x² + 8x + 5 in vertex form.
Step 1: Factor out the coefficient of x² from the first two terms: y = 2(x² + 4x) + 5.
Step 2: Inside the bracket, add and subtract (4/2)² = 4: y = 2(x² + 4x + 4 − 4) + 5.
Step 3: Rewrite as 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3.
Vertex is (−2, −3).

例题：将 y = 2x² + 8x + 5 化为顶点式。
第 1 步：从前两项提取 x² 的系数：y = 2(x² + 4x) + 5。
第 2 步：在括号内加上并减去 (4/2)² = 4：y = 2(x² + 4x + 4 − 4) + 5。
第 3 步：改写为 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3。
顶点为 (−2, −3)。

When the coefficient a ≠ 1, always factor it out first. This technique also helps derive the quadratic formula.

当系数 a ≠ 1 时，务必先将其提取出来。这一技巧也有助于推导求根公式。

4. Quadratic Formula and Roots | 求根公式与根

The solutions (roots) of the quadratic equation ax² + bx + c = 0 are given by the quadratic formula:

二次方程 ax² + bx + c = 0 的解（根）由求根公式给出：

x = [ −b ± √(b² − 4ac) ] / (2a)

The symbol ± indicates there are generally two roots, corresponding to the points where the parabola crosses the x‑axis. The expression under the square root, b² − 4ac, is called the discriminant.

符号 ± 表示通常有两个根，对应抛物线与 x 轴相交的两个点。根号下的表达式 b² − 4ac 称为判别式。

You must be able to use the formula in surd form and to approximate decimal answers when required. For example, to solve 3x² − 4x − 2 = 0:
a = 3, b = −4, c = −2 → x = [4 ± √(16 + 24)] / 6 = [4 ± √40] / 6 = [4 ± 2√10] / 6 = (2 ± √10) / 3.

你必须能够以根式形式使用该公式，并在需要时求出近似小数答案。例如解 3x² − 4x − 2 = 0：
a = 3, b = −4, c = −2 → x = [4 ± √(16 + 24)] / 6 = [4 ± √40] / 6 = [4 ± 2√10] / 6 = (2 ± √10) / 3。

5. The Discriminant (Δ) | 判别式 (Δ)

The discriminant is defined as Δ = b² − 4ac. It reveals the nature and number of real roots without solving the equation:

判别式定义为 Δ = b² − 4ac。它无需解方程即可揭示实根的性质与个数：

Δ > 0	Two distinct real roots	两个不相等的实根
Δ = 0	One repeated real root (the graph touches the x‑axis)	一个重根（图像与 x 轴相切）
Δ < 0	No real roots (the graph does not cross the x‑axis)	无实根（图像不与 x 轴相交）

Examiners often ask for the range of values of a parameter that yields a certain root condition. For instance, find k such that x² + kx + 9 = 0 has equal roots: set Δ = k² − 36 = 0 → k = ±6.

考官常要求找出使方程具有特定根条件的参数取值范围。例如求 k 使 x² + kx + 9 = 0 有等根：令 Δ = k² − 36 = 0 → k = ±6。

When a > 0 and Δ < 0, the entire parabola lies above the x‑axis, meaning the function is always positive. Conversely, a < 0 and Δ < 0 means it is always negative.

当 a > 0 且 Δ < 0 时，整条抛物线在 x 轴上方，即函数值恒正。反之，a < 0 且 Δ < 0 时函数值恒负。

6. Factorisation Methods | 因式分解法

Factorising is often the quickest way to solve quadratics when the roots are rational. Common techniques include:

因式分解在根为有理数时通常是最快的求解方法。常用技巧包括：

1. Common factor: 3x² − 6x = 3x(x − 2).

1. 提取公因式：3x² − 6x = 3x(x − 2)。

2. Difference of two squares: x² − 9 = (x − 3)(x + 3).

2. 平方差公式：x² − 9 = (x − 3)(x + 3)。

3. Trinomials (cross method): x² + 5x + 6 = (x + 2)(x + 3). For non‑monic quadratics like 2x² + 7x + 3, we look for factors of 2×3=6 that add to 7: 6 and 1 → 2x² + 6x + x + 3 = 2x(x+3)+1(x+3) = (2x+1)(x+3).

3. 二次三项式（十字相乘法）：x² + 5x + 6 = (x + 2)(x + 3)。对于非首一的二次式如 2x² + 7x + 3，寻找使 2×3=6 相加为 7 的因数：6 和 1 → 2x² + 6x + x + 3 = 2x(x+3)+1(x+3) = (2x+1)(x+3)。

Once factorised, set each bracket to zero to find the roots. Always check by expanding.

因式分解后，令每个括号为零即可求得根。务必通过展开验证。

7. Finding Intercepts | 求截距

The y‑intercept is found by setting x = 0, giving the point (0, c). The x‑intercepts (if they exist) are the real roots of ax² + bx + c = 0.

y 轴截距通过令 x = 0 求得，即点 (0, c)。x 轴截距（若存在）是方程 ax² + bx + c = 0 的实根。

To sketch a quadratic accurately, label the vertex, the intercepts and the axis of symmetry. Knowing the discriminant tells you at a glance whether the graph cuts, touches or misses the x‑axis entirely.

