Quadratic Functions: Solving Equations and Graphing | 二次函数：方程求解与图像

📚 Quadratic Functions: Solving Equations and Graphing | 二次函数：方程求解与图像

Quadratic functions are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, and many real-world problems. This lesson will help you understand the standard form, solve quadratic equations using different methods, and sketch their graphs accurately.

二次函数是IGCSE数学中最重要的内容之一。它出现在代数、图像以及许多实际问题中。本课将帮助你理解标准形式，运用不同方法求解二次方程，并准确画出它们的图像。

1. Standard Form of a Quadratic Function | 二次函数的标准形式

A quadratic function is any function that can be written in the form f(x) = ax² + bx + c, where a, b, and c are real numbers and a ≠ 0. The highest power of x is 2, which gives the graph its characteristic U-shape.

二次函数是任何可以写成 f(x) = ax² + bx + c 形式的函数，其中 a、b、c 是实数且 a ≠ 0。x 的最高次数是2，这使得图像呈现特有的U形。

The coefficient a controls the direction and width of the parabola. If a > 0, the graph opens upward, like a cup. If a < 0, it opens downward, like a cap. The larger the value of |a|, the narrower the parabola.

系数 a 控制抛物线的开口方向和宽度。如果 a > 0，图像向上开口，像杯子一样；如果 a < 0，图像向下开口，像帽子一样。|a| 的值越大，抛物线越窄。

2. Key Features: Vertex, Axis of Symmetry, and Roots | 关键特征：顶点、对称轴和根

Every quadratic graph has a vertex, which is the maximum or minimum point. The axis of symmetry is the vertical line through the vertex. The roots (or x-intercepts) are the solutions of the equation ax² + bx + c = 0.

每一个二次图像都有一个顶点，也就是最大值或最小值点。对称轴是经过顶点的竖直直线。根（或x截距）是方程 ax² + bx + c = 0 的解。

Axis of symmetry: x = −b ⁄ (2a)

对称轴：x = −b ⁄ (2a)

Vertex: ( −b ⁄ (2a), f(−b ⁄ (2a)) )

顶点：( −b ⁄ (2a), f(−b ⁄ (2a)) )

For example, in f(x) = x² − 4x + 3, we have a = 1, b = −4, c = 3. The axis of symmetry is x = 2. The vertex is at (2, −1). The roots are x = 1 and x = 3.

例如，在 f(x) = x² − 4x + 3 中，a = 1，b = −4，c = 3。对称轴是 x = 2，顶点在 (2, −1)，根是 x = 1 和 x = 3。

3. Solving Quadratic Equations by Factorization | 用因式分解法解二次方程

Factorization is the fastest method when the quadratic has simple integer roots. We rearrange the equation into the form ax² + bx + c = 0, then write the expression as a product of two binomials. For a monic quadratic (a = 1), we look for two numbers that multiply to give c and add to give b.

当二次方程有简单的整数根时，因式分解是最快的方法。我们将方程整理成 ax² + bx + c = 0 的形式，然后把表达式写成两个二项式的乘积。对于首一二次多项式（a = 1），我们寻找两个数，它们相乘等于 c，相加等于 b。

Solve: x² + 5x + 6 = 0. We need two numbers that multiply to 6 and add to 5. These are 2 and 3. So (x + 2)(x + 3) = 0. Therefore x = −2 or x = −3.

解：x² + 5x + 6 = 0。我们需要两个数相乘得6，相加得5。这两个数是2和3。所以 (x + 2)(x + 3) = 0。因此 x = −2 或 x = −3。

For non-monic quadratics, we can use the method of grouping. For example, 2x² + 5x + 2 = 0. Multiply a and c: 2 × 2 = 4. Find two numbers that multiply to 4 and add to 5: 1 and 4. Rewrite the middle term: 2x² + x + 4x + 2. Group: x(2x + 1) + 2(2x + 1) = (2x + 1)(x + 2) = 0. Thus x = −1⁄2 or x = −2.

