📚 Mastering Quadratic Functions: The Vertex Form and Completing the Square | 掌握二次函数：顶点式与配方法
Quadratic functions appear everywhere in IGCSE Mathematics, from solving equations to sketching parabolas and solving real-world optimisation problems. One of the most powerful algebraic tools for understanding a quadratic is completing the square, because it reveals the vertex of the parabola directly. This article explains the standard form, the vertex form, the method of completing the square, and how these ideas help you solve equations and analyse graphs with confidence.
在 IGCSE 数学中，二次函数无处不在，从解方程到绘制抛物线，再到求解现实中的最优化问题。理解二次函数最有力的代数工具之一就是配方法，因为它可以直接揭示抛物线的顶点。本文将详细讲解标准形式、顶点式、配方法的步骤，以及这些概念如何帮助你自信地解方程和分析图像。
1. The Standard Form of a Quadratic Function | 二次函数的标准形式
At IGCSE level, you will usually meet a quadratic function written in the standard form:
在 IGCSE 阶段，你通常见到的二次函数写成标准形式：
y = ax² + bx + c
Here a, b and c are constants. The most important restriction is that a ≠ 0; if a = 0, the expression becomes linear rather than quadratic. The value of a controls the width and direction of the parabola, b affects the horizontal position of the curve, and c gives the y-intercept, the point where the graph crosses the y-axis.
其中 a、b、c 为常数。最重要的条件是 a ≠ 0；如果 a = 0，表达式就变成了一次函数而非二次函数。a 的值控制抛物线的开口方向和宽窄，b 影响曲线的水平位置，c 给出 y 轴截距，即图像与 y 轴相交的点的坐标。
For example, in y = 2x² – 4x + 1, we have a = 2, b = -4 and c = 1. The y-intercept is (0, 1). However, the standard form does not immediately tell us the vertex or the axis of symmetry. To find those, we need to rewrite the function in vertex form.
例如，在 y = 2x² – 4x + 1 中，我们有 a = 2，b = -4，c = 1，y 轴截距为 (0, 1)。但是，标准形式不能直接告诉我们顶点或对称轴的位置。要找到它们，我们需要将函数改写为顶点式。
2. The Vertex Form and the Vertex | 顶点式与顶点
The vertex form of a quadratic function is:
二次函数的顶点式为：
y = a(x – h)² + k
In this form, the vertex is simply (h, k), and the axis of symmetry is the vertical line x = h. If a > 0, the parabola opens upward, so the vertex gives the minimum point of the curve. If a < 0, the parabola opens downward, and the vertex gives the maximum point.
在这个形式中，顶点就是 (h, k)，对称轴是竖直线 x = h。如果 a > 0，抛物线开口向上，顶点就是曲线的最小点；如果 a < 0，抛物线开口向下，顶点就是最大点。
How do the two forms relate? Expanding (x – h)² gives:
两种形式之间有什么关系呢？展开 (x – h)² 可得：
(x – h)² = x² – 2hx + h²
Therefore y = a(x – h)² + k = ax² – 2ahx + ah² + k. Comparing coefficients with y = ax² + bx + c, we see that b = -2ah, which means h = -b/(2a). This is exactly the horizontal coordinate of the vertex. The constant k = c – ah² = c – b²/(4a).
因此，y = a(x – h)² + k = ax² – 2ahx + ah² + k。将系数与 y = ax² + bx + c 比较，可得 b = -2ah，即 h = -b/(2a)。这正是顶点的横坐标。常数 k = c – ah² = c – b²/(4a)。
Rather than memorising these formulas, it is usually more reliable to complete the square. This method rewrites any quadratic expression in vertex form step by step, and it also helps solve equations without using the quadratic formula.
与其死记这些公式，通常更可靠的方法是使用配方法。这个方法可以把任何二次表达式一步步改写成顶点式，同时也有助于在不使用求根公式的情况下解方程。
3. Completing the Square: A Step-by-Step Guide | 配方法：分步指南
Let us complete the square for a simple quadratic expression, x² + 6x + 5. Follow these steps with any expression where a = 1.
