📚 IB Math: Quadratic Functions – Key Concepts & Exam Tips | IB 数学：二次函数考点精讲
Quadratic functions form a fundamental pillar in the IB Mathematics curriculum, appearing across both Analysis & Approaches (AA) and Applications & Interpretation (AI) courses. A deep understanding of their algebraic structure, graphical behaviour, and real‑world applications is essential for success in papers 1, 2, and the Internal Assessment. This article distills the core concepts, common pitfalls, and effective problem‑solving strategies to help you master quadratics.
二次函数是 IB 数学课程的基石，无论是分析与方法（AA）还是应用与解释（AI）都频繁考查。透彻掌握其代数结构、图像特征和实际应用，对应对卷一、卷二和内部评估至关重要。本文将提炼核心知识点、常见陷阱与高效解题策略，助你攻克二次函数难关。
1. Definition and the Three Standard Forms | 定义与三种标准形式
A quadratic function is any function of the form f(x) = ax² + bx + c, where a, b, c are real constants and a ≠ 0. Recognizing the three equivalent forms – standard, vertex, and factored – is the first step to flexible problem solving.
二次函数是形如 f(x) = ax² + bx + c 的函数，其中 a、b、c 为实数常数且 a ≠ 0。识别标准式、顶点式和因式分解式三种等价形式是灵活解题的第一步。
Standard form: f(x) = ax² + bx + c. It immediately gives the y‑intercept (0, c) and the sign of a tells the direction of opening.
标准式：f(x) = ax² + bx + c，能直接看出纵截距 (0, c)，a 的符号决定了开口方向。
Vertex form: f(x) = a(x − h)² + k, with vertex (h, k) and axis of symmetry x = h. You obtain it by completing the square, a skill tested heavily in IB.
顶点式：f(x) = a(x − h)² + k，顶点为 (h, k)，对称轴为 x = h。通过配方法化得，这是 IB 常考的重要技能。
Factored form: f(x) = a(x − p)(x − q), where p and q are the roots (x‑intercepts). In IB, questions may ask you to find roots by factoring or by using the quadratic formula.
因式分解式：f(x) = a(x − p)(x − q)，p 和 q 为函数的根（与 x 轴的交点）。IB 常要求通过因式分解或求根公式求根。
2. Direction of Opening and the Parameter a | 开口方向与系数 a
If a > 0, the parabola opens upward and has a minimum point at its vertex. If a < 0, it opens downward and the vertex is a maximum point. The absolute value |a| controls the width – larger |a| gives a narrower graph, while smaller |a| (|a|<1) makes it wider.
当 a > 0 时，抛物线开口向上，顶点为最小值点；当 a < 0 时，开口向下，顶点为最大值点。|a| 的大小决定图像的宽窄——|a| 越大图像越窄，|a| 越小（|a|<1）图像越宽。
In real‑world contexts, a negative a often models a projectile or a profit function that peaks and then falls. Always relate the sign of a to the nature of the stationary point.
在实际问题中，负的 a 常描述达到峰值后下降的抛射或利润函数。务必将 a 的符号与驻点的性质联系起来。
IB tip: When a question asks to describe the effect of changing a, mention both direction and steepness.
应试技巧：当题目要求描述改变 a 的影响时，要同时提及开口方向和陡峭程度。
3. Axis of Symmetry and Vertex | 对称轴与顶点
For f(x) = ax² + bx + c, the axis of symmetry is the vertical line x = −b/(2a). The vertex’s x‑coordinate is also −b/(2a), and the y‑coordinate is found by plugging this x into the function.
对于 f(x) = ax² + bx + c，对称轴为直线 x = −b/(2a)。顶点的 x 坐标也是 −b/(2a)，再代入函数即得 y 坐标。
Using vertex form, the vertex (h, k) is read directly. Many IB questions ask you to find the coordinates of the vertex and state its nature (maximum or minimum).
利用顶点式可直接读出顶点 (h, k)。IB 频繁要求求顶点坐标并说明其性质（最大值或最小值）。
Example: For y = 2x² − 8x + 5, the axis is x = −(−8)/(2×2) = 2, vertex: (2, 2(2)² − 8×2 + 5) = (2, −3).