要准确地画出二次函数图像，需标出顶点、截距和对称轴。知道了判别式，就能一眼看出图像是与 x 轴相交、相切，还是完全不相交。

Example: For y = −x² + 4x − 3, c = −3 gives y‑intercept (0, −3). Factorise: −(x² − 4x + 3) = −(x − 1)(x − 3) → x‑intercepts at (1,0) and (3,0). Vertex: x = −4/(2·(−1)) = 2, y = −(2)² + 8 − 3 = 1 → (2,1).

例题：对于 y = −x² + 4x − 3，c = −3 得 y 截距 (0, −3)。因式分解：−(x² − 4x + 3) = −(x − 1)(x − 3) → x 截距为 (1,0) 和 (3,0)。顶点：x = −4/(2·(−1)) = 2，y = −(2)² + 8 − 3 = 1 → (2,1)。

8. Transformations of Quadratic Graphs | 二次函数图像的变换

Starting from the parent graph y = x², we can apply rigid and non‑rigid transformations to obtain any quadratic. Understanding these mappings is a common exam requirement.

从母图 y = x² 出发，我们可以通过刚性和非刚性变换得到任意二次函数图像。理解这些映射是考试常见要求。

y = x² + k → vertical translation by k units (up if k>0).

y = x² + k → 垂直平移 k 个单位（k>0 向上平移）。
y = (x − h)² → horizontal translation by h units to the right (left if h negative).

y = (x − h)² → 向右水平平移 h 个单位（h 为负则向左）。
y = a x² → vertical stretch by factor |a|; if a<0, also a reflection in the x‑axis.

y = a x² → 沿 y 轴拉伸 |a| 倍；若 a<0，并关于 x 轴反射。
y = (kx)² → horizontal stretch by factor 1/|k|.

y = (kx)² → 沿 x 轴拉伸 1/|k| 倍。

The vertex form y = a(x − h)² + k neatly captures all these: stretch by a, shift right by h, shift up by k. Always transform in the correct order: horizontal shifts, stretch/reflection, then vertical shifts.

顶点式 y = a(x − h)² + k 清晰地综合了所有这些变换：拉伸 a 倍，向右移 h，向上移 k。务必按正确顺序变换：先水平移动，再拉伸/反射，最后垂直移动。

9. Solving Quadratic Inequalities | 解二次不等式

To solve an inequality like x² − 4x − 5 ≤ 0, first find the critical values by solving the equality: (x − 5)(x + 1) = 0 → x = 5 or x = −1. Sketch the parabola (here a=1>0, happy face). The portion ≤ 0 lies between the roots, so −1 ≤ x ≤ 5.

要解如 x² − 4x − 5 ≤ 0 的不等式，首先通过解等式求出临界值：(x − 5)(x + 1) = 0 → x = 5 或 x = −1。画出抛物线草图（此处 a=1>0 开口向上）。≤ 0 的部分位于两根之间，因此 −1 ≤ x ≤ 5。

If the inequality had been > 0, the solution would be x < −1 or x > 5. Always check the direction of the inequality and whether the parabola opens up or down.

如果不等式是 > 0，解将是 x < −1 或 x > 5。始终检查不等号方向和抛物线开口方向。

For a quadratic that does not factorise, use the quadratic formula to find essential critical values. Remember that if Δ < 0 and a > 0, then ax² + bx + c > 0 for all real x.

若二次式无法因式分解，可使用求根公式求出关键临界值。记住，若 Δ < 0 且 a > 0，则对所有实数 x 均有 ax² + bx + c > 0。

10. Modelling with Quadratic Functions | 二次函数建模应用

A‑Level exam questions regularly present real‑world scenarios modelled by quadratics, such as projectile motion (height vs time), area optimisation, or revenue/cost problems. The vertex often represents a maximum or minimum quantity.

A‑Level 试题中经常出现用二次函数模拟的现实场景，如抛体运动（高度与时间）、面积优化或收入/成本问题。顶点通常代表最大或最小值。

Example: A ball is thrown upwards. Its height h (m) at time t (s) is given by h = −5t² + 20t + 1. Find the maximum height and when it occurs.
Here a = −5, b = 20, c = 1. Vertex at t = −b/(2a) = −20/(−10) = 2 s. Maximum height = h(2) = −5(4) + 40 + 1 = 21 m. The time when the ball hits the ground is found by solving −5t² + 20t + 1 = 0, keeping the positive root.

例题：一个球向上抛出。其高度 h（米）与时间 t（秒）的关系为 h = −5t² + 20t + 1。求最大高度及其发生时间。
此处 a = −5, b = 20, c = 1。顶点位于 t = −b/(2a) = −20/(−10) = 2 秒。最大高度 = h(2) = −5(4) + 40 + 1 = 21 米。球落地的时间可通过解 −5t² + 20t + 1 = 0 求得，取正根。

When modelling, always check the domain (e.g

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A-Level Maths: Quadratic Functions Exam Guide | A-Level 数学：二次函数考点精讲