📚 IB CCEA Mathematics: Quadratic Functions Key Concepts | IB CCEA 数学：二次函数 考点精讲
Quadratic functions form a cornerstone of the IB CCEA Mathematics curriculum, appearing in everything from algebraic manipulation to real-world modelling. A deep understanding of their forms, graphs, and solution methods is essential for success in both paper-based assessments and applied problem-solving tasks. This revision guide breaks down every critical point, providing a clear, bilingual walkthrough.
二次函数是 IB CCEA 数学课程的核心内容，贯穿代数运算和实际建模。深入理解其不同形式、图像特征及求解方法，是应对笔试和应用题的关键。本考点精讲将逐一拆解每个重要知识点，提供清晰的双语讲解。
1. What is a Quadratic Function? | 什么是二次函数？
A quadratic function is a polynomial function of degree two, characterised by its highest exponent being 2. It can be expressed in the standard form y = ax² + bx + c, where a, b, and c are real constants and a ≠ 0. If a were zero, the function would degenerate into a linear function. The graph of a quadratic function is always a smooth curve called a parabola.
二次函数是一个最高次项为 2 的多项式函数，其标准形式为 y = ax² + bx + c，其中 a、b、c 为实数常数且 a ≠ 0。若 a = 0，函数将退化为一次函数。二次函数的图像总是一条光滑的曲线，称为抛物线。
The term ‘quadratic’ comes from the Latin word ‘quadratus’, meaning square. This reflects the fact that the variable is squared. In IB CCEA exams, you will often be required to identify the coefficient a, b, and c correctly from a given expression, even when the terms are not written in order.
“二次” 一词源于拉丁语 “quadratus”，意为平方，反映了变量平方的特征。在 IB CCEA 考试中，常要求你从给定的表达式中正确识别出系数 a、b、c，即使各项的顺序被打乱。
2. Standard Form and Coefficient Identification | 标准形式与系数识别
The standard form is the foundation for all subsequent analysis. Given y = ax² + bx + c, the coefficient a determines the width and direction of the parabola, b influences the position of the vertex horizontally, and c gives the y-intercept. For example, in y = 3x² – 5x + 2, we have a = 3, b = -5, c = 2.
标准形式是所有后续分析的基础。给定 y = ax² + bx + c，系数 a 决定了抛物线的开口方向和宽度，b 影响顶点在水平方向的位置，c 则是图像与 y 轴的截距。例如，在 y = 3x² – 5x + 2 中，a = 3，b = -5，c = 2。
Always pay attention to the sign of each coefficient. A missing linear term (e.g., y = 2x² + 4) implies b = 0, and a missing constant term (e.g., y = -x² + 6x) means c = 0. Misidentifying these values is a common source of error when substituting into the quadratic formula or completing the square.
务必注意各系数的符号。若缺失一次项（如 y = 2x² + 4），则意味着 b = 0；若缺失常数项（如 y = -x² + 6x），则 c = 0。在代入求根公式或进行配方时，错误识别这些值是常见的错误来源。
3. Graphing the Parabola: Vertex and Axis of Symmetry | 抛物线图像：顶点与对称轴
The graph of a quadratic function is a parabola that opens upward if a > 0 and downward if a < 0. The turning point of the parabola is called the vertex, and it represents either the minimum or maximum value of the function. The vertical line passing through the vertex is the axis of symmetry, which divides the parabola into two mirror-image halves.
二次函数的图像是一条抛物线，当 a > 0 时开口向上，a < 0 时开口向下。抛物线的转折点称为顶点，代表了函数的最小值或最大值。经过顶点的竖直线是对称轴，它将抛物线分成左右对称的两部分。
The x-coordinate of the vertex can be found using the formula h = -b/(2a). Once you have h, the y-coordinate k is obtained by substituting h back into the function: k = f(h). The axis of symmetry is then the line x = h. For example, for y = 2x² – 8x + 3, h = -(-8)/(2×2) = 2, so the axis is x = 2 and the vertex is (2, -5).
