The Complete Guide to Quadratic Functions | 二次函数完全指南

📚 The Complete Guide to Quadratic Functions | 二次函数完全指南

Quadratic functions are one of the most important topics in the IGCSE Mathematics syllabus. They appear in algebra, coordinate geometry, and even in problem-solving questions. Mastering quadratics is essential for achieving a high grade.

二次函数是 IGCSE 数学大纲中最重要的话题之一。它出现在代数、坐标几何甚至应用题中。掌握二次函数是取得高分的关键。

1. The General Form of a Quadratic | 二次函数的一般形式

A quadratic function is any function that can be written in the general form:

二次函数是可以写成以下一般形式的函数：

f(x) = ax² + bx + c, where a ≠ 0

The coefficient a determines the shape and orientation of the parabola. If a > 0, the parabola opens upward (a “U” shape). If a < 0, the parabola opens downward (an "∩" shape). The coefficient c represents the y-intercept — the point where the graph crosses the y-axis.

系数 a 决定抛物线的形状和开口方向。如果 a > 0，抛物线开口向上（“U”形）。如果 a < 0，抛物线开口向下（“∩”形）。系数 c 表示 y 轴截距——即图像与 y 轴相交的点。

Example: f(x) = 2x² − 3x + 5 has a = 2, b = −3, and c = 5. Since a > 0, this parabola opens upward and crosses the y-axis at (0, 5).

例如：f(x) = 2x² − 3x + 5 中 a = 2，b = −3，c = 5。因为 a > 0，此抛物线开口向上，并在 (0, 5) 处穿过 y 轴。

2. Expanding Quadratic Expressions | 展开二次表达式

To work with quadratics fluently, you must be able to expand products of two linear expressions. This is often called the FOIL method (First, Outer, Inner, Last).

要熟练处理二次函数，你必须能够展开两个一次表达式的乘积。这通常称为 FOIL 方法（首项、外项、内项、末项）。

(x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15

Key expansion formulas you must memorise:

你必须记住的关键展开公式：

(x + a)² = x² + 2ax + a²
(x − a)² = x² − 2ax + a²
(x + a)(x − a) = x² − a²（平方差）

(x + a)² = x² + 2ax + a²
(x − a)² = x² − 2ax + a²
(x + a)(x − a) = x² − a²（difference of squares）

Higher-tier students must also expand expressions like 2(x + 3)² − 5 or (2x + 1)(x − 4) carefully. Always check your signs — a common mistake is losing a negative sign during expansion.

Higher 级别学生还必须小心展开如 2(x + 3)² − 5 或 (2x + 1)(x − 4) 等表达式。始终检查符号——常见错误是在展开过程中漏掉负号。

3. Factorising Quadratics | 二次因式分解

Factorising is the reverse process of expanding. To factorise x² + bx + c, find two numbers that multiply to give c and add to give b.

因式分解是展开的逆过程。要对 x² + bx + c 进行因式分解，找到两个数，它们的乘积等于 c，和等于 b。

Example: Factorise x² + 7x + 12. We need two numbers that multiply to 12 and add to 7: these are 3 and 4. Therefore:

示例：分解 x² + 7x + 12。我们需要两个数乘积为 12、和为 7：这两个数是 3 和 4。因此：

x² + 7x + 12 = (x + 3)(x + 4)

When the coefficient of x² is not 1 (i.e., a ≠ 1), you must use a different strategy. For 2x² + 7x + 3, multiply a × c = 6. Find two numbers that multiply to 6 and add to 7: 1 and 6. Then split the middle term:

当 x² 的系数不为 1（即 a ≠ 1）时，你必须使用不同策略。对于 2x² + 7x + 3，先计算 a × c = 6。找到两个数乘积为 6、和为 7：即 1 和 6。然后拆分中间项：

2x² + 7x + 3 = 2x² + 1x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)

Special case: the difference of two squares. x² − 25 = (x + 5)(x − 5). This is very common in IGCSE exams.

