Mastering Quadratic Functions | 掌握二次函数

📚 Mastering Quadratic Functions | 掌握二次函数

Quadratic functions form one of the most important pillars of IGCSE Mathematics. Understanding their properties, graphs, and applications is essential for success in both Paper 2 and Paper 4. This revision guide will walk you through every key concept with worked examples, common pitfalls, and exam-style questions.

二次函数是IGCSE数学最重要的基石之一。理解其性质、图像和应用，是在Paper 2和Paper 4中取得好成绩的关键。本复习指南将带你逐一掌握所有核心概念，包括典型例题、常见错误以及考试风格习题。

1. The Standard Form of a Quadratic Function | 二次函数的标准形式

A quadratic function is any function that can be written in the form f(x) = ax² + bx + c, where a, b, and c are constants, and a ≠ 0. The coefficient a determines the shape of the parabola, b affects the position of the axis of symmetry, and c gives the y-intercept.

二次函数是指可以用 f(x) = ax² + bx + c 形式表示的函数，其中 a、b、c 为常数，且 a ≠ 0。系数 a 决定抛物线的开口方向与形状，b 影响对称轴的位置，c 为 y 轴截距。

f(x) = ax² + bx + c, where a ≠ 0

f(x) = ax² + bx + c，其中 a ≠ 0

If a > 0, the parabola opens upward and has a minimum point. If a < 0, the parabola opens downward and has a maximum point. Note that b or c can be zero, but a can never be zero, otherwise the function would become linear.

当 a > 0 时，抛物线开口向上，有最小值点；当 a < 0 时，抛物线开口向下，有最大值点。注意 b 或 c 可以为 0，但 a 绝不能为 0，否则函数就变成了一次函数。

2. The Vertex Form and Completing the Square | 顶点式与配方法

Every quadratic function can be expressed in vertex form: f(x) = a(x – h)² + k, where (h, k) is the vertex of the parabola. To obtain this form, we use the method of completing the square.

每个二次函数都可以表示为顶点式：f(x) = a(x – h)² + k，其中 (h, k) 为抛物线的顶点。要将一般式化为顶点式，我们需要使用配方法。

To complete the square for f(x) = x² + 6x + 5, we first halve the coefficient of x (which is 6), giving 3. We write (x + 3)² = x² + 6x + 9. Since our original expression has +5, we adjust: (x + 3)² – 4.

例如，对 f(x) = x² + 6x + 5 进行配方：先将 x 的系数 6 除以 2 得 3，写成 (x + 3)² = x² + 6x + 9。由于原式是 +5，我们做如下调整：(x + 3)² – 4。

f(x) = (x + 3)² – 4

Therefore, the vertex is at (-3, -4). Note the sign change: inside the bracket it is x + 3, which means h = -3. The minimum value of this function is -4, occurring when x = -3.

因此，顶点为 (-3, -4)。注意符号变化：括号内是 x + 3，意味着 h = -3。该函数的最小值为 -4，在 x = -3 时取得。

3. Roots and Solutions | 根与解

The roots of a quadratic equation f(x) = 0 represent the x-coordinates where the parabola crosses the x-axis. There are three possible scenarios: two distinct real roots, one repeated root, or no real roots.

二次方程 f(x) = 0 的根表示抛物线与 x 轴交点的 x 坐标。可能出现的三种情况：两个不同实数根、一个重根、或没有实数根。

To find the roots, we can use factorisation, the quadratic formula, or completing the square. The quadratic formula is:

求根的方法包括因式分解、求根公式或配方法。求根公式为：

x = (-b ± √(b² – 4ac)) / 2a

For example, solve 2x² – 5x – 3 = 0 using the formula. Here a = 2, b = -5, c = -3. We substitute: x = (5 ± √(25 + 24)) / 4 = (5 ± √49) / 4 = (5 ± 7) / 4. Thus x = 3 or x = -1/2.

例如，用公式求解 2x² – 5x – 3 = 0。这里 a = 2，b = -5，c = -3。代入得：x = (5 ± √(25 + 24)) / 4 = (5 ± √49) / 4 = (5 ± 7) / 4。因此 x = 3 或 x = -1/2。

4. The Discriminant | 判别式

The expression b² – 4ac under the square root in the quadratic formula is called the discriminant, denoted by Δ (delta). It tells us the nature of the roots without solving the full equation.

求根公式中根号下的表达式 b² – 4ac 称为判别式，用 Δ（德尔塔）表示。它无需解完整方程就能告诉我们根的性质。

If Δ > 0, there are two distinct real roots. If Δ = 0, there is exactly one repeated real root. If Δ < 0, there are no real roots (the curve does not cross the x-axis).

若 Δ > 0，则有两个不同的实数根；若 Δ = 0，则恰好有一个重根；若 Δ < 0，则没有实数根（曲线与 x 轴不相交）。

Discriminant 判别式	Nature of Roots 根的性质	Graph Interpretation 图像含义
Δ > 0	Two distinct real roots 两个不同实数根	Crosses x-axis twice 与 x 轴有两个交点
Δ = 0	One repeated root 一个重根	Touches x-axis once 与 x 轴相切
Δ < 0	No real roots 无实数根	Does not touch x-axis 不与 x 轴相交

For example, for 3x² – 2x + 1 = 0, the discriminant is (-2)² – 4(3)(1) = 4 – 12 = -8 < 0. Therefore, this equation has no real roots.

例如，对于 3x² – 2x + 1 = 0，判别式为 (-2)² – 4(3)(1) = 4 – 12 = -8 < 0。因此，该方程没有实数根。

5. Solving by Factorisation | 因式分解法求解

Factorisation is often the fastest method when the quadratic has simple integer roots. We look for two numbers that multiply to give ac and add to give b.

当二次方程有简单的整数根时，因式分解通常是最快的方法。我们需要找两个数，它们的乘积为 ac，和为 b。

Solve x² – 5x – 14 = 0. We need two numbers that multiply to -14 and add to -5. These are -7 and 2. So we can write (x – 7)(x + 2) = 0. Therefore, x = 7 or x = -2.

求解 x² – 5x – 14 = 0。我们需要两个数，乘积为 -14，和为 -5。这两个数是 -7 和 2。因此可以写成 (x – 7)(x + 2) = 0。所以，x = 7 或 x = -2。

When the coefficient a is not 1, such as 2x² + 7x + 3 = 0, we multiply a and c to get 6, then find two numbers multiplying to 6 and adding to 7. These are 6 and 1. We rewrite: 2x² + 6x + x + 3 = 0, then factor by grouping: 2x(x + 3) + 1(x + 3) = 0, giving (2x + 1)(x + 3) = 0. Thus x = -1/2 or x = -3.

当首项系数 a 不为 1 时，例如 2x² + 7x + 3 = 0，我们将 a 和 c 相乘得 6，然后找两个数乘积为 6、和为 7。这两个数是 6 和 1。改写：2x² + 6x + x + 3 = 0，然后分组因式分解：2x(x + 3) + 1(x + 3) = 0，得到 (2x + 1)(x + 3) = 0。因此 x = -1/2 或 x = -3。

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

The graph of a quadratic function is a smooth, symmetric curve called a parabola. Every parabola has an axis of symmetry that passes through its vertex.

二次函数的图像是一条平滑的对称曲线，称为抛物线。每条抛物线都有一条通过顶点的对称轴。

The axis of symmetry is given by the equation x = -b / 2a. The x-coordinate of the vertex is also -b/2a. To find the y-coordinate, substitute this x-value back into the original function.

对称轴的方程为 x = -b / 2a。顶点的 x 坐标同样为 -b/2a。要求 y 坐标，只需将 x 值代回原函数即可。

For example, for f(x) = 2x² – 8x + 3, the axis of symmetry is x = -(-8) / (2 × 2) = 8/4 = 2. Substituting x = 2 into the function: f(2) = 2(4) – 8(2) + 3 = 8 – 16 + 3 = -5. So the vertex is at (2, -5).

例如，对 f(x) = 2x² – 8x + 3，对称轴为 x = -(-8) / (2 × 2) = 8/4 = 2。将 x = 2 代入函数：f(2) = 2(4) – 8(2) + 3 = 8 – 16 + 3 = -5。因此顶点坐标为 (2, -5)。

7. Sketching Quadratic Graphs | 绘制二次函数草图

To sketch the graph of a quadratic function accurately, you need to identify four key features: the y-intercept, the roots (if any), the vertex, and the axis of symmetry.

要准确地绘制二次函数的草图，你需要确定四个关键特征：y 轴截距、根（若有）、顶点和对称轴。

The y-intercept is easy to find: for y = ax² + bx + c, the y-intercept is simply c, so the point is (0, c). The roots are found by setting f(x) = 0. The vertex is found using the method described in the previous section.

y 轴截距很容易找到：对于 y = ax² + bx + c，y 轴截距就是 c，即点 (0, c)。根通过令 f(x) = 0 求得。顶点用上一节描述的方法求得。

Sketch the graph of f(x) = x² – 4x + 3. The y-intercept is (0, 3). Factorising: (x – 1)(x – 3) = 0, so the roots are x = 1 and x = 3. The axis of symmetry is x = -(-4)/(2) = 2, and f(2) = 4 – 8 + 3 = -1, so the vertex is (2, -1). Since a = 1 > 0, the parabola opens upward. Now plot these points and draw a smooth curve through them.

