📚 Mastering IB Math Exercise B.2: Quadratic Functions | 攻克 IB 数学练习 B.2：二次函数
In many IB Mathematics textbooks, Exercise B.2 is dedicated to quadratic functions - a cornerstone of the Functions topic. This exercise typically challenges students to master algebraic manipulation, graph interpretation and real-world applications, all of which are essential for both Analysis and Approaches (AA) and Applications and Interpretation (AI) courses. Whether you are factoring, completing the square, or analysing parabolas, Exercise B.2 provides a comprehensive workout. In this article we break down the key concepts and common problem types found in such an exercise, equipping you with the strategies you need to succeed.
在许多 IB 数学教材中，练习 B.2 专门围绕二次函数展开——这是函数主题的核心内容。该练习通常涵盖代数运算、图像分析以及实际应用问题，所有这些都是分析数学（AA）和应用数学（AI）课程中必备的技能。无论是因式分解、配方法还是分析抛物线，练习 B.2 都提供了一次全面的训练。本文将拆解此类练习中的关键概念和常见题型，帮助你掌握取得高分的策略。
1. Recognising Quadratic Functions | 识别二次函数
Any function that can be written in the form below, where a, b and c are real constants and a is not zero, is a quadratic function. The highest power of the variable is 2, so the polynomial has degree 2, and the graph of a quadratic function is a parabola.
任何可以写成下面这种形式的函数都是二次函数，其中 a、b、c 为实常数且 a 不等于 0。变量的最高次数是 2，因此这是二次多项式，其图像是一条抛物线。
f(x) = ax² + bx + c, a ≠ 0
If the word quadratic appears in a question, or if you can rearrange an expression into the form ax² + bx + c, you immediately know it is quadratic. In Exercise B.2, you will often need to identify quadratic relationships from tables of values or from word problems rather than from an explicit equation.
如果题目中出现了 quadratic（二次）这个词，或者你能把表达式整理成 ax² + bx + c 的形式，那么立即可以判断它是二次函数。在练习 B.2 中，你常常需要从数值表或应用题中识别二次关系，而不是直接给出方程。
When only a table of values is given, calculate the first differences, then the second differences. If the second differences are constant, the relationship is quadratic and the constant value equals 2a.
当题目只给出数值表时，先算一阶差分，再算二阶差分。如果二阶差分是常数，则该关系是二次关系，且这个常数等于 2a。
| x | 0 | 1 | 2 | 3 | 4 |
| f(x) | 3 | 6 | 11 | 18 | 27 |
| 1st difference | - | 3 | 5 | 7 | 9 |
| 2nd difference | - | - | 2 | 2 | 2 |
c = 3, and from f(1) = 6 we get 1 + b + 3 = 6, so b = 2. The model is f(x) = x² + 2x + 3.
二阶差分恒为 2，所以 2a = 2，a = 1。由 f(0) = 3 得 c = 3，再由 f(1) = 6 得 1 + b + 3 = 6，所以 b = 2。模型为 f(x) = x² + 2x + 3。
2. Three Standard Forms of a Quadratic | 二次函数的三种标准形式
Exercise B.2 questions usually switch between three equivalent forms. Each form reveals different information instantly, and choosing the right one saves valuable exam time.
练习 B.2 的题目通常在三种等价形式之间切换。每种形式都能直接暴露不同的信息，选对形式可以节省大量考试时间。
| Form | Expression | Reveals |
| General | f(x) = ax² + bx + c | y-intercept c, direction of opening |
| Vertex | f(x) = a(x − h)² + k | vertex (h, k), maximum or minimum value |
| Factored | f(x) = a(x − p)(x − q) | x-intercepts p and q, axis of symmetry |
The coefficient a controls both the direction and the width of the parabola: if a > 0 the parabola opens upwards and has a minimum; if a < 0 it opens downwards and has a maximum. The larger the absolute value of a, the narrower the parabola.
系数 a 同时控制抛物线的开口方向和宽窄：a > 0 时开口向上且有最小值；a < 0 时开口向下且有最大值。|a| 越大，抛物线越窄。
3. Factorising and the Zero Product Property | 因式分解与零积性质
Factorising is usually the fastest route to the roots when the numbers are friendly. Look first for a common factor, then for a difference of two squares, then try the standard trinomial pattern.
当数字比较友好时，因式分解通常是求根最快的途径。先提取公因式，再看是否是平方差，最后尝试标准二次三项式模式。
- Common factor: 3x² − 12x = 3x(x − 4). 提公因式：3x² − 12x = 3x(x − 4)。
- Difference of squares: 4x² − 25 = (2x − 5)(2x + 5). 平方差：4x² − 25 = (2x − 5)(2x + 5)。
- Trinomial: x² − 5x + 6 = (x − 2)(x − 3). 二次三项式：x² − 5x + 6 = (x − 2)(x − 3)。
Once factorised, the zero product property applies: if the product of two factors is zero, at least one factor must be zero. Setting each bracket to zero gives the roots.
完成因式分解后即可使用零积性质：若两个因子的乘积为零，则至少有一个因子为零。令每个括号等于零即可求出根。
(x − 2)(x − 3) = 0 → x = 2 or x = 3
Always expand your factors mentally to check the middle term and the constant term before moving on - a single sign error here costs the whole question.
在继续之前，务必在脑中展开检验一次一次项和常数项——这里一个符号错误就会让整道题失分。
4. Completing the Square and the Vertex | 配方法与顶点
Completing the square converts the general form into the vertex form, which immediately gives the coordinates of the turning point. This is one of the most heavily examined skills in Exercise B.2.
