Quadratic Functions | 二次函数

📚 Quadratic Functions | 二次函数

A quadratic function is one of the most fundamental topics in A-Level CIE Mathematics. It describes a polynomial of degree 2, typically expressed as f(x) = ax² + bx + c where a ≠ 0. Understanding quadratic functions enables you to model parabolic paths, solve optimisation problems, and analyse the nature of roots using the discriminant. This article provides a comprehensive revision of all key concepts, formulas, and techniques required for the Cambridge International Examinations syllabus.

二次函数是A-Level CIE数学中最基础的课题之一。它描述了一个最高次为2的多项式，通常表示为f(x) = ax² + bx + c，其中a ≠ 0。理解二次函数能帮助你建立抛物线轨迹模型、解决最优化问题，并利用判别式分析根的性质。本文全面梳理了剑桥国际考试大纲要求的所有关键概念、公式与解题技巧。

1. Standard Form and Terminology | 标准式与术语

A quadratic function is usually written in standard form: f(x) = ax² + bx + c, with a, b, c ∈ ℝ and a ≠ 0. The coefficient ‘a’ determines the direction and ‘width’ of the parabola; ‘b’ affects the position of the axis of symmetry; and ‘c’ gives the y-intercept. The term ax² is the quadratic term, bx is the linear term, and c is the constant term.

二次函数通常写成标准形式：f(x) = ax² + bx + c，其中a、b、c为实数且a ≠ 0。系数a决定了抛物线的开口方向和“宽度”；b影响对称轴的位置；c给出y轴截距。ax²项为二次项，bx为一次项，c为常数项。

If a > 0, the parabola opens upwards (∪ shape).
若a > 0，抛物线开口向上（∪形）。
If a < 0, the parabola opens downwards (∩ shape).
若a < 0，抛物线开口向下（∩形）。
The larger |a|, the ‘steeper’ or narrower the graph.
|a|越大，图像越“陡”或越窄。

2. The Graph of a Quadratic: The Parabola | 二次函数的图像：抛物线

The graph of any quadratic function is called a parabola. It is symmetric about a vertical line called the axis of symmetry. The highest or lowest point is the vertex (turning point). The parabola can intersect the x-axis at zero, one, or two points, which are the real roots of the equation f(x) = 0.

任何二次函数的图像都叫作抛物线。它关于一条称为对称轴的竖直线对称。最高点或最低点是顶点（转折点）。抛物线与x轴可以有零个、一个或两个交点，这些就是方程f(x) = 0的实根。

Understanding the shape helps in sketching graphs accurately. For a > 0, the vertex is the minimum point; for a < 0, it is the maximum point. The y-coordinate of the vertex gives the minimum or maximum value of the function.

理解图像形状有助于准确作图。当a > 0时，顶点是最小值点；当a < 0时，顶点是最大值点。顶点的y坐标给出了函数的最小值或最大值。

3. Factorised Form and Roots | 因式分解形式与根

A quadratic can often be written as f(x) = a(x − p)(x − q), where p and q are the x-intercepts (roots). If the quadratic crosses the x-axis at x = p and x = q, the factorised form is immediate. Finding roots by factorising involves expressing the quadratic as a product of two linear factors.

二次函数常可写成f(x) = a(x − p)(x − q)的形式，其中p和q是x轴截距（根）。如果抛物线与x轴在x = p和x = q处相交，则因式分解形式即可得到。通过因式分解求根需要将二次式表示为两个一次因式的乘积。

Example: x² − 5x + 6 = (x − 2)(x − 3), so roots are x = 2, 3. The sign of a does not affect the roots themselves but flips the parabola.

例如：x² − 5x + 6 = (x − 2)(x − 3)，因此根为x = 2, 3。a的符号不影响根本身，但会翻转抛物线。

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

The discriminant Δ (delta) is given by Δ = b² − 4ac. It determines the number and type of roots of the quadratic equation ax² + bx + c = 0. This is a core concept tested frequently in CIE exams, especially for questions on intersection of curves and line.

判别式Δ由Δ = b² − 4ac给出。它决定了二次方程ax² + bx + c = 0的根的数量和类型。这是CIE考试中经常考查的核心概念，尤其在曲线与直线交点的问题中。

Δ > 0: two distinct real roots (graph cuts x-axis at two points).
Δ > 0：两个不等实根（图像与x轴相交于两点）。
Δ = 0: one repeated real root (graph touches x-axis; tangent).
Δ = 0：一个重实根（图像与x轴相切）。
Δ < 0: no real roots, two complex conjugate roots (graph does not meet x-axis).
Δ < 0：无实根，有两个共轭复根（图像不与x轴相交）。

Also, with rational coefficients, if Δ is a perfect square, the roots are rational; if not, they are irrational.

另外，当系数为有理数时，若Δ为完全平方数，则根为有理数；否则为无理数。

5. Quadratic Formula | 求根公式

When factorisation is difficult, the quadratic formula is used to find the roots: x = (−b ± √(b² − 4ac)) / (2a). The expression √(b² − 4ac) is the square root of the discriminant. The formula works for all quadratic equations, including those with irrational or complex roots.

当因式分解困难时，使用求根公式求根：x = (−b ± √(b² − 4ac)) / (2a)。表达式√(b² − 4ac)是判别式的平方根。该公式适用于所有二次方程，包括具有无理根或复根的方程。

It is essential to use the formula carefully, especially the ‘−b’ term. Remember to divide the entire numerator by 2a. In CIE papers, exact values in surd form are often required.

必须仔细使用该公式，特别是“−b”这一项。要记住将整个分子除以2a。在CIE试卷中，通常要求以根式形式给出精确值。

x = (−b ± √(b² − 4ac)) / (2a)

6. Completing the Square and Vertex Form | 配方法与顶点式

Completing the square transforms the standard form into vertex form: f(x) = a(x − h)² + k, where (h, k) is the vertex. The method involves taking half the coefficient of x, squaring it, and adjusting the constant term. This form immediately reveals the vertex, the axis of symmetry x = h, and the maximum/minimum value k.

配方法将标准形式转化为顶点式：f(x) = a(x − h)² + k，其中(h, k)为顶点。该方法包括取x系数的一半、平方，再调整常数项。这种形式立即揭示出顶点、对称轴x = h以及最大值/最小值k。

For f(x) = ax² + bx + c:
Firstly, factor a from the x² and x terms: f(x) = a[x² + (b/a)x] + c.
Then add and subtract (b/(2a))² inside the bracket: f(x) = a[(x + b/(2a))² − (b/(2a))²] + c.
Finally, expand to get f(x) = a(x + b/(2a))² + (c − b²/(4a)). So h = −b/(2a) and k = c − b²/(4a).

对于 f(x) = ax² + bx + c：
首先，从x²和x项中提取a：f(x) = a[x² + (b/a)x] + c。
然后在括号内加上并减去(b/(2a))²：f(x) = a[(x + b/(2a))² − (b/(2a))²] + c。
最后展开得到 f(x) = a(x + b/(2a))² + (c − b²/(4a))。因此 h = −b/(2a)，k = c − b²/(4a)。

7. Axis of Symmetry and Vertex | 对称轴与顶点

The axis of symmetry is the vertical line x = h, where h = −b/(2a). The vertex is the point (h, k). These can be found directly from the standard form using the formulas or by completing the square. In many word problems, the vertex provides the optimal value (e.g., maximum profit, minimum cost).

对称轴是竖直线x = h，其中h = −b/(2a)。顶点为点(h, k)。你可以使用这些公式直接从标准形式求得，或通过配方法求得。在许多文字题中，顶点提供了最优值（例如，最大利润、最小成本）。

For example, the function f(x) = 2x² − 8x + 5 has axis of symmetry x = −(−8)/(2·2) = 2. Substitute x = 2 to get y = 2(4) − 16 + 5 = −3, so vertex is (2, −3). Since a = 2 > 0, the point is a minimum.

例如，函数f(x) = 2x² − 8x + 5的对称轴为x = −(−8)/(2·2) = 2。代入x = 2得到y = 2(4) − 16 + 5 = −3，因此顶点是(2, −3)。因为a = 2 > 0，该点为最小值。

8. Intercepts: x- and y-intercepts | 截距：x截距与y截距

To find the y-intercept, set x = 0 in f(x). This gives f(0) = c, so the y-intercept is (0, c). For x-intercepts, solve f(x) = 0, which may give 0, 1, or 2 real solutions as discussed. When sketching, always label intercepts clearly.

为求y截距，令x = 0代入f(x)。得到f(0) = c，因此y截距为(0, c)。为求x截距，解方程f(x) = 0，可能得到0个、1个或2个实数解，如前所述。作图时，务必清晰标出截距。

In vertex form f(x) = a(x − h)² + k, the y-intercept is found by x = 0: f(0) = a(0 − h)² + k = ah² + k. The x-intercepts (if any) satisfy a(x − h)² + k = 0 → (x − h)² = −k/a → x = h ± √(−k/a), provided −k/a ≥ 0.

在顶点式f(x) = a(x − h)² + k中，令x = 0求y截距：f(0) = a(0 − h)² + k = ah² + k。x截距（如有）满足a(x − h)² + k = 0 → (x − h)² = −k/a → x = h ± √(−k/a)，前提是 −k/a ≥ 0。

9. Quadratic Inequalities | 二次不等式

Solving quadratic inequalities like ax² + bx + c > 0 (or < 0) requires a sketch or sign analysis. First, find the real roots of the corresponding equation. These roots divide the x-axis into intervals. Test a value from each interval in the inequality, or use the parabola shape to determine where the function is positive/negative.

解二次不等式如ax² + bx + c > 0（或 < 0）需要借助草图或符号分析。首先，求出相应方程的实根。这些根将x轴分成若干区间。在每个区间中选取一个值代入不等式检验，或利用抛物线形状判断函数在何处为正/负。

Key points: If a > 0 (∪ shape), ax² + bx + c > 0 outside the roots (x < smaller root or x > larger root); ax² + bx + c < 0 between the roots. If a < 0 (∩ shape), the inequalities reverse. Always express final solution in set notation or interval notation as required.

要点：若a > 0（∪形），则ax² + bx + c > 0在两根之外（x < 较小根或 x > 较大根）；ax² + bx + c < 0在两根之间。若a < 0（∩形），不等式方向相反。最终答案务必按要求用集合符号或区间符号表示。

Example: Solve x² − 5x + 6 ≥ 0. Roots are x = 2, 3. The parabola opens upward. So solution: x ≤ 2 or x ≥ 3.

例：解不等式x² − 5x + 6 ≥ 0。根为x = 2, 3。抛物线开口向上。因此解为：x ≤ 2 或 x ≥ 3。

10. Applications and Word Problems | 应用与文字题

Quadratic functions model many real-world situations: projectile motion, area optimisation, revenue/cost functions, and bridge arches. Typically, you need to formulate a quadratic from given information, then find its maximum/minimum using the vertex, or solve for specific values using the quadratic formula.

二次函数可对许多现实情境建模：抛体运动、面积最优化、收入/成本函数以及桥梁拱形。通常，你需要根据所给信息建立二次函数，然后利用顶点求其最大值/最小值，或利用求根公式解出特定值。

A standard problem: A farmer has 100 m of fencing to enclose a rectangular area against a wall. Let the side perpendicular to the wall be x m. Then the parallel side is (100 − 2x) m. Area A = x(100 − 2x) = −2x² + 100x. Maximum area occurs at vertex x = −b/(2a) = −100/(2·(−2)) = 25 m, giving A_max = 25×(50) = 1250 m².

一个标准问题：一位农民有100米围栏，想靠墙围成一个矩形区域。设垂直于墙的边长为x米。则平行于墙的边长为(100 − 2x)米。面积A = x(100 − 2x) = −2x² + 100x。最大面积在顶点x = −b/(2a) = −100/(2·(−2)) = 25米处取得，此时A_max = 25×50 = 1250平方米。

Examiners expect clear definitions of variables, a correct quadratic expression, and a concluding statement. Always check that your solution makes sense in the context (e.g., length cannot be negative).

考官期望清晰定义变量、列出正确的二次表达式并给出结论性陈述。务必检验解在实际情境中的合理性（例如，长度不能为负）。

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Quadratic Functions | 二次函数