📚 Quadratic Functions Key Points | GCSE CIE 数学：二次函数 考点精讲
A quadratic function is one of the most fundamental topics in GCSE CIE Mathematics. It describes a curved graph called a parabola, which appears in countless real-world contexts – from the trajectory of a ball to the shape of a satellite dish. Understanding quadratics is essential for success in algebra, coordinate geometry and problem-solving. In this article, we will break down every key point you need to master: forms of quadratic expressions, solving equations, sketching graphs, using the discriminant, and tackling inequalities. The explanations are paired in English and Chinese to help bilingual learners grasp both the concepts and the precise terminology. Let us begin our in-depth tour of quadratic functions, designed to support your revision and boost your confidence for the exam.
二次函数是 GCSE CIE 数学中最基础的主题之一。它描绘的曲线图像名为抛物线，在现实世界中随处可见——从球的飞行轨迹到卫星天线的形状。理解二次函数对代数、坐标几何和问题求解都至关重要。本文将分解你需要掌握的所有核心考点：二次表达式的各种形式、解方程、画图像、运用判别式以及处理不等式。每段解释都以英文和中文配对呈现，帮助双语学习者同时掌握概念和准确术语。让我们开始深入探讨二次函数，助力你的复习并提升应试信心。
1. Standard Form of Quadratic Functions | 二次函数的标准形式
A quadratic function can be written in the standard form: y = ax2 + bx + c, where a, b and c are constants and a ≠ 0. The term ax2 is the quadratic term, bx is the linear term, and c is the constant term. The shape of the graph is a parabola. If a > 0, the parabola opens upwards (U-shaped); if a < 0, it opens downwards (∩-shaped). The coefficient a controls the 'width' of the parabola: the larger |a|, the narrower the graph.
二次函数可以写成标准形式：y = ax2 + bx + c，其中 a、b、c 是常数且 a ≠ 0。ax2 是二次项，bx 是一次项，c 是常数项。其图像是一条抛物线。若 a > 0，抛物线开口向上（U 形）；若 a < 0，则开口向下（∩ 形）。系数 a 控制抛物线的“宽度”：|a| 越大，图像越窄。
For example, y = 2x2 + 3x − 5 has a = 2, b = 3, c = −5, and because a > 0, the parabola opens upward. The y-intercept is given by c, so the graph crosses the y-axis at (0, −5). Recognising standard form quickly allows you to identify these key features even before detailed calculations.
例如，y = 2x2 + 3x − 5 中 a = 2, b = 3, c = −5，由于 a > 0，抛物线开口向上。y 轴截距由 c 给出，因此图像与 y 轴交于 (0, −5)。迅速识别标准形式能让你在进行详细计算之前就判断出这些关键特征。
2. Factorised Form and Roots | 因式分解形式与根
A quadratic can often be expressed in factorised form: y = a(x − p)(x − q). The values p and q are the roots (or x-intercepts) of the equation y = 0. This form reveals the points where the parabola cuts the x-axis. If the quadratic does not factorise over integers, it may still have real roots that can be found by other methods, but the factorised form is extremely useful for quick sketching and solving.
二次函数常可表示为因式分解形式：y = a(x − p)(x − q)。其中 p 和 q 是方程 y = 0 的根（即 x 轴截距）。这一形式直接显示了抛物线与 x 轴的交点。如果二次式无法在整数范围内因式分解，它仍可能有实数根，可通过其他方法求得，但因式分解形式对于快速画图和求解极为有用。
For instance, y = x2 − 5x + 6 factorises to y = (x − 2)(x − 3). The roots are x = 2 and x = 3, so the graph crosses the x-axis at (2,0) and (3,0). Always start by checking whether a common factor can be taken out first, as this simplifies the expression. Remember that the order of subtraction matters: (x + 1) corresponds to a root of −1, not +1.
例如，y = x2 − 5x + 6 可分解为 y = (x − 2)(x − 3)。其根为 x = 2 和 x = 3，因此图像与 x 轴交于 (2,0) 和 (3,0)。务必先检查是否存在公因子可提取，这能简化表达式。注意相减的顺序：(x + 1) 对应的根为 −1，而非 +1。
3. Completing the Square | 配方法
Completing the square transforms a quadratic from standard form into vertex form. For an expression x2 + bx, we add (b/2)2 to create a perfect square trinomial, then adjust by subtracting the same quantity. The method is essential for deriving the quadratic formula, finding the turning point, and solving equations that cannot be factorised easily.
配方法将二次式从标准形式转化为顶点形式。对于表达式 x2 + bx，我们加上 (b/2)2 以构成完全平方三项式，再减去同样的数以进行调整。该方法对推导二次公式、求转折点以及解不易因式分解的方程都至关重要。
Example: Write y = x2 + 6x + 5 in completed square form. Halve the coefficient of x: 6/2 = 3. Square it: 32 = 9. Then y = (x2 + 6x + 9) + 5 − 9 = (x + 3)2 − 4. The turning point is at (−3, −4). When a ≠ 1, factor out a from the x-terms first, then complete the square inside the bracket and remember to multiply the subtracted constant correctly.
示例：将 y = x2 + 6x + 5 写成完全平方形式。将 x 的系数取半：6/2 = 3。求平方：32 = 9。则 y = (x2 + 6x + 9) + 5 − 9 = (x + 3)2 − 4。其转折点为 (−3, −4)。当 a ≠ 1 时，需先将 x 各项提出 a，再在括号内配平方，并注意正确处理减去的常数乘以 a。
4. Vertex Form and the Turning Point | 顶点形式与转折点
The vertex form of a quadratic is y = a(x − h)2 + k, where (h, k) is the vertex (turning point) of the parabola. If a > 0, the vertex is a minimum point; if a < 0, it is a maximum point. The axis of symmetry is the vertical line x = h. This form is the most convenient for identifying the graph's peak or trough and for solving optimisation problems.
二次函数的顶点形式为 y = a(x − h)2 + k，其中 (h, k) 为抛物线的顶点（转折点）。若 a > 0，顶点是最小值点；若 a < 0，则是最大值点。对称轴为直线 x = h。这种形式最便于识别图形的峰或谷，并解决最优化问题。
From the completed square example y = (x + 3)2 − 4, we can read h = −3, k = −4. So the vertex is (−3, −4) and since a = 1 > 0, it is a minimum. If the equation were y = −2(x − 1)2 + 8, the vertex would be (1, 8) with a maximum value of 8. The sign of h inside the bracket is reversed: (x − 1) gives h = +1, whereas (x + 3) gives h = −3.
由配方法实例 y = (x + 3)2 − 4 可得 h = −3, k = −4。所以顶点为 (−3, −4)，又因 a = 1 > 0，是最小值点。若方程为 y = −2(x − 1)2 + 8，其顶点为 (1, 8)，且最大值是 8。注意括号内 h 的符号相反：(x − 1) 得出 h = +1，而 (x + 3) 得出 h = −3。
5. Symmetry and Axis of Symmetry | 对称性与对称轴
Every parabola is symmetric about a vertical line called the axis of symmetry. For a quadratic in standard form y = ax2 + bx + c, this line has equation x = −b/(2a). In vertex form y = a(x − h)2 + k, the axis is simply x = h. The symmetry means that points equidistant horizontally from the axis have the same y-coordinate. This property is extremely helpful when sketching the graph or finding a missing root if one root is known.
每一条抛物线都关于一条称为对称轴的竖直线对称。对于标准形式 y = ax2 + bx + c，该直线方程为 x = −b/(2a)。在顶点形式 y = a(x − h)2 + k 中，对称轴即为 x = h。对称性意味着水平方向上与对称轴等距的两个点具有相同的 y 坐标。这一性质在画图或已知一根求另一根时非常有用。
For example, the quadratic y = x2 − 4x + 1 has axis of symmetry x = −(−4)/(2×1) = 2. The roots are x = 2 − √3 and x = 2 + √3, which are symmetric about x = 2. When the discriminant is zero, the vertex lies exactly on the x-axis and the axis of symmetry passes through the single repeated root.