顶点的 x 坐标可用公式 h = -b/(2a) 求得。得出 h 后，再将 h 代入函数即可得到 y 坐标 k = f(h)。对称轴就是直线 x = h。例如，对于 y = 2x² – 8x + 3，h = -(-8)/(2×2) = 2，因此对称轴为 x = 2，顶点为 (2, -5)。
4. Finding the Vertex: Completing the Square | 求顶点：配方法
Completing the square is a technique that rewrites the standard form into vertex form: y = a(x – h)² + k. This form is extremely useful because the vertex (h, k) can be read directly from the equation without any further calculation. It also clearly shows the transformations applied to the parent function y = x².
配方法是一种将标准形式转化为顶点式 y = a(x – h)² + k 的技巧。这种形式非常有用，因为无需额外计算即可直接从方程中读出顶点 (h, k)。它也清晰地展示了对母函数 y = x² 所做的变换。
To complete the square for y = x² + 6x + 5, take half the coefficient of x (which is 3), square it (9), and add and subtract it: y = (x² + 6x + 9) – 9 + 5 = (x + 3)² – 4. Thus the vertex is (-3, -4). When a ≠ 1, factor out a from the x-terms first, then complete the square inside the bracket, and remember to balance the constant term accordingly.
对 y = x² + 6x + 5 进行配方：取 x 项系数的一半 (3)，将其平方 (9)，然后在式中加减这个数：y = (x² + 6x + 9) – 9 + 5 = (x + 3)² – 4。因此顶点为 (-3, -4)。当 a ≠ 1 时，需先将 a 从含 x 的项中提出，再在括号内配方，并记得相应调整常数项。
5. The Quadratic Formula and Discriminant | 求根公式与判别式
The quadratic formula provides the roots of any quadratic equation ax² + bx + c = 0. It is derived by completing the square on the general form and is given by:
求根公式可求出任何二次方程 ax² + bx + c = 0 的根。它是通过对一般式进行配方推导而得：
x = [-b ± √(b² – 4ac)] / (2a)
The expression under the square root, b² – 4ac, is called the discriminant, often denoted by the Greek letter delta (Δ). The discriminant determines how many real solutions a quadratic equation has, without actually solving it.
根号内的表达式 b² – 4ac 称为判别式，常用希腊字母 Δ 表示。判别式无需真正求解方程，就能判断二次方程有多少个实数解。
Always ensure the equation is set to zero before identifying a, b, and c for the formula. For example, to solve 3x² = 2x + 1, first rewrite as 3x² – 2x – 1 = 0, giving a = 3, b = -2, c = -1.
务必在识别 a、b、c 之前将方程设为零。例如，求解 3x² = 2x + 1，应先改写为 3x² – 2x – 1 = 0，此时 a = 3, b = -2, c = -1。
6. Factoring Quadratic Expressions | 因式分解二次表达式
Factoring is a fundamental skill that breaks a quadratic expression down into the product of two linear expressions. For a simple monic case where a = 1, we look for two numbers that multiply to c and add up to b. For example, x² – 5x + 6 factors as (x – 2)(x – 3) because (-2) × (-3) = 6 and (-2) + (-3) = -5.
因式分解是一项基础技能，它将二次表达式分解为两个一次式的乘积。对于首项系数 a = 1 的简单情况，我们需要找到两个数，其乘积为 c，且其和为 b。例如，x² – 5x + 6 可分解为 (x – 2)(x – 3)，因为 (-2) × (-3) = 6 而 (-2) + (-3) = -5。
When a ≠ 1, the process involves more trial and error. One method is to find two numbers that multiply to a × c and add to b, then split the middle term and factor by grouping. For 2x² + 7x + 3, a×c = 6, and numbers 6 and 1 give sum 7; rewrite as 2x² + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3).
当 a ≠ 1 时，分解过程需要更多的尝试。一种方法是找到两个数，其积为 a × c，其和为 b，然后拆分中项并分组分解。以 2x² + 7x + 3 为例，a×c = 6，6 和 1 的和为 7；于是改写为 2x² + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3)。
7. Solving Quadratic Equations | 解二次方程
Once a quadratic equation is set to zero, you have three primary solution methods: factoring and applying the zero product property, using the quadratic formula, or completing the square. The zero product property states that if A × B = 0, then either A = 0 or B = 0. This is why factoring is so effective when it works neatly.
