📚 Worked Example 2.2.1: Completing the Square and Quadratic Graphs | 例题 2.2.1：配方法与二次函数图像
form, and use the result to identify the vertex, axis of symmetry and key intercepts for sketching the graph. This technique appears frequently in AQA A-level Mathematics, especially in questions on quadratics, transformations and coordinate geometry.
在本例题中，我们将对二次函数 f(x) = 2x² + 8x + 5 进行配方，将其写成顶点式，并利用该结果确定顶点、对称轴和关键截距，以便绘制图像。这一技巧在 AQA A-level 数学中经常出现，尤其是在二次函数、图像变换和坐标几何相关题目中。
1. Understanding the Problem Statement | 理解题目要求
5 in the form a(x + h)² + k, where a, h and k are constants. The completed square form is useful because it reveals the vertex of a parabola directly: for a(x + h)² + k, the vertex is at (-h, k).
题目要求我们将 f(x) = 2x² + 8x + 5 改写为 a(x + h)² + k 的形式，其中 a、h 和 k 为常数。配方法的顶点式非常有用，因为它直接给出抛物线的顶点：对于 a(x + h)² + k，顶点坐标为 (-h, k)。
In this example, the coefficient of x² is not 1, so we must factor out the leading coefficient before completing the square. This is the most common source of error in AQA exam scripts.
在本例中，x² 的系数不是 1，因此我们必须先提出首项系数，再完成配方。这是 AQA 考试答卷中最常见的错误来源。
2. Step 1: Factor Out the Leading Coefficient | 第一步：提出首项系数
t of x² is 2, we factor 2 from the first two terms only, leaving the constant 5 outside the bracket:
从原始二次函数开始：f(x) = 2x² + 8x + 5。由于 x² 的系数是 2，我们只对前两项提取公因数 2，将常数项 5 保留在括号外：
f(x) = 2(x² + 4x) + 5
This step keeps the original function unchanged. The bracket now has a leading coefficient of 1, which allows us to complete the square inside the bracket cleanly.
这一步保持原函数不变。括号内二次项的系数现在变为 1，这样我们就能在括号内清晰地完成配方。
3. Step 2: Completing the Square Inside the Bracket | 第二步：在括号内完成配方
ake half the coefficient of x, which is 4 ÷ 2 = 2, and square it to get 2² = 4. Therefore:
括号内有 x² + 4x。为了配方，取 x 的系数的一半，即 4 ÷ 2 = 2，再将其平方得到 2² = 4。因此：
x² + 4x = (x + 2)² - 4
This identity is true because (x + 2)² expands to x² + 4x + 4, so we subtract 4 to leave x² + 4x unchanged. Substitute this back into the expression from Step 1:
这个恒等式成立，因为 (x + 2)² 展开为 x² + 4x + 4，所以我们减去 4，使 x² + 4x 保持不变。将其代回第一步的表达式：
f(x) = 2[(x + 2)² - 4] + 5
4. Step 3: Simplify to Vertex Form | 第三步：化简为顶点式
Distribute the 2 across the bracket and combine the constant terms:
将 2 分配到括号内，并合并常数项：
f(x) = 2(x + 2)² - 8 + 5
f(x) = 2(x + 2)² - 3
f(x) = 2(x + 2)² - 3. Comparing this with a(x + h)² + k, we have a = 2, h = 2 and k = -3.
配方法的顶点式现在为 f(x) = 2(x + 2)² - 3。将其与 a(x + h)² + k 比较，得到 a = 2，h = 2，k = -3。
5. Identifying the Vertex and Axis of Symmetry | 确定顶点与对称轴
vertex form a(x + h)² + k, the vertex is (-h, k). Here h = 2 and k = -3, so the vertex is:
对于顶点式 a(x + h)² + k，顶点为 (-h, k)。这里 h = 2，k = -3，因此顶点为：
Vertex = (-2, -3)
Because a = 2 is positive, the parabola opens upwards, and the vertex is a minimum point.
对称轴是经过顶点的竖直直线，因此其方程为 x = -2。由于 a = 2 为正数，抛物线开口向上，顶点为最低点。
6. Finding the Intercepts | 求解截距
original function: f(0) = 2(0)² + 8(0) + 5 = 5. So the graph crosses the y-axis at (0, 5).
为了求 y 轴截距，令原函数中 x = 0：f(0) = 2(0)² + 8(0) + 5 = 5。因此图像与 y 轴相交于 (0, 5)。
To find the x-intercepts, set f(x) = 0 and solve the completed square equation:
为了求 x 轴截距，令 f(x) = 0，并解顶点式方程：
2(x + 2)² - 3 = 0
2(x + 2)² = 3
(x + 2)² = 3/2
x + 2 = ±√3/2
x = -2 ± √3/2
.225. These roots confirm that the vertex lies symmetrically between the two x-intercepts.
因此 x 轴截距大约为 x ≈ -0.775 和 x ≈ -3.225。这些根证实了顶点位于两个 x 轴截距的对称中心。
7. Sketching the Quadratic Graph | 绘制二次函数图像
Draw a smooth U-shaped parabola through these points, with the axis of symmetry at x = -2.
绘制曲线时，先标出顶点 (-2, -3)，再标出 y 轴截距 (0, 5)，如有需要再标出 x 轴截距。经过这些点画出一条光滑的 U 形抛物线，对称轴为 x = -2。
is translated 2 units left, stretched vertically by factor 2, and translated 3 units down.