特殊情况：平方差。x² − 25 = (x + 5)(x − 5)。这在 IGCSE 考试中非常常见。

4. Solving Quadratics by Factorisation | 用因式分解法解二次方程

If a quadratic equation can be factorised, solving it is straightforward. Set each bracket equal to zero because if the product of two factors is zero, at least one factor must be zero.

如果一个二次方程可以被因式分解，求解就很简单。令每个括号等于零，因为如果两个因数的乘积为零，则至少有一个因数为零。

Example: Solve x² − 5x + 6 = 0.

示例：解方程 x² − 5x + 6 = 0。

x² − 5x + 6 = (x − 2)(x − 3) = 0

Therefore x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3. Always verify your solutions by substituting back into the original equation.

因此 x − 2 = 0 或 x − 3 = 0，得到 x = 2 或 x = 3。始终通过代入原方程来验证你的解。

Remember that some quadratic equations have repeated roots, such as x² − 6x + 9 = (x − 3)² = 0, giving only x = 3. Others may have no real roots and cannot be solved by factorisation.

请记住有些二次方程有重根，如 x² − 6x + 9 = (x − 3)² = 0，只有 x = 3 一个解。其他方程可能没有实数根，无法通过因式分解求解。

5. The Quadratic Formula | 求根公式

For any quadratic equation ax² + bx + c = 0 where factorisation is difficult or impossible, use the quadratic formula:

对于任何难以或无法因式分解的二次方程 ax² + bx + c = 0，使用求根公式：

x = (−b ± √(b² − 4ac)) ÷ (2a)

This formula is provided in the IGCSE formula sheet for both Core and Extended levels. Substitute the values of a, b, and c carefully, paying close attention to negative signs.

此公式在 IGCSE 公式表中提供，适用于 Core 和 Extended 两个级别。仔细代入 a、b 和 c 的值，特别留意负号。

Example: Solve 2x² − 4x − 3 = 0 using the quadratic formula. Here a = 2, b = −4, c = −3.

示例：使用求根公式解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。

x = (4 ± √(16 + 24)) ÷ 4 = (4 ± √40) ÷ 4

Thus x ≈ 2.58 or x ≈ −0.58 (to 2 decimal places). When the question asks for solutions “correct to 2 decimal places” or “in surd form,” follow the instruction precisely.

因此 x ≈ 2.58 或 x ≈ −0.58（保留两位小数）。当题目要求答案“精确到两位小数”或“以根式形式表示”时，请严格按照要求作答。

6. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. This form is extremely useful for finding the vertex (turning point) of a parabola.

配方法将二次函数改写为 a(x + p)² + q 的形式。这种形式对于找到抛物线的顶点（转折点）非常有用。

For x² + bx, add and subtract (b/2)². For example, x² + 6x = (x + 3)² − 9.

对于 x² + bx，加上并减去 (b/2)²。例如，x² + 6x = (x + 3)² − 9。

Worked example: Write x² − 8x + 3 in the form (x − p)² + q.

示例：将 x² − 8x + 3 写成 (x − p)² + q 的形式。

x² − 8x + 3 = (x − 4)² − 16 + 3 = (x − 4)² − 13

When the coefficient of x² is not 1, factor it out first. For 2x² + 12x + 5, first factor out 2: 2(x² + 6x) + 5 = 2[(x + 3)² − 9] + 5 = 2(x + 3)² − 13.

当 x² 的系数不为 1 时，先把它提取出来。对于 2x² + 12x + 5，先提取 2：2(x² + 6x) + 5 = 2[(x + 3)² − 9] + 5 = 2(x + 3)² − 13。

From the completed square form, the vertex is at (−p, q). For the example above, the vertex is at (−3, −13) since the expression is 2(x + 3)² − 13.

从配方法形式中，顶点坐标为 (−p, q)。在上面例子中，因为表达式为 2(x + 3)² − 13，顶点在 (−3, −13)。

7. The Discriminant | 判别式

The expression b² − 4ac is called the discriminant. It tells us how many real roots a quadratic equation has, without solving it.