让我们对一个简单的二次表达式 x² + 6x + 5 进行配方。当 a = 1 时，任何表达式都可以按以下步骤处理。
Step 1: Identify the coefficient of x, which is 6 in this case.
步骤 1：确定 x 的系数，这里为 6。
Step 2: Take half of 6, giving 3, and write (x + 3)².
步骤 2：取 6 的一半，得到 3，写下 (x + 3)²。
Step 3: Expand (x + 3)² = x² + 6x + 9. Since the original expression only has x² + 6x, we must subtract the extra 9.
步骤 3：展开 (x + 3)² = x² + 6x + 9。由于原表达式只有 x² + 6x，我们必须减去多余的 9。
Step 4: Combine the constant terms: 5 – 9 = -4.
步骤 4：合并常数项：5 – 9 = -4。
The final result is:
最终结果为：
x² + 6x + 5 = (x + 3)² – 4
For a general expression x² + bx + c, the same method gives:
对于一般表达式 x² + bx + c，同样的方法给出：
x² + bx + c = (x + b/2)² – (b/2)² + c
Now consider an example with a ≠ 1, such as 2x² + 8x + 3. First factor out 2 from the first two terms: 2(x² + 4x) + 3. Then complete the square inside the bracket: x² + 4x = (x + 2)² – 4. Therefore:
现在考虑一个 a ≠ 1 的例子，例如 2x² + 8x + 3。首先从前两项中提取 2：2(x² + 4x) + 3。然后在括号内配方：x² + 4x = (x + 2)² – 4。因此：
2x² + 8x + 3 = 2[(x + 2)² – 4] + 3 = 2(x + 2)² – 8 + 3 = 2(x + 2)² – 5
Notice how important it is to multiply the subtracted square by the factor 2 before adding the constant. In this case, the vertex is (-2, -5), and since a = 2 > 0, the curve opens upward.
注意，在加上常数之前，必须把减去的平方项乘以系数 2。本例中顶点为 (-2, -5)，由于 a = 2 > 0，曲线开口向上。
4. Solving Quadratic Equations by Completing the Square | 用配方法解二次方程
Completing the square is not only useful for graphing; it also provides a direct way to solve quadratic equations. Suppose we want to solve x² + 6x + 5 = 0. The completed-square form gives us:
配方法不仅对画图有用，它还为解二次方程提供了直接途径。假设我们要解 x² + 6x + 5 = 0。配方式给出：
(x + 3)² – 4 = 0
Add 4 to both sides: (x + 3)² = 4. Then take the square root of both sides, remembering the ± sign:
两边同时加 4：(x + 3)² = 4。然后对方程两边开平方根，记得加上 ± 号：
x + 3 = ±2
Solving x + 3 = 2 gives x = -1. Solving x + 3 = -2 gives x = -5. Hence the solutions are x = -1 and x = -5.
解 x + 3 = 2 得 x = -1；解 x + 3 = -2 得 x = -5。因此方程的解为 x = -1 和 x = -5。
For an equation like 2x² + 8x + 3 = 0, we use the vertex form 2(x + 2)² – 5 = 0. Then (x + 2)² = 5/2, so x + 2 = ±√(5/2), and finally x = -2 ± √(5/2). These are two irrational roots, which are perfectly valid answers at IGCSE level.
对于方程 2x² + 8x + 3 = 0，我们使用顶点式 2(x + 2)² – 5 = 0。于是 (x + 2)² = 5/2，所以 x + 2 = ±√(5/2)，最终得到 x = -2 ± √(5/2)。这两个是无理根，在 IGCSE 阶段是完全有效的答案。
This method works for any quadratic equation, even when the quadratic formula would be messy. It also shows clearly why the discriminant matters: if the right-hand side is negative, the equation has no real solutions.