例如：y = 2x² − 8x + 5，对称轴 x = −(−8)/(2×2) = 2，顶点为 (2, 2(2)² − 8×2 + 5) = (2, −3)。
4. The Discriminant and Nature of Roots | 判别式与根的性质
The discriminant Δ = b² − 4ac determines the number and type of real roots (solutions to f(x)=0). This is a crucial concept for IB Paper 1 (non‑calculator) and Paper 2.
判别式 Δ = b² − 4ac 决定实根的个数和类型，是卷一（无计算器）和卷二的重要考点。
If Δ > 0, there are two distinct real roots. If Δ = 0, there is one repeated real root (tangent to the x‑axis). If Δ < 0, there are no real roots (the parabola does not cross the x‑axis).
Δ > 0 时，有两个不相等的实根；Δ = 0 时，有一个重根（切于 x 轴）；Δ < 0 时，无实根（抛物线不与 x 轴相交）。
IB often asks: “Show that the quadratic has two distinct real roots” or “Find k such that the graph touches the x‑axis”. In the latter case, set Δ = 0 and solve for k.
IB 常考：“证明该二次方程有两个不等实根”或“求 k 值使图像与 x 轴相切”。对于后者，设 Δ = 0 并解出 k。
Discriminant analysis also feeds into quadratic inequalities and sign diagrams, which appear regularly in both AA and AI.
判别式分析也与二次不等式和符号表直接相关，在 AA 和 AI 试卷中频繁出现。
5. Factoring and Finding Roots | 因式分解与求根
Finding roots is at the heart of quadratic problem solving. The quadratic formula x = [−b ± √(b² − 4ac)]/(2a) is provided in the IB data booklet, but you must be able to apply it accurately.
求根是二次函数解题的核心。求根公式 x = [−b ± √(b² − 4ac)]/(2a) 在 IB 公式手册中给出，但你必须能准确运用。
Simple quadratics can be factorised by inspection, e.g., x² − 5x + 6 = (x − 2)(x − 3) so roots are 2 and 3. Always check factoring by expanding.
简单的二次式可通过观察因式分解，如 x² − 5x + 6 = (x − 2)(x − 3)，根为 2 和 3。分解后务必展开验证。
When factoring with a ≠ 1, the “ac method” or splitting the middle term is preferred. Also, remember to take out a common factor first if possible.
当 a ≠ 1 时，常使用“十字相乘法”或分解中项。若存在公因子，务必先提取公因子。
IB often mixes factoring with the discriminant: a question may ask you to factorise a quadratic and hence state the x‑intercepts, then verify using the formula.
IB 常将因式分解与判别式结合：要求分解二次式并给出 x 轴截距，再用求根公式验证。
6. Sum and Product of Roots (Vieta’s Formulas) | 根与系数的关系（韦达定理）
If the roots of ax² + bx + c = 0 are r₁ and r₂, then r₁ + r₂ = −b/a and r₁ r₂ = c/a. This is extremely useful when the roots are not obvious or when asked to form an equation from given roots.
若 ax² + bx + c = 0 的两根为 r₁ 和 r₂，则 r₁ + r₂ = −b/a，r₁ r₂ = c/a。当根不明显或需要由给定根构造方程时，该关系极为有用。
Example: If the roots are 3+√5 and 3−√5, then the sum is 6, product is (3)² − (√5)² = 4, so the quadratic is x² − 6x + 4 = 0.
例：若根为 3+√5 与 3−√5，则和 6，积 4，故二次方程为 x² − 6x + 4 = 0。
IB also uses Vieta’s formulas in symmetric expressions such as root²₁ + root²₂ = (sum)² − 2×product, which avoids finding the roots explicitly.
IB 还会考查对称表达式如 r₁² + r₂² = (和)² − 2×积，无需具体求出根即可计算。
Tip: On Paper 1, you may be given a sum and product then asked to sketch the sign diagram — knowing the roots helps locate intercepts immediately.
技巧：在卷一，可能给和与积然后要求画符号图——已知根可立即定位截距。
7. Completing the Square | 配方法
Completing the square converts standard form to vertex form. The process is: f(x) = ax² + bx + c = a(x² + (b/a)x) + c = a[(x + b/(2a))² − (b/(2a))²] + c.
配方法将标准式化为顶点式。步骤为：f(x) = ax² + bx + c = a(x² + (b/a)x) + c = a[(x + b/(2a))² − (b/(2a))²] + c。
Simplify to get vertex (h, k) with h = −b/(2a) and k = c − b²/(4a). Mastery of this technique is essential for graphing, solving, and proving that a quadratic is always positive or negative.
化简后得到顶点 (h, k)，其中 h = −b/(2a)，k = c − b²/(4a)。掌握配方法对作图、求解和证明二次式恒正或恒负至关重要。
IB frequently asks: “Express f(x) in the form a(x−h)² + k, and hence find the minimum value.” Always show the algebraic steps.
IB 常考：“将 f(x) 写成 a(x−h)² + k 的形式，并由此求最小值。”务必展示代数步骤。
Common mistake: forgetting to factor out a from the first two terms before completing the square inside the bracket. Check your expansion after each step.
常见错误：括号内配方前忘了提取系数 a。每步完成后都应展开检验。
8. Transformations of Quadratic Graphs | 二次函数图像变换
Knowing how changes in the function equation translate to graph transformations is tested in the topic of functions. For y = a(x − h)² + k:
理解函数方程变化如何导致图像变换，是函数专题的考查重点。对于 y = a(x − h)² + k：
– h > 0 shifts the parabola right by h units; h < 0 shifts left.
– k > 0 shifts upward; k < 0 downward.
– a > 1 gives a vertical stretch; 0<a<1 a vertical compression; a<0 reflects in the x‑axis.
– h > 0 图像向右平移 h 单位，h < 0 向左平移。
– k > 0 向上平移，k < 0 向下平移。
– a > 1 垂直拉伸，0<a<1 垂直压缩，a<0 关于 x 轴反射。
Horizontal scaling is linked to the coefficient of x inside the square: y = (bx − h)² + k. IB expects you to distinguish the effect of a horizontal stretch from a shift.
水平缩放与括号内 x 的系数有关：y = (bx − h)² + k。IB 要求能区分水平拉伸与平移的效果。
Apply transformations in the correct order: horizontal shifts and stretches first, then reflections, then vertical shifts and stretches. Using function notation f(x) = a f(b(x − h)) + k can help.
按正确顺序进行变换：先水平平移与伸缩，再反射，最后垂直平移与伸缩。使用函数符号 f(x) = a f(b(x − h)) + k 会很有帮助。
9. Quadratic Inequalities | 二次不等式
Solving inequalities like 2x² − 5x − 3 ≤ 0 requires finding the roots first, then constructing a sign diagram. IB questions often ask for the solution set in interval notation.
解诸如 2x² − 5x − 3 ≤ 0 的不等式需先求出根，然后制作符号表。IB 常要求用区间记号表示解集。
Steps: (1) Solve 2x² − 5x − 3 = 0 → roots: −1/2 and 3. (2) Sketch the parabola (a=2>0, opening up). (3) The quadratic is ≤ 0 between the roots, so solution: [−1/2, 3].
步骤：(1) 解 2x² − 5x − 3 = 0 → 根为 −1/2 和 3。(2) 画出抛物线草图（a=2>0，开口向上）。(3) 二次式在两根之间 ≤ 0，故解为 [−1/2, 3]。
Watch for strict vs. non‑strict inequalities: ≥ or ≤ include the roots; > or < exclude them. Sign diagrams also work for rational inequalities involving quadratics.
注意严格与不严格不等式的区别：≥ 或 ≤ 包含根，> 或 < 不包含。符号表同样适用于含二次的分式不等式。
When a < 0, the inequality f(x) > 0 is satisfied between the roots. Always test a value from each interval to confirm your sign diagram.
当 a < 0 时，不等式 f(x) > 0 在两根之间成立。务必在每个区间取一个测试值验证符号表。
10. Real‑life Applications and Optimization | 实际应用与最优化问题
IB loves contextual problems where you must formulate a quadratic model and then use vertex or discriminant analysis to find maximum height, maximum profit, minimum area, etc.