绘制 f(x) = x² – 4x + 3 的草图。y 轴截距为 (0, 3)。因式分解：(x – 1)(x – 3) = 0，所以根为 x = 1 和 x = 3。对称轴为 x = -(-4)/(2) = 2，f(2) = 4 – 8 + 3 = -1，所以顶点为 (2, -1)。由于 a = 1 > 0，抛物线开口向上。现在标出这些点并画出一条平滑曲线。

8. Solving Quadratic Inequalities | 解二次不等式

Quadratic inequalities are solved using the roots we find from the corresponding equation. The key is to determine the sign of the quadratic expression in different intervals separated by the roots.

解二次不等式需要先求出对应方程的根，关键在于判断被根分割的各区间内二次表达式的正负号。

Solve x² – x – 6 > 0. First, factorise: (x – 3)(x + 2) > 0. The roots are x = -2 and x = 3. These two roots divide the number line into three intervals: x < -2, -2 < x < 3, and x > 3.

解 x² – x – 6 > 0。首先因式分解：(x – 3)(x + 2) > 0。根为 x = -2 和 x = 3。这两个根将数轴分为三个区间：x < -2、-2 < x < 3 和 x > 3。

Test a value in each interval. For x = -3: (-3 – 3)(-3 + 2) = (-6)(-1) = 6 > 0 ✓. For x = 0: (0 – 3)(0 + 2) = (-3)(2) = -6 < 0 ✗. For x = 4: (4 - 3)(4 + 2) = (1)(6) = 6 > 0 ✓. Therefore, the solution is x < -2 or x > 3.

在每个区间内取一个值测试。当 x = -3：(-3 – 3)(-3 + 2) = (-6)(-1) = 6 > 0 ✓。当 x = 0：(0 – 3)(0 + 2) = (-3)(2) = -6 < 0 ✗。当 x = 4：(4 - 3)(4 + 2) = (1)(6) = 6 > 0 ✓。因此，解集为 x < -2 或 x > 3。

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

Quadratic functions model numerous real-world scenarios, including projectile motion, area optimisation, profit maximisation, and population growth patterns.

二次函数可用来建模许多现实场景，包括抛体运动、面积优化、利润最大化和人口增长模式等。

For example, a ball thrown upward has height h(t) = -5t² + 20t + 1, where t is time in seconds and h is height in metres. To find the maximum height, we find the vertex. The axis of symmetry is t = -20 / (2 × -5) = 2 seconds. Then h(2) = -5(4) + 20(2) + 1 = -20 + 40 + 1 = 21 metres.

例如，向上抛出的球的高度为 h(t) = -5t² + 20t + 1，其中 t 为时间（秒），h 为高度（米）。要求最大高度，我们找顶点。对称轴为 t = -20 / (2 × -5) = 2 秒。则 h(2) = -5(4) + 20(2) + 1 = -20 + 40 + 1 = 21 米。

To find when the ball hits the ground, we set h(t) = 0: -5t² + 20t + 1 = 0. Using the quadratic formula with a = -5, b = 20, c = 1: t = (-20 ± √(400 + 20)) / (-10) = (-20 ± √420) / (-10). Since t must be positive, we take the positive root: t = (20 + √420) / 10 ≈ 4.05 seconds.

要求球何时落地，令 h(t) = 0：-5t² + 20t + 1 = 0。用求根公式，a = -5，b = 20，c = 1：t = (-20 ± √(400 + 20)) / (-10) = (-20 ± √420) / (-10)。由于 t 必须为正，取正根：t = (20 + √420) / 10 ≈ 4.05 秒。

10. The Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0 with roots α and β, we have two useful relationships: the sum of roots α + β = -b/a, and the product of roots αβ = c/a.

对于二次方程 ax² + bx + c = 0，设其两根为 α 和 β，我们有两个有用的关系式：两根之和 α + β = -b/a，两根之积 αβ = c/a。

α + β = -b/a, αβ = c/a

These relationships allow us to find the sum and product of roots without solving the equation. For example, for 2x² + 5x – 3 = 0, the sum of roots is -5/2 and the product is -3/2.

这些关系使我们无需解方程即可求出根的和与积。例如，对于 2x² + 5x – 3 = 0，两根之和为 -5/2，两根之积为 -3/2。

We can also construct a quadratic equation from given roots. If the roots are α and β, the equation is x² – (α + β)x + αβ = 0. For example, if roots are 3 and -4, then the sum is -1 and the product is -12, giving x² + x – 12 = 0, which indeed equals (x – 3)(x + 4).

我们还可以根据已知的根构造二次方程。若根为 α 和 β，则方程为 x² – (α + β)x + αβ = 0。例如，若两根为 3 和 -4，则和为 -1，积为 -12，得到 x² + x – 12 = 0，它确实等于 (x – 3)(x + 4)。

11. Common Pitfalls and Exam Tips | 常见错误与考试技巧

Many students lose marks on quadratic questions due to avoidable mistakes. Here are the most common errors to watch out for.