对于非首一二次方程，我们可以使用分组法。例如：2x² + 5x + 2 = 0。将 a 和 c 相乘：2 × 2 = 4。找到相乘得4且相加得5的两个数：1和4。重写中间项：2x² + x + 4x + 2。分组：x(2x + 1) + 2(2x + 1) = (2x + 1)(x + 2) = 0。因此 x = −1⁄2 或 x = −2。

4. Solving by Completing the Square | 用配方法解方程

Completing the square is a powerful technique that works for any quadratic equation. It transforms ax² + bx + c into a perfect square plus a constant. For a = 1, we take half of b and square it.

配方是一种强大的技巧，适用于任何二次方程。它把 ax² + bx + c 转化为一个完全平方加上一个常数。当 a = 1 时，我们取 b 的一半并平方。

Solve x² + 6x + 2 = 0. First write x² + 6x. Half of 6 is 3, and 3² = 9. So we add and subtract 9: (x² + 6x + 9) − 9 + 2 = 0. This becomes (x + 3)² − 7 = 0. Then (x + 3)² = 7, so x + 3 = ±√7. Thus x = −3 ± √7.

解 x² + 6x + 2 = 0。先写 x² + 6x。6的一半是3，3² = 9。所以我们加上再减去9：(x² + 6x + 9) − 9 + 2 = 0。这变成 (x + 3)² − 7 = 0。然后 (x + 3)² = 7，所以 x + 3 = ±√7。因此 x = −3 ± √7。

If a ≠ 1, first factor out a from the first two terms. For 2x² − 8x + 5 = 0, write 2(x² − 4x) + 5 = 0. Complete the square inside: 2[(x − 2)² − 4] + 5 = 0. Expand: 2(x − 2)² − 8 + 5 = 0. Then 2(x − 2)² = 3, so (x − 2)² = 3⁄2. Therefore x = 2 ± √(3⁄2).

如果 a ≠ 1，先从前三项中提取 a。对于 2x² − 8x + 5 = 0，写为 2(x² − 4x) + 5 = 0。在括号内配方：2[(x − 2)² − 4] + 5 = 0。展开：2(x − 2)² − 8 + 5 = 0。然后 2(x − 2)² = 3，所以 (x − 2)² = 3⁄2。因此 x = 2 ± √(3⁄2)。

5. The Quadratic Formula | 二次求根公式

For any quadratic equation ax² + bx + c = 0, the solutions are given by the quadratic formula. This formula works for all cases, including when factorization is difficult or impossible.

对于任何二次方程 ax² + bx + c = 0，解都可以由二次求根公式给出。这个公式适用于所有情况，包括因式分解困难或无法分解的情形。

x = [ −b ± √(b² − 4ac) ] ⁄ (2a)

Let us solve 3x² + 2x − 5 = 0. Here a = 3, b = 2, c = −5. Substitute into the formula: x = [ −2 ± √(4 − 4 × 3 × (−5)) ] ⁄ 6 = [ −2 ± √(4 + 60) ] ⁄ 6 = [ −2 ± √64 ] ⁄ 6 = (−2 ± 8) ⁄ 6. So x = 6⁄6 = 1 or x = −10⁄6 = −5⁄3.

我们来解 3x² + 2x − 5 = 0。这里 a = 3，b = 2，c = −5。代入公式：x = [ −2 ± √(4 − 4 × 3 × (−5)) ] ⁄ 6 = [ −2 ± √(4 + 60) ] ⁄ 6 = [ −2 ± √64 ] ⁄ 6 = (−2 ± 8) ⁄ 6。所以 x = 6⁄6 = 1 或 x = −10⁄6 = −5⁄3。

Remember to simplify your surds and fractions fully. In an exam, always check whether you have written the answer in the required form, such as exact values or decimals.

记得要完全化简根式和分数。考试中，始终检查答案是否以要求的形式书写，例如精确值或小数。

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

The expression b² − 4ac is called the discriminant, usually written as Δ. It tells us how many and what type of roots the quadratic equation has, without solving it fully.

表达式 b² − 4ac 称为判别式，通常记为 Δ。它能告诉我们二次方程有多少个根以及根的类型，而不需要完整求解。

If Δ > 0, the equation has two distinct real roots. The graph crosses the x-axis at two points.
如果 Δ > 0，方程有两个不相等的实数根，图像与x轴有两个交点。
If Δ = 0, the equation has one repeated real root. The graph touches the x-axis at the vertex.
如果 Δ = 0，方程有一个重复的实数根，图像在顶点处与x轴相切。
If Δ < 0, the equation has no real roots. The graph does not touch the x-axis.
如果 Δ < 0，方程没有实数根，图像不与x轴相交。

For example, x² + 4x + 5 = 0 has Δ = 16 − 20 = −4 < 0, so it has no real solutions. Its graph lies entirely above the x-axis because a > 0.

例如，x² + 4x + 5 = 0 中 Δ = 16 − 20 = −4 < 0，所以没有实数解。由于 a > 0，它的图像完全位于x轴上方。

7. Graphing Quadratic Functions | 二次函数图像的绘制

To sketch a quadratic graph, follow these steps: determine the direction (using a), find the axis of symmetry and vertex, find the y-intercept (c), and find the roots if they exist. Then plot these points and draw a smooth curve.

绘制二次图像时按以下步骤：确定开口方向（看a）、找对称轴和顶点、找y截距（c）、找根（如果存在）。然后描出这些点并画一条平滑的曲线。

Take f(x) = x² − 2x − 3. Here a = 1 > 0, so it opens upward. The axis of symmetry is x = −(−2)⁄(2×1) = 1. The vertex is at x = 1, f(1) = 1 − 2 − 3 = −4. The y-intercept is (0, −3). Solving x² − 2x − 3 = 0 gives (x − 3)(x + 1) = 0, so roots are x = 3 and x = −1.

取 f(x) = x² − 2x − 3。这里 a = 1 > 0，所以开口向上。对称轴是 x = −(−2)⁄(2×1) = 1。顶点在 x = 1，f(1) = 1 − 2 − 3 = −4。y截距是 (0, −3)。解 x² − 2x − 3 = 0 得 (x − 3)(x + 1) = 0，所以根是 x = 3 和 x = −1。

Plot these four points: (−1, 0), (3, 0), (0, −3), and (1, −4). Connect them with a smooth U-shaped curve. The vertex is the minimum point.

描出这四个点：(−1, 0)、(3, 0)、(0, −3) 和 (1, −4)。用一条平滑的U形曲线连接它们。顶点是最小值点。

8. Transformations of Quadratic Graphs | 二次函数图像的变换

Graphs of quadratic functions can be transformed by changing parameters. If we write f(x) = a(x − h)² + k, the vertex is at (h, k). This is called the vertex form.

二次函数的图像可以通过改变参数来变换。如果写成 f(x) = a(x − h)² + k，顶点在 (h, k)。这称为顶点式。

Transformation	Effect on Graph
Change a (e.g., negative, larger)	Flips or stretches the parabola vertically
Change h in (x − h)²	Shifts the graph horizontally (left/right)
Change k	Shifts the graph vertically (up/down)

For example, y = x² is shifted 3 units left and 2 units up to give y = (x + 3)² + 2. Its vertex moves from (0, 0) to (−3, 2).

例如，y = x² 向左平移3个单位、向上平移2个单位得到 y = (x + 3)² + 2。顶点从 (0, 0) 移动到 (−3, 2)。

9. Applications in Real Life | 在实际生活中的应用

Quadratic functions model many real-world situations. The path of a thrown ball, the shape of a suspension bridge cable, and the profit of a business can often be described by quadratic equations.

二次函数可以模拟许多现实情境。投掷球的轨迹、吊桥缆绳的形状以及企业的利润，通常可以用二次方程来描述。

For instance, if a stone is thrown upward with height h(t) = −5t² + 20t + 1 (in metres), then the maximum height can be found by locating the vertex. Here t = −b⁄(2a) = −20⁄(2 × (−5)) = 2 seconds. The maximum height is h(2) = −5(4) + 40 + 1 = 21 metres.