表达式 b² − 4ac 称为判别式。它告诉我们一个二次方程有多少个实数根，而无需实际求解。

If b² − 4ac > 0: two distinct real roots (the parabola crosses the x-axis at two points)
If b² − 4ac = 0: one repeated real root (the parabola touches the x-axis at one point)
If b² − 4ac < 0: no real roots (the parabola does not intersect the x-axis)

如果 b² − 4ac > 0：两个不同的实数根（抛物线与 x 轴相交于两点）
如果 b² − 4ac = 0：一个重根（抛物线与 x 轴相切于一点）
如果 b² − 4ac < 0：没有实数根（抛物线与 x 轴不相交）

Example: For the equation 3x² + 2x − 5 = 0, the discriminant is 2² − 4×3×(−5) = 4 + 60 = 64 > 0, so there are two distinct real roots.

示例：对于方程 3x² + 2x − 5 = 0，判别式为 2² − 4×3×(−5) = 4 + 60 = 64 > 0，因此有两个不同的实数根。

In exam problems, you may be asked to “find the range of values of k for which the equation has no real roots.” This requires setting up an inequality using the discriminant.

在考试中，你可能会被要求“找出使方程没有实数根时 k 的取值范围”。这需要利用判别式建立不等式。

8. The Graph of a Quadratic Function | 二次函数的图像

The graph of a quadratic function is a smooth curve called a parabola. It has a vertical axis of symmetry and one turning point (vertex), which is either a maximum or a minimum.

二次函数的图像是一条平滑曲线，称为抛物线。它有一条垂直对称轴和一个转折点（顶点），这个点要么是最大值要么是最小值。

To sketch a parabola accurately, you need:

要准确地画出抛物线草图，你需要：

The y-intercept: substitute x = 0, which gives c
The x-intercepts (roots): solve ax² + bx + c = 0
The vertex: use the formula x = −b ÷ (2a) to find the x-coordinate, then substitute to find y

y 轴截距：代入 x = 0，即得到 c
x 轴截距（根）：解方程 ax² + bx + c = 0
顶点：使用公式 x = −b ÷ (2a) 求 x 坐标，然后代入求 y

Example: Sketch f(x) = x² − 4x + 3. The y-intercept is 3. Factorising gives (x − 1)(x − 3), so roots are x = 1 and x = 3. The vertex x-coordinate is −(−4) ÷ (2×1) = 2, and y = 4 − 8 + 3 = −1. The axis of symmetry is x = 2, and the minimum point is (2, −1).

示例：画出 f(x) = x² − 4x + 3 的草图。y 轴截距为 3。因式分解得 (x − 1)(x − 3)，所以根为 x = 1 和 x = 3。顶点 x 坐标为 −(−4) ÷ (2×1) = 2，y = 4 − 8 + 3 = −1。对称轴为 x = 2，最小值为 (2, −1)。

9. Transformations of Quadratics | 二次函数的变换

Understanding how changes to the equation affect the graph is a key skill. In the form a(x − h)² + k:

理解方程的变化如何影响图像是关键技能。在形式 a(x − h)² + k 中：

a controls the vertical stretch/compression and direction (negative = reflected in x-axis)
h translates the graph horizontally (opposite direction: x − h moves right)
k translates the graph vertically

a 控制纵向拉伸/压缩和方向（负值 = 关于 x 轴对称反射）
h 使图像水平平移（方向相反：x − h 向右移动）
k 使图像垂直平移

For example, f(x) = (x − 3)² + 2 is the graph of y = x² shifted 3 units right and 2 units up. Its vertex is at (3, 2).

例如，f(x) = (x − 3)² + 2 是 y = x² 的图像向右移动 3 个单位、向上移动 2 个单位的结果。其顶点在 (3, 2)。

If the question asks for the image of a point after a transformation, apply the same shift to the coordinates. For instance, the point (1, 1) on y = x² becomes (4, 3) on y = (x − 3)² + 2.

如果题目要求某个点在变换后的像，对坐标施加相同的平移。例如，y = x² 上的点 (1, 1) 在 y = (x − 3)² + 2 上变成 (4, 3)。

10. Solving Quadratic Inequalities | 解二次不等式

Quadratic inequalities appear in Extended-level papers. Solve them by first finding the roots of the corresponding equation, then using a sketch or a number line to determine the correct interval(s).