配方法把一般式转化为顶点式，从而直接给出转折点的坐标。这是练习 B.2 中考查频率最高的技能之一。
t of x, square it, and adjust the constant: x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8.
对于首项系数为 1 的二次式，取一次项系数的一半并平方，再调整常数项：x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8。
f(x) = x² + 6x + 1 = (x + 3)² − 8
, then multiply back: 2x² − 8x + 3 = 2(x² − 4x) + 3 = 2[(x − 2)² − 4] + 3 = 2(x − 2)² − 5.
当 a 不等于 1 时，先把 a 从前两项中提出，在括号内配方，再乘回去：2x² − 8x + 3 = 2(x² − 4x) + 3 = 2[(x − 2)² − 4] + 3 = 2(x − 2)² − 5。
aximum value of the function is k. If a > 0, k is the minimum; if a < 0, k is the maximum.
顶点为 (h, k)，对称轴为 x = h，函数的最小值或最大值就是 k。若 a > 0，k 是最小值；若 a < 0，k 是最大值。
5. The Quadratic Formula and the Discriminant | 求根公式与判别式
When a quadratic does not factorise with integers, use the quadratic formula. It works for every quadratic equation and is printed in the IB formula booklet.
当二次式无法用整数因式分解时，就使用求根公式。它适用于所有二次方程，并且收录在 IB 公式手册中。
x = (−b ± √b² − 4ac) / (2a)
you how many real roots the equation has and how many times the parabola cuts the x-axis.
根号下的表达式称为判别式，记作 Δ = b² − 4ac。它的符号告诉你方程有多少个实根，以及抛物线与 x 轴有几个交点。
| Discriminant | Number of real roots | Graph |
| Δ > 0 | Two distinct real roots | Cuts the x-axis twice |
| Δ = 0 | One repeated real root | Touches the x-axis at the vertex |
| Δ < 0 | No real roots | Never meets the x-axis |
oots are discriminant questions in disguise: set Δ = 0 and solve for the unknown constant.
凡是问"求使方程有两个相等实根的 k 值"这类题目，本质上都是判别式问题：令 Δ = 0 并解出未知常数即可。
6. Vertex, Axis of Symmetry and Intercepts | 顶点、对称轴与截距
= −b / (2a) in general form, or x = h in vertex form, or x = (p + q) / 2 in factored form.
每条抛物线都有一条经过顶点的竖直对称轴：一般式中为 x = −b / (2a)，顶点式中为 x = h，因式分解式中为 x = (p + q) / 2。
ys gives c in the general form. The x-intercepts are the roots, found by solving f(x) = 0.
令 x = 0 可求 y 轴截距，在一般式中它永远等于 c。x 轴截距即方程的根，通过解 f(x) = 0 得到。
Axis of symmetry: x = −b / (2a), vertex: (−b / (2a), f(−b / (2a)))
y substituting the axis of symmetry back into the function, or by using k = c − b² / (4a).
一个实用的捷径：把对称轴代回函数即可得到顶点的纵坐标，也可以使用 k = c − b² / (4a)。
7. Sketching Graphs and Reading Tables | 绘制草图与读取数值表
IB examiners do not usually want a perfect plot; they want a clear sketch showing the shape and the key features. Label the vertex, the axis of symmetry, both intercepts and the y-intercept.
IB 考官通常并不要求画出完美的图像，而是要求一张清晰、体现形状与关键特征的草图。请标注顶点、对称轴、两个截距以及 y 轴截距。
- State the coordinates of the vertex and whether it is a maximum or minimum. 写出顶点坐标，并说明它是最大值还是最小值。
- Mark the x-intercepts (if any) and the y-intercept on the axes. 在坐标轴上标出 x 轴截距（若有）与 y 轴截距。
- Draw a smooth curve - never join points with straight line segments. 画一条光滑曲线——绝不要用直线段连接各点。
- Show the correct end behaviour: both arms up when a > 0, both arms down when a < 0. 体现正确的两端走势：a > 0 时两端向上，a < 0 时两端向下。
When a question gives a table of values instead of an equation, plot the given points first, then use symmetry about the vertex to complete the missing side of the parabola.
当题目给出数值表而非方程时，先描出已知点，再利用关于顶点的对称性补全抛物线的另一侧。
8. Transformations of Quadratic Graphs | 二次函数图像的变换
Exercise B.2 frequently asks how a graph changes when the equation is modified. Working from vertex form makes every transformation easy to read off directly.
练习 B.2 经常考查方程改变后图像如何变化。使用顶点式可以让每一种变换都一目了然。
| Transformation | Equation | Effect on vertex |
| Vertical translation | y = f(x) + k | (h, k′ + k) |
| Horizontal translation | y = f(x − h) | (h′ + h, k) |
| Vertical stretch | y = a·f(x) | y-coordinate multiplied by a |
| Reflection in x-axis | y = −f(x) | Opens downwards |
h moves 3 units to the right, not to the left, because the bracket equals zero when x = 3.
一个常见陷阱是水平平移：在 y = (x − 3)² 中，图像向右平移 3 个单位而非向左，因为当 x = 3 时括号内为零。
9. Solving Quadratic Inequalities | 解二次不等式
A quadratic inequality asks for a range of values rather than single points. The reliable method is to find the roots, sketch the parabola, then read the solution set off the sketch.
二次不等式要求的是一个取值范围，而不是单个点。最可靠的方法是先求根，再画抛物线草图，然后从草图上读出解集。
Solve x² − x − 6 < 0. Factorising gives (x − 3)(
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