例如，如果一块石头向上抛出，其高度为 h(t) = −5t² + 20t + 1（单位：米），则可以通过找到顶点来确定最大高度。这里 t = −b⁄(2a) = −20⁄(2 × (−5)) = 2 秒。最大高度是 h(2) = −5(4) + 40 + 1 = 21 米。

Similarly, to find when the stone hits the ground, set h(t) = 0 and solve the quadratic equation. Only positive t values are meaningful.

类似地，要找出石头何时落地，令 h(t) = 0 并解二次方程。只有正的 t 值才有意义。

10. Common Exam Mistakes and Tips | 常见考试错误与提示

One common mistake is forgetting that a quadratic equation can have two solutions. When you take the square root of both sides, you must include the plus-or-minus sign (±).

一个常见错误是忘记二次方程可能有两个解。当你对方程两边同时开平方时，必须加上正负号（±）。

Another mistake is misidentifying the values of a, b, and c when the equation is not written in standard form. Always rearrange into ax² + bx + c = 0 before applying the formula.

另一个错误是当方程没有写成标准形式时，错误识别 a、b、c 的值。在应用公式之前，务必整理成 ax² + bx + c = 0 的形式。

Always check if a = 1 before attempting simple factorization.
求解前先检查是否 a = 1，再尝试简单因式分解。
When completing the square, remember to add or subtract the same value inside and outside the bracket.
配方时，记住在括号内外加上或减去相同的值。
Read the question carefully: it may ask for roots, turning point, or intercepts.
仔细读题：可能会要求根、顶点或截距。

In graphing questions, label the vertex and axis of symmetry clearly. Use a ruler to draw the axis and write the coordinates of key points.

在作图题中，请清楚标注顶点和对称轴。用直尺画出对称轴，并写出关键点的坐标。

11. Practice Questions | 练习题目

Try these exam-style questions to test your understanding. Solutions are shown after each question.

尝试以下考试风格题目来测试你的理解。每个题目后面给出解答。

Question 1: Solve x² − 3x − 10 = 0 by factorization.

题目1：用因式分解法解 x² − 3x − 10 = 0。

Solution: x² − 3x − 10 = (x − 5)(x + 2) = 0, so x = 5 or x = −2.

解答：x² − 3x − 10 = (x − 5)(x + 2) = 0，所以 x = 5 或 x = −2。

Question 2: Find the discriminant of 2x² + 3x + 1 = 0 and state the nature of the roots.

题目2：求 2x² + 3x + 1 = 0 的判别式，并说明根的性质。

Solution: Δ = 3² − 4 × 2 × 1 = 9 − 8 = 1 > 0, so there are two distinct real roots.

解答：Δ = 3² − 4 × 2 × 1 = 9 − 8 = 1 > 0，所以有两个不相等的实数根。

Question 3: Write y = x² − 6x + 11 in the form y = (x − h)² + k and state the vertex.

题目3：将 y = x² − 6x + 11 写成 y = (x − h)² + k 的形式，并指出顶点。

Solution: y = (x² − 6x + 9) + 2 = (x − 3)² + 2. The vertex is (3, 2).

解答：y = (x² − 6x + 9) + 2 = (x − 3)² + 2。顶点是 (3, 2)。

12. Summary | 总结

In this lesson, we explored quadratic functions from every angle. We learned the standard form, the vertex, the axis of symmetry, and the roots. We practiced solving quadratics by factorization, completing the square, and using the quadratic formula. We also examined the discriminant and the effect of transformations.

本课我们从各个角度探索了二次函数。我们学习了标准形式、顶点、对称轴和根。我们练习了因式分解法、配方法和二次公式来解二次方程。我们还研究了判别式和图像变换的影响。

Mastering these skills will help you solve many problems in IGCSE Mathematics. Remember to practise regularly and check your answers with a graph or substitution.

掌握这些技巧将帮助你在IGCSE数学中解决许多问题。记住要定期练习，并通过图像或代入法检查你的答案。

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Quadratic Functions: Solving Equations and Graphing | 二次函数：方程求解与图像