二次不等式出现在 Extended 级别试卷中。先求对应方程的根，然后使用草图或数轴确定正确的区间来求解。

Example: Solve x² − 4x − 5 > 0.

示例：解不等式 x² − 4x − 5 > 0。

x² − 4x − 5 = (x − 5)(x + 1) = 0 → x = 5 or x = −1

Since the coefficient of x² is positive, the parabola opens upward. The graph is above the x-axis when x < −1 or x > 5. Therefore the solution is x < −1 or x > 5.

由于 x² 的系数为正，抛物线开口向上。当 x < −1 或 x > 5 时图像在 x 轴上方。因此解为 x < −1 或 x > 5。

If the inequality were x² − 4x − 5 ≤ 0, the solution would be −1 ≤ x ≤ 5. Remember: “greater than” gives the outside regions, “less than” gives the inside region for a positive quadratic.

如果不等式是 x² − 4x − 5 ≤ 0，则解为 −1 ≤ x ≤ 5。请记住：对于开口向上的二次函数，“大于”取外部区域，“小于”取内部区域。

11. Applications in Real-World Problems | 在实际问题中的应用

Quadratic functions model many real-world situations, including projectile motion, profit maximisation, and area optimisation. These application problems are increasingly common in IGCSE exams.

二次函数可以模拟许多现实世界的情境，包括抛体运动、利润最大化以及面积优化。这种应用类问题在 IGCSE 考试中越来越常见。

Projectile motion: the height h (in metres) of a ball after t seconds is given by h(t) = 20t − 5t². The maximum height occurs at the vertex. Using t = −b ÷ (2a) = −20 ÷ (2×(−5)) = 2 seconds, so h(2) = 20×2 − 5×4 = 20 metres.

抛体运动：球在 t 秒后的高度 h（米）由 h(t) = 20t − 5t² 给出。最大高度出现在顶点处。利用 t = −b ÷ (2a) = −20 ÷ (2×(−5)) = 2 秒，所以 h(2) = 20×2 − 5×4 = 20 米。

Area optimisation: A farmer has 60 metres of fencing and wants to enclose a rectangle against a wall. If the width is x, the area is A = x(60 − 2x) = 60x − 2x². The maximum area is at x = −60 ÷ (2×(−2)) = 15 metres, giving area = 15 × 30 = 450 m².

面积优化：一位农民有 60 米围栏，想靠围墙围出一个矩形区域。如果宽为 x，面积为 A = x(60 − 2x) = 60x − 2x²。最大面积出现在 x = −60 ÷ (2×(−2)) = 15 米处，面积为 15 × 30 = 450 平方米。

When solving word problems, always define your variable clearly, write the quadratic model, identify whether you need the vertex or the roots, and check that your answer makes sense in context.

解文字题时，请务必明确设定变量、写出二次模型、确定你需要的是顶点还是根，并检查答案在情境中是否合理。

12. Summary of Key Formulas | 关键公式总结

The table below summarises the essential formulas and techniques for quadratic functions in the IGCSE syllabus.

下表总结了 IGCSE 大纲中二次函数必备的公式和技巧。

Concept \| 概念	Formula \| 公式
General form \| 一般形式	ax² + bx + c = 0 (a ≠ 0)
Roots \| 根	Sum = −b⁄a, Product = c⁄a
Vertex x-coordinate \| 顶点 x 坐标	x = −b ÷ (2a)
Axis of symmetry \| 对称轴	x = −b ÷ (2a)
Quadratic formula \| 求根公式	x = (−b ± √(b² − 4ac)) ÷ (2a)
Discriminant \| 判别式	Δ = b² − 4ac
Completed square \| 配方法形式	a(x − h)² + k, vertex at (h, k)

Practice is the key to mastering quadratics. Work through past paper questions on factorisation, the quadratic formula, completing the square, and graph sketching. The more problems you solve, the more automatic these techniques become.

练习是掌握二次函数的关键。系统地做历年真题中关于因式分解、求根公式、配方法和图像绘制的题目。你解决的问题越多，这些技巧就越得心应手。

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The Complete Guide to Quadratic Functions | 二次函数完全指南