一旦将二次方程设为零，你就有三种主要的求解方法：因式分解并运用零积性质、使用求根公式，或配方法。零积性质指出，若 A × B = 0，则 A = 0 或 B = 0。这就是因式分解在能顺利分解时如此有效的原因。
If factoring seems difficult or impossible, the quadratic formula is a reliable fallback. It always works and is especially handy when the coefficients are large or irrational. Always check whether your solutions satisfy the original equation, and present exact answers (in surd form) unless otherwise instructed.
若因式分解看似困难或不可能，求根公式是一个可靠的备用方法。它总能奏效，且当系数较大或为无理数时尤其方便。务必检验所得解是否满足原方程，除特别说明外，答案应保留确切值（根式形式）。
8. The Discriminant and Nature of Roots | 判别式与根的性质
The discriminant Δ = b² – 4ac reveals the nature of the roots without solving the equation. The table below summarises the possibilities:
判别式 Δ = b² – 4ac 无需解方程即可揭示根的性质。下表总结了各种可能：
| Discriminant Δ | Nature of Roots / 根的性质 |
|---|---|
| Δ > 0 | Two distinct real roots 两个相异实根 |
| Δ = 0 | Exactly one real root (repeated) 一个实根（重根） |
| Δ < 0 | No real roots; two complex conjugate roots 无实根；两个共轭复根 |
When analysing quadratic graphs, these cases correspond to the parabola intersecting the x-axis at two points, touching it at exactly one point (the vertex lies on the x-axis), or floating entirely above or below the x-axis. This geometric interpretation is a favourite topic in IB CCEA exam questions.
在分析二次函数图像时，这些情况分别对应于抛物线与 x 轴交于两点、恰好交于一点（顶点在 x 轴上），或完全位于 x 轴上方或下方。这种几何解释是 IB CCEA 考试中常见的考点。
9. Quadratic Inequalities | 二次不等式
Solving a quadratic inequality such as x² – 4x – 5 > 0 involves more than just finding roots. The key is to sketch a quick graph or use a sign chart. First find the roots by solving the corresponding equation, then determine on which intervals the parabola lies above or below the x-axis.
解诸如 x² – 4x – 5 > 0 的二次不等式，不仅仅是求根。关键在于快速画出示意图或使用符号表。首先解相应的方程求出根，然后判断抛物线在 x 轴上方或下方的区间。
For a > 0, the parabola opens upward, so the expression is positive outside the roots and negative between them. Thus for (x – 5)(x + 1) > 0, the solution is x < -1 or x > 5. Remember that multiplying or dividing an inequality by a negative number reverses the direction, which is crucial when the coefficient a is negative.
当 a > 0 时，抛物线开口向上，因此表达式在两根之外为正，在两根之间为负。所以对于 (x – 5)(x + 1) > 0，解为 x < -1 或 x > 5。记住，当乘以或除以一个负数时，不等号方向要反转，这在系数 a 为负时至关重要。
10. Applications and Word Problems | 应用与文字题
Quadratic functions model a wide range of real-world quantities, such as the path of a projectile, area optimisation, and revenue maximisation. The key to solving these problems is to define a variable, write a quadratic expression for the quantity to be optimised, and then use the vertex to find the maximum or minimum value.
二次函数可以模拟许多现实世界中的量，如抛物体的轨迹、面积优化以及收入最大化。解决这类问题的关键是定义变量，为待优化的量写出二次表达式，然后利用顶点求出最大值或最小值。
For example, if a farmer wants to enclose a rectangular area along a river using 100 m of fencing on three sides, letting the width be x, the area A = x(100 – 2x) = -2x² + 100x. The maximum area occurs at the vertex, x = -b/(2a) = 100/4 = 25, giving an area of 1250 m². Always check that your solution makes practical sense within the context.
例如，一个农民想用 100 米长的篱笆沿河围一个三面矩形区域，设宽为 x，则面积 A = x(100 – 2x) = -2x² + 100x。最大面积出现在顶点处，x = -b/(2a) = 100/4 = 25，此时面积为 1250 平方米。务必检查你的解在实际情境中是否合理。
11. Transformations of Quadratic Functions | 二次函数的变换
Starting from the parent function y = x², every quadratic function can be viewed as a series of transformations. The vertex form y = a(x – h)² +
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