例如，二次函数 y = x2 − 4x + 1 的对称轴为 x = −(−4)/(2×1) = 2。其根为 x = 2 − √3 和 x = 2 + √3，二者关于 x = 2 对称。当判别式为零时，顶点恰好位于 x 轴上，对称轴穿过那个唯一的重根。
6. Graphing Quadratic Functions | 二次函数图像绘制
To sketch a quadratic graph, follow a systematic procedure. First, determine whether the parabola opens upwards (a > 0) or downwards (a < 0). Find the y-intercept by setting x = 0, which gives the point (0, c). Calculate the x-intercepts by solving ax2 + bx + c = 0; these may be found by factorising, using the quadratic formula, or completing the square. Locate the vertex either by completing the square or by using x = −b/(2a) and then substituting to find y. Plot these key points and draw a smooth, symmetric curve through them. Label the vertex and intercepts. If the discriminant is negative, the graph does not cross the x-axis; it floats entirely above or below it depending on the sign of a.
画二次函数图像可遵循系统步骤。首先确定抛物线开口向上 (a > 0) 还是向下 (a < 0)。令 x = 0 求 y 轴截距，得点 (0, c)。通过解 ax2 + bx + c = 0 计算 x 轴截距；可通过因式分解、二次公式或配方法求得。用配方法或 x = −b/(2a) 代入求 y 来确定顶点。标出这些关键点，并用平滑对称的曲线连接它们。标注顶点和截距。若判别式为负，图像不与 x 轴相交；它会完全位于 x 轴上方或下方，取决于 a 的符号。
Consider y = −x2 + 2x + 3. Here a = −1 so it opens downwards. y-intercept (0,3). Factorise: −(x2 − 2x − 3) = −(x − 3)(x + 1) so x-intercepts at 3 and −1. Vertex x = −2/(2×(−1)) = 1, y = −1 + 2 + 3 = 4, giving (1,4). Sketch the parabola peaking at (1,4) and crossing the axes at (0,3), (3,0), (−1,0).
考虑 y = −x2 + 2x + 3。这里 a = −1，因此开口向下。y 截距 (0,3)。因式分解：−(x2 − 2x − 3) = −(x − 3)(x + 1)，故 x 截距为 3 和 −1。顶点 x = −2/(2×(−1)) = 1，y = −1 + 2 + 3 = 4，得 (1,4)。画出以 (1,4) 为最高点，经过 (0,3)、(3,0)、(−1,0) 的抛物线。
7. Solving Quadratic Equations by Factorising | 因式分解解二次方程
When a quadratic equation ax2 + bx + c = 0 can be written as (px + q)(rx + s) = 0, we set each factor equal to zero: px + q = 0 or rx + s = 0, giving solutions x = −q/p and x = −s/r. This method is quick and reveals the roots directly. Always rearrange the equation so that one side is zero before attempting to factorise. If a ≠ 1, use the splitting-the-middle-term or trial method to factorise.
当二次方程 ax2 + bx + c = 0 可写成 (px + q)(rx + s) = 0 时，令每个因式等于零：px + q = 0 或 rx + s = 0，得到解 x = −q/p 和 x = −s/r。这种方法快捷且直接显示根。务必先将方程整理成一边为零再尝试因式分解。若 a ≠ 1，可采用十字相乘法或尝试法进行因式分解。
Solve 2x2 + 7x + 3 = 0. Multiply a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: 6 and 1. Rewrite: 2x2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0. Solutions: x = −1/2 and x = −3. Checking by substitution confirms their correctness. If the equation has a common factor, extract it first to keep numbers small.
解 2x2 + 7x + 3 = 0。将 a 与 c 相乘：2 × 3 = 6。找出两数相乘得 6、相加得 7：6 和 1。重写：2x2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0。解为 x = −1/2 和 x = −3。代入验证可确认其正确性。若方程含有公因子，应首先提取，使数字保持较小。
8. Quadratic Formula | 二次公式
The quadratic formula gives the solutions of ax2 + bx + c = 0 as:
x = (−b ± √(b2 − 4ac)) / (2a)
This formula works for all quadratic equations, even when factorising is difficult or impossible. The expression under the square root, b2 − 4ac, is called the discriminant. It determines the number and nature of the roots. Remember to write the equation in standard form with zero on one side before identifying a, b, c. Pay close attention to signs when substituting negative values.
二次公式给出 ax2 + bx + c = 0 的解为：
x = (−b ± √(b2 − 4ac)) / (2a)
该公式适用于所有二次方程，即使因式分解困难或不可能时也有效。平方根下的表达式 b2 − 4ac 称为判别式，它决定根的个数和性质。使用前务必先将方程写成一边为零的标准形式，再确定 a、b、c。代入负值时特别注意符号。
Example: Solve 3x2 − 5x − 2 = 0. Here a = 3, b = −5, c = −2. Discriminant = (−5)2 − 4(3)(−2) = 25 + 24 = 49. Then x = (5 ± √49) / (6) = (5 ± 7)/6. The two solutions are x = 12/6 = 2 and x = −2/6 = −1/3. The quadratic formula always yields correct answers; just be systematic with your arithmetic.
示例：解 3x2 − 5x − 2 = 0。这里 a = 3, b = −5, c = −2。判别式 = (−5)2 − 4(3)(−2) = 25 + 24 = 49。则 x = (5 ± √49) / (6) = (5 ± 7)/6。两个解为 x = 12/6 = 2 和 x = −2/6 = −1/3。二次公式总能给出正确答案，只需一步一步细心运算。
9. The Discriminant | 判别式
The discriminant Δ = b2 − 4ac reveals crucial information about the roots without solving the equation fully. If Δ > 0, the quadratic has two distinct real roots (the graph crosses the x-axis at two points). If Δ = 0, there is exactly one real root (a repeated root; the vertex touches the x-axis). If Δ < 0, there are no real roots (the graph does not meet the x-axis). In CIE exams, you may be asked to determine the number of roots or find unknown coefficients that give a specific number of solutions.
判别式 Δ = b2 − 4ac 无需完全解方程即可揭示根的关键信息。若 Δ > 0，二次方程有两个不同的实数根（图像与 x 轴相交于两点）。若 Δ = 0，恰好有一个实数根（重根；顶点触及 x 轴）。若 Δ < 0，则无实数根（图像不与 x 轴相交）。在 CIE 考试中，你可能需要判断根的个数或求取使特定解数成立的未知系数。
For example, find k such that x2 + kx + 9 = 0 has exactly one real root. Set discriminant = 0: k2 − 4(1)(9) = 0 ⇒ k2 = 36 ⇒ k = ±6. So k = 6 or k = −6. Another typical question: show that x2 + x + 1 = 0 has no real roots. Here Δ = 12 − 4(1)(1) = −3 < 0, hence no real roots. The discriminant is also used when proving that a quadratic is always positive or always negative.
例如，求 k 使得 x2 + kx + 9 = 0 恰有一个实数根。令判别式 = 0：k2 − 4(1)(9) = 0 ⇒ k2 = 36 ⇒ k = ±6。故 k = 6 或 k = −6。另一个典型问题：证明 x2 + x + 1 = 0 无实数根。这里 Δ = 12 − 4(1)(1) = −3 < 0，因此无实数根。判别式也用于证明二次函数恒正或恒负。
10. Quadratic Inequalities | 二次不等式
Solving a quadratic inequality such as x2 − x − 6 > 0 involves first solving the corresponding quadratic equation to find the critical values where the expression equals zero. Here x2 − x − 6 = (x − 3)(x + 2) = 0 gives x = 3 and x = −2. These values divide the number line into intervals. Test a point from each interval in the inequality to see whether the expression is positive. For this example, when x < −2, say x = −3, (−32 − (−3) − 6) = 9+3−6 = 6 > 0, so true. When −2 < x < 3, say x = 0, 0 − 0 − 6 = −6 < 0, false. When x > 3, say x = 4, 16 − 4 − 6 = 6 > 0, true. Therefore the solution is x < −2 or x > 3. Always pay attention to whether the inequality is strict (> or
解二次不等式如 x2 − x − 6 > 0，首先解对应的二次方程求出表达式等于零的临界值。此处 x2 − x − 6 = (x − 3)(x + 2) = 0 得 x = 3 和 x = −2。这些值将数轴分成若干区间。在每个区间选取测试点代入不等式，观察表达式是否为正。在本例中，当 x < −2，取 x = −3，(−32 − (−3) − 6) = 9+3−6 = 6 > 0，为真。当 −2 < x < 3，取 x = 0，0 − 0 − 6 = −6 < 0，为假。当 x > 3，取 x = 4，16 − 4 − 6 = 6 > 0，为真。因此解为 x < −2 或 x > 3。务必注意不等式是严格不等（> 或
If the quadratic has a positive leading coefficient and the inequality is < 0, the solution typically lies between the roots. A sketch of the parabola helps visualise which region satisfies the condition. For repeated-root or no-real-root discriminants, the parabola does not cross the x-axis, so the inequality may be true for all real x or for none.
