Quadratic Functions and Graphs | 二次函数与图像

📚 Quadratic Functions and Graphs | 二次函数与图像

Quadratic functions form a core part of the IGCSE Mathematics syllabus. Understanding their algebraic forms, graphical features, and real-world applications is essential for achieving high marks in both Paper 2 and Paper 4. This article provides a comprehensive, exam-focused revision guide.

二次函数是 IGCSE 数学课程的核心内容。掌握其代数形式、图像特征以及实际应用，是在 Paper 2 和 Paper 4 中取得高分的关键。本文提供一份全面、紧扣考点的复习指南。

1. Standard Form and General Form | 标准形式与一般形式

A quadratic function is a polynomial of degree 2. The general form is written as y = ax² + bx + c, where a, b, and c are constants, and a ≠ 0. The coefficient a determines the shape and orientation of the parabola.

二次函数是次数为 2 的多项式函数，其一般形式写作 y = ax² + bx + c，其中 a、b、c 为常数，且 a ≠ 0。系数 a 决定了抛物线的形状与开口方向。

The standard form is also called the vertex form: y = a(x – h)² + k. Here, (h, k) is the vertex of the parabola. Converting between these two forms is a key skill tested in IGCSE.

标准形式也称为顶点式：y = a(x – h)² + k，其中 (h, k) 为抛物线的顶点。在这两种形式之间进行转换，是 IGCSE 的重要考点。

2. Shape and Orientation of the Parabola | 抛物线的形状与开口方向

If a > 0, the parabola opens upward and has a minimum point. If a < 0, the parabola opens downward and has a maximum point. The magnitude of a affects the width: larger |a| gives a narrower parabola, smaller |a| gives a wider one.

当 a > 0 时，抛物线开口向上，存在最小值点；当 a < 0 时，抛物线开口向下，存在最大值点。|a| 的大小影响抛物线的宽窄：|a| 越大，抛物线越窄；|a| 越小，抛物线越宽。

系数 a	开口方向	极值
a > 0	向上（U 形）	最小值
a < 0	向下（∩ 形）	最大值

3. Vertex and Axis of Symmetry | 顶点与对称轴

The vertex is the turning point of the parabola. Its x-coordinate is found by the formula x = -b / (2a). Substitute this x-value into the equation to find the y-coordinate. The axis of symmetry is the vertical line x = -b / (2a).

顶点是抛物线的转折点，其 x 坐标由公式 x = -b / (2a) 求得。将 x 值代入原方程即可得到 y 坐标。对称轴是竖直线 x = -b / (2a)。

For example, for y = 2x² – 8x + 5, we have a = 2, b = -8. Therefore, x = -(-8) / (2 × 2) = 8/4 = 2. Then y = 2(2)² – 8(2) + 5 = 8 – 16 + 5 = -3. So the vertex is (2, -3), and the axis of symmetry is x = 2.

例如，对于 y = 2x² – 8x + 5，有 a = 2，b = -8。因此 x = -(-8) / (2 × 2) = 8/4 = 2。代入得 y = 2(2)² – 8(2) + 5 = 8 – 16 + 5 = -3。所以顶点为 (2, -3)，对称轴为 x = 2。

4. Finding Roots by Factorisation | 因式分解求根

To solve a quadratic equation ax² + bx + c = 0, factorisation is often the quickest method. This requires expressing the quadratic as a product of two linear factors.

解二次方程 ax² + bx + c = 0 时，因式分解通常是最快捷的方法。这需要将二次式表示为两个一次因式的乘积。

Example: Solve x² – 5x + 6 = 0. We look for two numbers that multiply to 6 and add to -5. These are -2 and -3. Hence (x – 2)(x – 3) = 0, so x = 2 or x = 3.

例：解 x² – 5x + 6 = 0。寻找两个数，其乘积为 6，和为 -5。它们是 -2 和 -3。因此 (x – 2)(x – 3) = 0，所以 x = 2 或 x = 3。

For quadratics with a ≠ 1, use the method of splitting the middle term or the box method. For example, 2x² + 5x + 2 = 0. Multiply a and c: 2 × 2 = 4. Find two numbers whose product is 4 and sum is 5: 1 and 4. Rewrite: 2x² + x + 4x + 2 = 0, then factor by grouping: x(2x + 1) + 2(2x + 1) = 0, so (x + 2)(2x + 1) = 0. Thus x = -2 or x = -1/2.

当 a ≠ 1 时，可使用十字相乘法或分组分解法。例如，2x² + 5x + 2 = 0。a × c = 2 × 2 = 4。找两个数乘积为 4 且和为 5：1 和 4。改写为 2x² + x + 4x + 2 = 0，分组：x(2x + 1) + 2(2x + 1) = 0，因此 (x + 2)(2x + 1) = 0，所以 x = -2 或 x = -1/2。

5. Finding Roots by the Quadratic Formula | 用求根公式求根

When factorisation is difficult or impossible, use the quadratic formula. For ax² + bx + c = 0, the roots are given by

当因式分解困难或无法进行时，使用求根公式。对于 ax² + bx + c = 0，其根由下式给出：

x = (-b ± √(b² – 4ac)) / (2a)

This formula works for all quadratic equations, including those with irrational or complex roots. In IGCSE, you are expected to use it confidently and substitute the values of a, b, and c correctly.

该公式适用于所有二次方程，包括有無理根或复数根的情形。在 IGCSE 中，你需要熟练运用公式，并正确代入 a、b、c 的值。

Example: Solve x² + 3x – 7 = 0. Here a = 1, b = 3, c = -7. Substituting gives

例：解 x² + 3x – 7 = 0。这里 a = 1，b = 3，c = -7。代入得：

x = (-3 ± √(3² – 4 × 1 × (-7))) / (2 × 1) = (-3 ± √37) / 2

So the two exact roots are (-3 + √37)/2 and (-3 – √37)/2.

因此两个精确根为 (-3 + √37)/2 和 (-3 – √37)/2。

6. Discriminant and Nature of Roots | 判别式与根的性质

The expression Δ = b² – 4ac is called the discriminant. It determines the nature of the roots without solving the equation.

式子 Δ = b² – 4ac 称为判别式。它无须解方程即可判断根的性质。

If Δ > 0: two distinct real roots.
若 Δ > 0： 有两个不相等的实数根。
If Δ = 0: two equal real roots (one repeated root).
若 Δ = 0： 有两个相等的实数根（一个重根）。
If Δ < 0: no real roots, two complex conjugate roots.
若 Δ < 0： 没有实数根，有两个共轭复数根。

The discriminant also tells us about the graph: if Δ > 0, the parabola crosses the x-axis twice; if Δ = 0, it touches the x-axis once; if Δ < 0, it never touches the x-axis.

判别式还能反映图像特征：若 Δ > 0，抛物线与 x 轴有两个交点；若 Δ = 0，抛物线与 x 轴相切；若 Δ < 0，抛物线不与 x 轴相交。

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

Completing the square is a powerful technique to convert ax² + bx + c into vertex form. When a = 1, the process is straightforward: rewrite x² + bx as (x + b/2)² – (b/2)².

配方法是将 ax² + bx + c 转换为顶点式的有力工具。当 a = 1 时，过程简单：将 x² + bx 改写为 (x + b/2)² – (b/2)²。

Example: Write y = x² + 6x + 5 in vertex form. First, y = (x + 3)² – 9 + 5 = (x + 3)² – 4. Hence the vertex is (-3, -4).

例：将 y = x² + 6x + 5 写成顶点式。首先 y = (x + 3)² – 9 + 5 = (x + 3)² – 4。因此顶点为 (-3, -4)。

When a ≠ 1, first factor out a from the x² and x terms. For example, y = 2x² – 8x + 3 becomes y = 2(x² – 4x) + 3 = 2[(x – 2)² – 4] + 3 = 2(x – 2)² – 8 + 3 = 2(x – 2)² – 5. The vertex is (2, -5).

