Quadratic Functions & Equations: A Complete IGCSE Guide | 二次函数与方程：IGCSE 完全指南

📚 Quadratic Functions & Equations: A Complete IGCSE Guide | 二次函数与方程：IGCSE 完全指南

Quadratic functions and equations form the backbone of IGCSE Mathematics, appearing in almost every exam paper. This guide covers every major technique you need, from factorisation to the quadratic formula, with exam-style examples and common pitfalls highlighted throughout.

二次函数与方程是 IGCSE 数学的核心内容，几乎每份试卷都会出现。本指南涵盖你需要的所有主要技巧，从因式分解到求根公式，并配有考试风格的例题和常见易错点提示。

1. Standard Form and Graphs | 标准形式与图像

A quadratic function is any function of the form f(x) = ax² + bx + c, where a ≠ 0. The graph of a quadratic function is called a parabola. When a > 0, the parabola opens upward (a ‘smile’ shape); when a < 0, it opens downward (a 'frown' shape).

二次函数是指任何形如 f(x) = ax² + bx + c 的函数，其中 a ≠ 0。二次函数的图像称为抛物线。当 a > 0 时，抛物线开口向上（微笑形）；当 a < 0 时，抛物线开口向下（皱眉形）。

The y-intercept is always at the point (0, c).
The axis of symmetry is the vertical line x = −b/(2a).
The turning point (vertex) has x-coordinate −b/(2a). To find the y-coordinate, substitute this x-value back into the function.

抛物线与 y 轴的交点总是在 (0, c) 处。对称轴是垂直线 x = −b/(2a)。顶点的 x 坐标为 −b/(2a)，将 x 值代回函数即可求出 y 坐标。

Vertex form: f(x) = a(x − h)² + k, where (h, k) is the turning point.

顶点式：f(x) = a(x − h)² + k，其中 (h, k) 为顶点坐标。

For example, f(x) = 2(x − 3)² + 5 has its vertex at (3, 5) and opens upward since a = 2 > 0.

例如，f(x) = 2(x − 3)² + 5 的顶点在 (3, 5)，且因 a = 2 > 0 而开口向上。

2. Solving by Factorisation | 因式分解法求解

Factorisation is the fastest method when the quadratic has simple integer roots. The key principle: if (x − p)(x − q) = 0, then either x = p or x = q.

当二次方程具有简单的整数根时，因式分解是最快速的方法。关键原理：如果 (x − p)(x − q) = 0，那么 x = p 或 x = q。

To factorise ax² + bx + c when a = 1, find two numbers that multiply to give c and add to give b.

当 a = 1 时，因式分解 ax² + bx + c 的方法是找到两个数，它们的乘积等于 c，且它们的和等于 b。

Example: Solve x² − 7x + 12 = 0. The numbers −3 and −4 multiply to 12 and add to −7. Therefore (x − 3)(x − 4) = 0, so x = 3 or x = 4.

例：解方程 x² − 7x + 12 = 0。数字 −3 和 −4 相乘得 12，相加得 −7。因此 (x − 3)(x − 4) = 0，所以 x = 3 或 x = 4。

When a ≠ 1, use the method of factors of ac that sum to b, then factor by grouping.

当 a ≠ 1 时，使用 ac 的因子中其和为 b 的方法，然后分组因式分解。

Example: Solve 2x² + 7x + 3 = 0. Multiply a × c = 2 × 3 = 6. The pair 1 and 6 sum to 7. Rewrite: 2x² + x + 6x + 3 = 0. Group: x(2x + 1) + 3(2x + 1) = 0. Thus (2x + 1)(x + 3) = 0, giving x = −½ or x = −3.

例：解方程 2x² + 7x + 3 = 0。将 a × c = 2 × 3 = 6。因子对 1 和 6 相加得 7。改写：2x² + x + 6x + 3 = 0。分组：x(2x + 1) + 3(2x + 1) = 0。因此 (2x + 1)(x + 3) = 0，得 x = −½ 或 x = −3。

3. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. This form reveals the vertex directly and is essential for solving when factorisation fails.

配方法将二次式改写为 a(x + p)² + q 的形式。这种形式直接揭示了顶点坐标，也是因式分解失败时求解的关键方法。

The algorithm for x² + bx + c:

x² + bx + c = (x + b/2)² − (b/2)² + c

配方法的步骤（针对 x² + bx + c）：

将 x² + bx + c 改写为 (x + b/2)² − (b/2)² + c

Example: Write x² + 6x + 2 in completed square form. Since b = 6, b/2 = 3. Then x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7. The vertex is (−3, −7).

例：将 x² + 6x + 2 写成配方法形式。由于 b = 6，b/2 = 3。则 x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7。顶点坐标为 (−3, −7)。

When the coefficient of x² is not 1, factor it out first: 2x² − 8x + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3.