若二次项系数为正且不等式为 < 0，则解通常位于两根之间。画出抛物线的草图有助于直观看出满足条件的区域。对于重根或无实数根的情况，抛物线不与 x 轴相交，此时不等式可能对所有实数 x 成立，也可能无解。
11. Applications and Word Problems | 应用与文字题
Quadratic functions frequently appear in real-life contexts such as projectile motion, area problems, and profit optimisation. A typical problem might state: “A ball is thrown upward; its height h (in metres) after t seconds is given by h = −5t2 + 20t + 1. Find the maximum height and the time when the ball hits the ground.” The maximum height occurs at the vertex. Rewrite in completed square form or use t = −b/(2a) = −20/(2×(−5)) = 2 seconds; then h = −5(2)2 + 20(2) + 1 = 21 m. To find when it hits the ground, set h = 0: −5t2 + 20t + 1 = 0, solve using the quadratic formula, and take the positive root. Another classic problem involves finding the dimensions of a rectangle with given perimeter and area, leading to a quadratic equation.
二次函数经常出现在现实情境中，如抛体运动、面积问题以及利润最优化。一个典型题目可能是：“向上抛出一个球；t 秒后的高度 h（米）由 h = −5t2 + 20t + 1 给出。求最大高度及球落地的时刻。”最大高度出现在顶点。可改写为完全平方形式，或使用 t = −b/(2a) = −20/(2×(−5)) = 2 秒；则 h = −5(2)2 + 20(2) + 1 = 21 米。求落地时刻，令 h = 0：−5t2 + 20t + 1 = 0，用二次公式求解，取正根。另一个经典问题涉及已知周长和面积求长方形的尺寸，从而导出二次方程。
Word problems require translating the description into a quadratic equation, selecting the appropriate mathematical method, and interpreting the solution in context. Always discard answers that don’t make sense in the real-world situation (e.g. negative time or length). Drawing a diagram and defining variables clearly should be your first steps.
文字题需要将描述转化为二次方程，选择合适的数学方法，并在情境中解释解的意义。务必舍去在现实情境中不合理的答案（例如时间为负或长度为负）。绘图并清晰地定义变量应是你解题的第一步。
12. Summary of Key Points | 考点总结
Mastering quadratic functions means being fluent with all three forms: standard (y = ax2 + bx + c), factorised (y = a(x − p)(x − q)), and vertex (y = a(x − h)2 + k). You should be able to move between them using factorisation, completing the square, and algebraic expansion. Know how to find the vertex and axis of symmetry both from the formula x = −b/(2a) and from the completed square. Recognise the role of the discriminant in determining the nature of roots. Be confident solving quadratics by factorising, by the quadratic formula, and by completing the square. Sketch graphs by identifying intercepts, vertex, and orientation. Finally, apply these skills to inequalities and real-world problems, always interpreting your mathematical results correctly. Regular practice with past paper questions will solidify these concepts and build exam speed.
掌握二次函数意味着要熟练运用三种形式：标准式 (y = ax2 + bx + c)、因式分解式 (y = a(x − p)(x − q)) 和顶点式 (y = a(x − h)2 + k)。你应当能够通过因式分解、配方法和代数展开在它们之间转换。要懂得如何通过公式 x = −b/(2a) 和配方法求得顶点及对称轴。认识判别式在判断根的性质中的作用。要能自信地通过因式分解、二次公式和配方法求解二次方程。通过确定截距、顶点和开口方向来画图像。最后，将这些技能应用于不等式和现实问题，并总是正确地解读你的数学结果。通过定期练习往年试卷，这些概念将得以巩固，并提升应试速度。
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