当 a ≠ 1 时，先将 a 从 x² 和 x 项中提取出来。例如，y = 2x² – 8x + 3 变形为 y = 2(x² – 4x) + 3 = 2[(x – 2)² – 4] + 3 = 2(x – 2)² – 8 + 3 = 2(x – 2)² – 5。顶点为 (2, -5)。

8. Graph Transformations | 图像变换

Quadratic graphs can be transformed by translating, reflecting, or stretching. The vertex form y = a(x – h)² + k makes these transformations easy to identify.

二次函数图像可以通过平移、反射和伸缩进行变换。顶点式 y = a(x – h)² + k 使这些变换易于辨认。

Horizontal shift: y = (x – h)² shifts h units right (if h > 0) or left (if h < 0).
水平平移： y = (x – h)² 表示向右平移 h 个单位（h > 0）或向左平移 |h| 个单位（h < 0）。
Vertical shift: y = x² + k shifts k units up (if k > 0) or down (if k < 0).
垂直平移： y = x² + k 表示向上平移 k 个单位（k > 0）或向下平移 |k| 个单位（k < 0）。
Reflection: y = -x² reflects the graph across the x-axis.
反射： y = -x² 将图像沿 x 轴翻转。
Vertical stretch: y = ax² with |a| > 1 stretches vertically; |a| < 1 compresses vertically.
垂直伸缩： y = ax²，当 |a| > 1 时纵向拉伸；当 |a| < 1 时纵向压缩。

9. Applications: Maximum and Minimum Values | 应用：最大值与最小值

Many real-world problems involve finding the maximum or minimum value of a quadratic function. This is done by locating the vertex. In physics, the height of a projectile follows a quadratic path; in business, profit functions can be quadratic.

许多实际问题涉及求二次函数的最大值或最小值，这可通过找到顶点来实现。在物理学中，抛体的高度遵循二次路径；在商业中，利润函数也可以是二次的。

Example: A ball is thrown upwards. Its height h in metres after t seconds is given by h(t) = -5t² + 20t + 1. Find the maximum height. The vertex occurs at t = -b / (2a) = -20 / (2 × (-5)) = 2 seconds. Then h(2) = -5(2)² + 20(2) + 1 = -20 + 40 + 1 = 21 metres.

例：小球向上抛出，其高度 h（米）与时间 t（秒）的关系为 h(t) = -5t² + 20t + 1。求最大高度。顶点出现在 t = -b / (2a) = -20 / (2 × (-5)) = 2 秒。代入得 h(2) = -5(2)² + 20(2) + 1 = -20 + 40 + 1 = 21 米。

Recall that if a < 0, the vertex gives a maximum; if a > 0, it gives a minimum. This qualitative check should always be performed.

注意，当 a < 0 时，顶点对应最大值；当 a > 0 时，顶点对应最小值。做这类题时一定要先判断开口方向。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Students often lose marks by forgetting to set the equation to zero before factorising, misusing the quadratic formula sign, or confusing roots with graph intersections. Below are key tips.

学生常因以下原因丢分：因式分解前忘记将方程化为零，公式求根时符号错误，或混淆根与图像交点的概念。以下是几个关键技巧。

Always write the equation in the form ax² + bx + c = 0 before solving.
解题前务必先将方程写为 ax² + bx + c = 0 的形式。
Check the discriminant first to know how many roots to expect.
先通过判别式判断根的数量，确保所求根的数量正确。
When sketching a curve, label the vertex, axis of symmetry, and roots clearly.
画草图时，务必标出顶点、对称轴与根。
In word problems, interpret the vertex correctly: time for maximum height vs. height value.
在应用题中，正确理解顶点含义：区分达到最大值的时间与最大值本身。
Use a calculator only for arithmetic; do not rely on it for algebraic manipulation.
计算器仅用于数值计算，不要依赖它进行代数变形。

By mastering these ten aspects of quadratic functions, you will be well-prepared for IGCSE exam questions involving equation solving, curve sketching, and optimisation problems. Practice converting forms, understanding the discriminant, and interpreting graphs to build confidence.

通过掌握二次函数的这十个方面，你将能够从容应对 IGCSE 中涉及解方程、画图像和最优化问题的题目。多加练习不同形式之间的转换、理解判别式含义并结合图像进行分析，定能提升解题信心。

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Quadratic Functions and Graphs | 二次函数与图像