这种方法适用于任何二次方程，即使求根公式很麻烦时也能使用。它还能清楚地说明判别式为什么重要：如果右边为负数，方程就没有实数解。
5. Graphing from the Vertex Form | 利用顶点式绘制图像
Once a quadratic is written in vertex form, sketching the graph becomes straightforward. You immediately know the vertex, the axis of symmetry, and the direction of opening. From y = a(x – h)² + k:
一旦二次函数写成顶点式，画图就变得非常简单。你可以立刻知道顶点、对称轴和开口方向。根据 y = a(x – h)² + k：
- The vertex is (h, k).
- The axis of symmetry is x = h.
- The parabola opens upward if a > 0, downward if a < 0.
- The graph is stretched or compressed compared to y = x² depending on |a|.
- 顶点是 (h, k)。
- 对称轴是 x = h。
- 如果 a > 0，抛物线开口向上；如果 a < 0，抛物线开口向下。
- 与 y = x² 相比，图像会根据 |a| 的值被拉伸或压缩。
To sketch the curve accurately, find the vertex first, then plot two or three extra points by choosing x-values around the axis of symmetry. Remember that the y-intercept is found by setting x = 0 in the original expression. Reflecting points across the axis of symmetry helps you draw a balanced parabola.
要准确画出曲线，先找到顶点，然后在对称轴两侧选择几个 x 值，代入求出几个额外的点。记住 y 轴截距可以通过在原始表达式中令 x = 0 来找到。将点关于对称轴反射，能帮助你画出一条对称的抛物线。
The table below summarises how the two forms help us describe the graph.
下表总结了两种形式如何帮助我们描述图像。
| Property | y = ax² + bx + c | y = a(x – h)² + k |
| Vertex | (-b/(2a), c – b²/(4a)) | (h, k) |
| Axis of symmetry | x = -b/(2a) | x = h |
| Direction | up if a > 0, down if a < 0 | up if a > 0, down if a < 0 |
| y-intercept | c | ah² + k |
| 性质 | y = ax² + bx + c | y = a(x – h)² + k |
| 顶点 | (-b/(2a), c – b²/(4a)) | (h, k) |
| 对称轴 | x = -b/(2a) | x = h |
| 开口方向 | a > 0 向上，a < 0 向下 | a > 0 向上，a < 0 向下 |
| y 轴截距 | c | ah² + k |
6. The Discriminant and the Nature of the Roots | 判别式与根的性质
The discriminant of a quadratic equation ax² + bx + c = 0 is defined as:
二次方程 ax² + bx + c = 0 的判别式定义为：
Δ = b² – 4ac
The sign of the discriminant tells us how many real roots the equation has.
判别式的符号告诉我们方程有多少个实数根。
- If Δ > 0, there are two distinct real roots.
- If Δ = 0, there is exactly one repeated real root.
- If Δ < 0, there are no real roots; the curve does not intersect the x-axis.
- 如果 Δ > 0，方程有两个不同的实数根。
- 如果 Δ = 0，方程恰有一个重根。
- 如果 Δ < 0，方程没有实数根；曲线与 x 轴不相交。
Completing the square provides another way to see this. In the vertex form y = a(x – h)² + k, if a > 0 and k > 0, the minimum point lies above the x-axis, so the graph never touches it. If k = 0, the vertex lies exactly on the x-axis, giving one repeated root. If k < 0, the curve crosses the axis twice.
配方法提供了另一种理解方式。在顶点式 y = a(x – h)² + k 中，如果 a > 0 且 k > 0，最小点在 x 轴上方，图像永远不会接触到 x 轴。如果 k = 0，顶点正好在 x 轴上，方程有一个重根。如果 k < 0，曲线与 x 轴相交两次。
For example, y = (x + 3)² – 4 has k = -4, so it has two x-intercepts. Meanwhile, y = (x – 1)² + 2 has k = 2, so it has no roots at all.
例如，y = (x + 3)² – 4 中 k = -4，因此它有两个 x 轴截点。而 y = (x – 1)² + 2 中 k = 2，所以它没有任何根。
7. Applications: Maximum and Minimum Values | 应用：最大值与最小值
One of the most common exam applications of completing the square is finding the maximum or minimum value of a quadratic function. This appears in problems about projectile motion, area, profit, and many other contexts.