顶点式还能帮助检查图像变换：y = x² 的图像先向左平移 2 个单位，再在竖直方向上拉伸为原来的 2 倍，最后向下平移 3 个单位。
8. Common Mistakes and Exam Tips | 常见错误与考试技巧
A common error is to add and subtract the same number inside the bracket without accounting for the factor outside. For example, writing (x + 2)² - 4 inside the bracket and then forgetting to multiply the -4 by 2 leads to an incorrect constant term.
一个常见错误是在括号内加上和减去同一个数，却没有考虑括号外的系数。例如，在括号内写成 (x + 2)² - 4，然后忘记将 -4 乘以 2，会导致常数项错误。
Another mistake is incorrectly identifying the vertex as (h, k) instead of (-h, k) from the form a(x + h)² + k. Always write the vertex as the value that makes the bracket equal to zero.
另一个错误是将顶点式 a(x + h)² + k 中的顶点误认为 (h, k) 而不是 (-h, k)。请始终将顶点写为使括号内表达式等于零的 x 值。
- Always factor the x² coefficient first if it is not 1.
- Check your completed square form by expanding it back.
- Use the vertex form to find the minimum or maximum value quickly.
- Present exact roots with surds unless the question asks for decimal approximations.
以上要点为：如果 x² 系数不是 1，请先提取该系数；通过展开检查你的顶点式；利用顶点式快速求出最小值或最大值；除非题目要求近似小数，否则保留带有根号的精确根。
9. Connections to AQA Exam Questions | 与 AQA 考试题型的联系
AQA A-level papers often ask candidates to complete the square and then use the result to solve a quadratic inequality, find the range of a function, or sketch the graph with coordinates of the turning point. This worked example provides the core method needed for those question types.
AQA A-level 试卷经常要求考生先配方，然后利用结果解二次不等式、求函数的值域，或绘制图像并标出驻点坐标。本例题提供了解决这些题型所需的核心方法。
For instance, if asked to solve 2x² + 8x + 5 > 0, you can write 2(x + 2)² - 3 > 0, then solve (x + 2)² > 3/2. This leads directly to the solution set x < -2 - √3/2 or x > -2 + √3/2.
例如，如果要求解 2x² + 8x + 5 > 0，你可以写成 2(x + 2)² - 3 > 0，然后解 (x + 2)² > 3/2。这直接得出解集 x < -2 - √3/2 或 x > -2 + √3/2。
| Concept | Result for f(x) = 2x² + 8x + 5 | 中文解释 |
|---|---|---|
| Vertex form | 2(x + 2)² - 3 | 顶点式 |
| Vertex | (-2, -3) | 顶点 |
| Axis of symmetry | x = -2 | 对称轴 |
| Minimum value | -3 | 最小值 |
10. Practice Variants and Extensions | 练习变式与拓展
7. You should first factor out 3, then complete the square inside the bracket, and finally simplify to find the vertex.
请尝试用相同方法处理 f(x) = 3x² - 12x + 7。你应当先提取公因数 3，再在括号内完成配方，最后化简求出顶点。
n of a is negative. The parabola would then open downwards and the vertex would be a maximum point instead of a minimum.
作为拓展，考虑如果 a 的符号为负，y = 2(x + 2)² - 3 的图像会如何变化。此时抛物线将开口向下，顶点将从最低点变为最高点。
x² + 3x + 1. The same principle applies: take half of 3, square it, and adjust the constant.
另一个有用的变式是对 x 的系数为奇数的二次函数配方，例如 f(x) = x² + 3x + 1。原理相同：取 3 的一半，平方后调整常数项。
11. Summary and Key Takeaways | 总结与关键要点
Completing the square converts a quadratic from standard form ax² + bx + c into vertex form a(x + h)² + k. The process involves factoring out a if a ≠ 1, adding and subtracting the square of half the x-coefficient inside the bracket, and simplifying.
配方法将二次函数从标准式 ax² + bx + c 转化为顶点式 a(x + h)² + k。其步骤包括：若 a ≠ 1，先提取 a；在括号内加上并减去 x 系数一半的平方；最后化简。
- Vertex: (-h, k)
- Axis of symmetry: x = -h
- Minimum if a > 0, maximum if a < 0
- Always expand to check your completed square form
以上要点为：顶点为 (-h, k)；对称轴为 x = -h；若 a > 0 则为最小值，若 a < 0 则为最大值；务必展开检查顶点式。
12. Final Worked Answer for Example 2.2.1 | 例题 2.2.1 的最终解答
e axis of symmetry is x = -2, the y-intercept is (0, 5), and the x-intercepts are x = -2 ± √3/2.
已知 f(x) = 2x² + 8x + 5，其顶点式为 f(x) = 2(x + 2)² - 3。因此顶点为 (-2, -3)，对称轴为 x = -2，y 轴截距为 (0, 5)，x 轴截距为 x = -2 ± √3/2。
This answer demonstrates all the essential skills required for AQA quadratic manipulation, including factorising, completing the square, identifying key features and linking algebraic results to graphical interpretation.
该解答展示了 AQA 二次函数处理所需的所有关键技能，包括因式分解、配方、确定图像关键特征，以及将代数结果与图形解释联系起来。
Published by TutorHao | Maths Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图