当 x² 的系数不为 1 时，先将其提出：2x² − 8x + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3。

4. The Quadratic Formula | 求根公式

The quadratic formula works for all quadratics, including those that cannot be factorised. It is derived from completing the square and is a guaranteed method for finding roots.

求根公式适用于所有二次方程，包括无法因式分解的方程。它由配方法推导而来，是求根的有力保证。

x = (−b ± √(b² − 4ac)) / 2a

Example: Solve 3x² − 5x − 2 = 0. Here a = 3, b = −5, c = −2.

x = (5 ± √(25 − 4(3)(−2))) / 6 = (5 ± √(25 + 24)) / 6 = (5 ± √49) / 6 = (5 ± 7) / 6

Therefore x = (5 + 7)/6 = 2 or x = (5 − 7)/6 = −⅓.

例：解方程 3x² − 5x − 2 = 0。这里 a = 3，b = −5，c = −2。

x = (5 ± √(25 − 4(3)(−2))) / 6 = (5 ± √(25 + 24)) / 6 = (5 ± √49) / 6 = (5 ± 7) / 6

因此 x = (5 + 7)/6 = 2 或 x = (5 − 7)/6 = −⅓。

Memorise the formula carefully — the most common mistakes are sign errors with −b and misplacing the denominator 2a. Always write down a, b, c explicitly before substituting.

请仔细记忆公式——最常见的错误是 −b 的符号错误以及分母 2a 的放置错误。代入前务必明确写出 a、b、c 的值。

5. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root: Δ = b² − 4ac. It tells us how many real roots a quadratic equation has, without actually solving it.

判别式是求根公式中根号下的部分：Δ = b² − 4ac。它可以告诉我们二次方程有多少个实数根，而无需实际求解。

Discriminant Value / 判别式值	Nature of Roots / 根的性质	Graph Interpretation / 图像意义
Δ > 0	Two distinct real roots	Parabola crosses the x-axis twice
Δ = 0	One repeated real root	Parabola touches the x-axis at the vertex
Δ < 0	No real roots	Parabola does not intersect the x-axis

判别式值	根的性质	图像意义
Δ > 0	两个不相等的实数根	抛物线与 x 轴相交于两点
Δ = 0	一个重根	抛物线在顶点处与 x 轴相切
Δ < 0	无实数根	抛物线与 x 轴不相交

Example: For which values of k does the equation x² + kx + 9 = 0 have two distinct real roots? We need Δ > 0, so k² − 36 > 0. Thus (k − 6)(k + 6) > 0, giving k > 6 or k < −6.

例：k 为何值时，方程 x² + kx + 9 = 0 有两个不相等的实数根？需要 Δ > 0，即 k² − 36 > 0。因此 (k − 6)(k + 6) > 0，得 k > 6 或 k < −6。

6. Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0 with roots α and β, the following relationships hold:

对于根为 α 和 β 的二次方程 ax² + bx + c = 0，有以下关系式：

Sum of roots: α + β = −b/a | Product of roots: αβ = c/a

根的和：α + β = −b/a | 根的积：αβ = c/a

These relationships allow us to find symmetric expressions in α and β without solving the equation. For instance, α² + β² = (α + β)² − 2αβ.

这些关系式使我们无需解方程就能求出关于 α 和 β 的对称表达式。例如，α² + β² = (α + β)² − 2αβ。

Example: If α and β are roots of 2x² − 6x + 3 = 0, find α² + β². Here α + β = 3 and αβ = 3/2. Therefore α² + β² = 3² − 2(3/2) = 9 − 3 = 6.

例：若 α 和 β 是方程 2x² − 6x + 3 = 0 的根，求 α² + β²。这里 α + β = 3，αβ = 3/2。因此 α² + β² = 3² − 2(3/2) = 9 − 3 = 6。

You can also construct a quadratic equation from given roots: x² − (sum of roots)x + (product of roots) = 0.

你还可以根据已知根构造二次方程：x² − (根的和)x + (根的积) = 0。

7. Solving Quadratic Inequalities | 解二次不等式

Solving quadratic inequalities combines factorisation with sign analysis. The general strategy: factorise, find the critical points, then test intervals on a number line.

解二次不等式将因式分解与符号分析相结合。一般策略：因式分解、找到临界点、在数轴上测试区间。

Example: Solve x² − x − 6 > 0. Factorise to get (x − 3)(x + 2) > 0. The critical points are x = 3 and x = −2. Test the three intervals:

例：解不等式 x² − x − 6 > 0。因式分解得到 (x − 3)(x + 2) > 0。临界点为 x = 3 和 x = −2。测试三个区间：

x < −2: both factors negative, product positive → solution
−2 < x < 3: one positive, one negative, product negative → not a solution
x > 3: both factors positive, product positive → solution

Therefore the solution is x < −2 or x > 3.

x < −2：两个因子均为负，乘积为正 → 解
−2 < x < 3：一正一负，乘积为负 → 非解
x > 3：两个因子均为正，乘积为正 → 解

因此解为 x < −2 或 x > 3。

If the inequality sign were x² − x − 6 < 0, the solution would be the middle interval: −2 < x < 3. Remember: when multiplying or dividing an inequality by a negative number, flip the inequality sign.