许多学生在二次函数题目上失分，往往是因为一些可以避免的错误。以下是需要特别留意的最常见错误。

Forgetting that a quadratic equation can have two solutions. Always check whether both values of x satisfy the original equation.
忘记二次方程可能有两个解。务必检查两个 x 值是否都满足原方程。
Errors when completing the square when a ≠ 1. Factor out a first before completing the square inside the bracket.
当 a ≠ 1 时配方出错。应先将 a 提取出来，再对括号内进行配方。
Mixing up the signs in the quadratic formula, especially when b is negative. Write down the formula carefully and substitute step by step.
求根公式中符号搞混，特别是当 b 为负数时。务必仔细写出公式并逐步代入。
Drawing a straight line instead of a smooth curve when sketching parabolas.
绘制抛物线时误画成直线，必须画平滑曲线。
Forgetting that if a > 0, the parabola opens up; if a < 0, it opens down.
忘记当 a > 0 时开口向上，当 a < 0 时开口向下。
When solving inequalities, not considering the correct intervals or reversing the inequality sign incorrectly.
解不等式时，没有考虑正确的区间或错误地颠倒不等号方向。

Remember to always check your solutions by substituting back into the original equation. In the exam, show your working clearly, as method marks are awarded even when the final answer is incorrect.

记住，始终通过代回原方程来检验答案。考试中要展示清晰的解题步骤，即使最终答案有误，也能获得步骤分。

12. Practice Problems | 练习题目

Now let us apply everything we have learned with some practice problems. Try each one before looking at the solution.

现在让我们通过练习来应用所学的一切。先尝试独立求解，再对照答案解析。

Problem 1: Solve x² – 2x – 15 = 0.
题目1：求解 x² – 2x – 15 = 0。

Factorise: (x – 5)(x + 3) = 0, so x = 5 or x = -3.

因式分解：(x – 5)(x + 3) = 0，所以 x = 5 或 x = -3。

Problem 2: Write f(x) = 2x² – 12x + 5 in the form a(x – h)² + k.

题目2：将 f(x) = 2x² – 12x + 5 化为 a(x – h)² + k 的形式。

First factor out 2: f(x) = 2(x² – 6x) + 5. Complete the square inside: x² – 6x = (x – 3)² – 9. Then f(x) = 2[(x – 3)² – 9] + 5 = 2(x – 3)² – 18 + 5 = 2(x – 3)² – 13. So h = 3 and k = -13.

先提取 2：f(x) = 2(x² – 6x) + 5。内部配方：x² – 6x = (x – 3)² – 9。则 f(x) = 2[(x – 3)² – 9] + 5 = 2(x – 3)² – 18 + 5 = 2(x – 3)² – 13。所以 h = 3，k = -13。

Problem 3: For what values of k does the equation x² + kx + 9 = 0 have exactly one real root?

题目3：当 k 取何值时，方程 x² + kx + 9 = 0 恰好有一个实数根？

For exactly one root, the discriminant must be zero: Δ = k² – 4(1)(9) = k² – 36 = 0. Therefore, k = 6 or k = -6.

恰好有一个根时，判别式必须为零：Δ = k² – 4(1)(9) = k² – 36 = 0。因此，k = 6 或 k = -6。

Problem 4: A rectangle has a length given by (2x + 1) cm and a width of (x + 3) cm. If the area is 55 cm², find the value of x.

题目4：一个矩形的长为 (2x + 1) cm，宽为 (x + 3) cm。若面积为 55 cm²，求 x 的值。

Area = (2x + 1)(x + 3) = 55. Expanding: 2x² + 6x + x + 3 = 55, which simplifies to 2x² + 7x – 52 = 0. Factorising: (2x + 13)(x – 4) = 0. Thus x = -13/2 or x = 4. Since x must be positive for lengths, x = 4 cm.

面积 = (2x + 1)(x + 3) = 55。展开：2x² + 6x + x + 3 = 55，化简得 2x² + 7x – 52 = 0。因式分解：(2x + 13)(x – 4) = 0。因此 x = -13/2 或 x = 4。由于长度的 x 必须为正数，所以 x = 4 cm。

Quadratic functions are a fundamental topic that connects algebra, geometry, and real-world applications. Master the techniques of expanding, factorising, completing the square, and using the discriminant, and you will be well prepared for any quadratic question in your IGCSE exam. Regular practice is the key to building confidence and speed.

二次函数是连接代数、几何与现实应用的基础课题。掌握展开、因式分解、配方法和判别式的运用，你将能够在IGCSE考试中从容应对任何二次函数题目。定期练习是增强信心和提升解题速度的关键。

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Mastering Quadratic Functions | 掌握二次函数