IB 偏爱情境题：需要建立二次模型，再用顶点或判别式分析求最大高度、最大利润、最小面积等。
Example: A projectile height h(t) = −5t² + 20t + 1. The maximum height occurs at t = −20/(2×−5) = 2 s, and h(2)= 21 m. The time to hit the ground is the positive root of h(t)=0.
例：抛射高度 h(t) = −5t² + 20t + 1。最高点对应 t = −20/(2×−5) = 2 秒，高度为 21 米。落地时间为 h(t)=0 的正根。
Optimization of revenue: R(x) = price × quantity. If price = 100 − 2x, revenue = (100 − 2x)x = −2x² + 100x, max at x = 25. Always describe the nature of the stationary point using the sign of a or second derivative.
收入最优化：若价格 = 100 − 2x，收入 R = (100 − 2x)x = −2x² + 100x，最大值在 x=25。务必用 a 的符号或二阶导数说明驻点性质。
In IB exam questions, clearly state the domain of x (often x ≥ 0 and integer constraints) and check whether the vertex lies within the feasible interval.
IB 考题中，明确给出 x 的定义域（常含 x ≥ 0 及整数限制）并检验顶点是否在可行区间内。
11. Intersections Between a Quadratic and a Line | 二次函数与直线的交点
To find intersection points, set the quadratic and linear expressions equal: ax² + bx + c = mx + d. Rearrange to ax² + (b−m)x + (c−d) = 0 and use discriminant or solve for x.
求交点时，令二次式和线性式相等：ax² + bx + c = mx + d。整理得 ax² + (b−m)x + (c−d) = 0，再用判别式或求 x 解。
The number of intersections depends on the discriminant of the resulting quadratic:
交点个数由所得二次方程的判别式决定：
– Δ > 0 → two intersection points (secant line).
– Δ = 0 → one intersection point (tangent line).
– Δ < 0 → no intersection.
– Δ > 0 → 两个交点（割线）。
– Δ = 0 → 一个交点（切线）。
– Δ < 0 → 无交点。
IB often asks: “Find the equation of the tangent to the curve at x = 2” or “Determine k such that the line is tangent to the parabola.” Set Δ = 0 and solve.
IB 常问：“求曲线在 x=2 处的切线方程”或“确定 k 使直线与抛物线相切”。设 Δ = 0 求解。
The y‑coordinate of the tangent point is found by plugging the x back into either original equation. Check consistency.
切点的 y 坐标通过将 x 代回原任一方程得到，确保数值一致。
12. Common Mistakes and Exam Strategies | 常见错误与考试策略
Mistake 1: Forgetting to set the quadratic to zero before using the quadratic formula. Always rearrange to Standard Form.
错误1：使用求根公式前忘记令二次方程等于零。务必先整理成标准形式。
Mistake 2: Losing a negative sign when computing −b in the formula. Write the formula and substitute carefully.
错误2：求根公式中 −b 时丢掉负号。写出公式，谨慎代入。
Mistake 3: Misinterpreting the vertex when a is negative – still (h, k) read from vertex form, but now gives a maximum.
错误3：a 为负时错误理解顶点——仍从顶点式读出 (h, k)，但它表示最大值。
Mistake 4: In transformations, applying shifts before horizontal stretches or vice versa. Use bracket notation to clarify order.
错误4：图像变换时平移和伸缩顺序颠倒。使用括号标记明确顺序。
Exam strategy: In Paper 2, use your GDC to sketch the graph, find roots and vertex. But show algebraic validation for method marks. In Paper 1, do all steps by hand; checking discriminant can often confirm the nature of a solution.
考试策略：卷二可利用图形计算器画图、求根和顶点，但仍需展示代数过程以获得方法分。卷一必须手解所有步骤；用判别式检验常能确认解的性质。
Always answer in the required form – interval notation, set notation, or inequality as specified. Clearly label your final answers.
始终按题目要求的形式作答——区间记号、集合记号或不等式。清晰标出最终答案。
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