如果不等式符号是 x² − x − 6 < 0，则解为中间区间：−2 < x < 3。记住：当不等式两边乘以或除以负数时，需要翻转不等号。

8. Quadratic Functions in Real-World Problems | 二次函数在实际问题中的应用

Quadratic models appear in physics (projectile motion), economics (profit maximisation), and geometry (area optimisation). IGCSE exams often present these as word problems where you must form the equation yourself.

二次模型出现在物理学（抛体运动）、经济学（利润最大化）和几何学（面积优化）中。IGCSE 考试通常以应用题形式出现，需要你自己建立方程。

Example: A rectangular field has length 10 m more than its width. Its area is 600 m². Find the width.

Let the width be w m. Then length is (w + 10) m. The area equation is:

w(w + 10) = 600 → w² + 10w − 600 = 0

Factorise: (w + 30)(w − 20) = 0. Therefore w = 20 or w = −30. Since width cannot be negative, w = 20 m.

例：一块矩形田地的长度比宽度多 10 米，面积为 600 平方米。求宽度。

设宽度为 w 米，则长度为 (w + 10) 米。面积方程为：

w(w + 10) = 600 → w² + 10w − 600 = 0

因式分解：(w + 30)(w − 20) = 0。因此 w = 20 或 w = −30。由于宽度不能为负数，w = 20 米。

Always check whether the physical context rejects any roots. Negative lengths, times, or quantities are usually invalid.

务必检查物理背景是否排除某些根。负的长度、时间或数量通常是无效的。

9. Graphical Transformations of Quadratics | 二次函数的图像变换

Understanding transformations helps you sketch graphs quickly and interpret data. For a base quadratic y = x²:

理解图像变换有助于快速画图和解读书面数据。以基础二次函数 y = x² 为例：

y = x² + k translates the graph vertically by k units.
y = (x − h)² translates the graph horizontally by h units (right if h > 0).
y = −x² reflects the graph in the x-axis.
y = ax² stretches the graph vertically by a factor of a.

y = x² + k 将图像垂直平移 k 个单位。
y = (x − h)² 将图像水平平移 h 个单位（h > 0 时向右移）。
y = −x² 将图像关于 x 轴反射。
y = ax² 将图像垂直拉伸 a 倍。

The vertex form y = a(x − h)² + k combines all these transformations in one expression: horizontal shift h, vertical shift k, vertical stretch a, and reflection when a < 0.

顶点式 y = a(x − h)² + k 将所有变换合并到一个表达式中：水平平移 h、垂直平移 k、垂直伸缩 a，当 a < 0 时还包含反射。

10. Intersection of a Line and a Parabola | 直线与抛物线的交点

To find where a line y = mx + c intersects a parabola y = ax² + bx + c, set the equations equal and solve the resulting quadratic in x.

要求直线 y = mx + c 与抛物线 y = ax² + bx + c 的交点，将两个方程相等后求解关于 x 的二次方程。

The discriminant of this resulting quadratic determines the number of intersection points:

所得二次方程的判别式决定交点的个数：

Discriminant / 判别式	Number of Intersection Points / 交点数
Δ > 0	Two points (secant line)
Δ = 0	One point (tangent line)
Δ < 0	No intersection

判别式	交点数
Δ > 0	两个交点（割线）
Δ = 0	一个交点（切线）
Δ < 0	无交点

Example: Does the line y = 2x + 1 intersect the parabola y = x² − x + 3? Setting equal: x² − x + 3 = 2x + 1, which simplifies to x² − 3x + 2 = 0. The discriminant is (−3)² − 4(1)(2) = 9 − 8 = 1 > 0. Therefore two intersection points exist, at x = 1 and x = 2.

例：直线 y = 2x + 1 与抛物线 y = x² − x + 3 相交吗？令两式相等：x² − x + 3 = 2x + 1，化简得 x² − 3x + 2 = 0。判别式为 (−3)² − 4(1)(2) = 9 − 8 = 1 > 0。因此存在两个交点，分别在 x = 1 和 x = 2 处。

11. Common Mistakes and Exam Tips | 常见错误与应试技巧

Even strong students lose marks on quadratics through avoidable errors. Here are the most frequent pitfalls and how to avoid them.

即使是优秀的学生也会在二次函数相关

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Quadratic Functions & Equations: A Complete IGCSE Guide | 二次函数与方程：IGCSE 完全指南