配方法最常见的考试应用之一就是求二次函数的最大值或最小值。这出现在抛体运动、面积、利润及其他许多情境中。
Consider a ball thrown upward with height given by h(t) = -5t² + 20t + 2, where t is time in seconds and h is height in metres. Complete the square:
考虑一个向上抛出的球，其高度为 h(t) = -5t² + 20t + 2，其中 t 是以秒为单位的时间，h 是以米为单位的高度。对其进行配方：
h(t) = -5(t² – 4t) + 2 = -5[(t – 2)² – 4] + 2 = -5(t – 2)² + 20 + 2 = -5(t – 2)² + 22
Since the term -5(t – 2)² is always non-positive, the largest possible value of h is 22, achieved when t = 2. Therefore the ball reaches its maximum height of 22 metres after 2 seconds.
因为 -5(t – 2)² 永远是非正数，所以 h 的最大可能值就是 22，在 t = 2 时取得。因此，球在 2 秒后达到最大高度 22 米。
In general, if a quadratic has a positive leading coefficient, completing the square gives the minimum value of the function. If a is negative, it gives the maximum value. This is a reliable and quick method that also tells you exactly where the optimum occurs.
一般来说，如果二次函数的首项系数为正，配方法给出函数的最小值；如果 a 为负，则给出最大值。这是一种可靠且快速的方法，同时能准确地告诉你最值出现在何处。
8. Common Mistakes to Avoid | 需要避免的常见错误
Students often lose marks on completing the square questions because of a few predictable mistakes. Knowing these will help you stay accurate under exam pressure.
学生在配方法题目中经常因为一些可以预测的错误而丢分。了解这些错误将帮助你在考试压力下保持准确。
- Forgetting to factor out a when a ≠ 1. You must extract a from the x² and x terms before completing the square.
- Sign errors on h. In y = (x + 1)² – 3, the vertex is (-1, -3), not (1, -3).
- Writing (x – 3)² = x² – 6x – 9 instead of x² – 6x + 9.
- Missing the ± sign when taking square roots during equation solving.
- Forgetting the factor a when squaring the half-coefficient, as in 2(x + 2)² – 4 instead of 2(x + 2)² – 8.
- 当 a ≠ 1 时忘记提取 a。在配方之前必须从 x² 项和 x 项中提取 a。
- h 的符号错误。在 y = (x + 1)² – 3 中，顶点是 (-1, -3)，而不是 (1, -3)。
- 把 (x – 3)² 错写为 x² – 6x – 9，而不是 x² – 6x + 9。
- 解方程开平方时漏掉 ± 号。
- 忘记系数 a 对平方项外常数的影响，例如写 2(x + 2)² – 4，而正确应为 2(x + 2)² – 8。
Always check your completed square by expanding it back to the original form. A five-second check can catch most sign errors.
一定要将配方式展开回去，检验是否等于原式。一个五秒钟的检查就能发现大多数符号错误。
9. Practice Problems and Solutions | 练习与解答
Work through these problems without a calculator if possible, then check your answers below.
如果可以，请尽量不用计算器完成以下问题，然后对照下面的答案检查。
Problem 1: Write x² – 8x + 15 in vertex form and find the coordinates of the vertex.
练习 1：将 x² – 8x + 15 写成顶点式，并求出顶点坐标。
Problem 2: Solve x² + 10x + 21 = 0 by completing the square.
练习 2：用配方法解 x² + 10x + 21 = 0。
Problem 3: Find the maximum value of y = -x² + 6x – 1 and the x-value at which it occurs.
练习 3：求 y = -x² + 6x – 1 的最大值及其对应的 x 值。
Problem 4: Show that 3x² – 12x + 13 has no real roots.
练习 4：证明 3x² – 12x + 13 没有实数根。
| Problem | Solution |
| 1 | x² – 8x + 15 = (x – 4)² – 1; vertex (4, -1) |
| 2 | (x + 5)² – 4 = 0 → x = -3 or x = -7 |
| 3 | y = -(x – 3)² + 8; maximum value 8 at x = 3 |
更多咨询请联系16621398022（同微信）
CommentsMore posts |
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply