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Quadratic Functions and Graphs | 二次函数与图像

2026-08-30 04:56 · 本文部分内容由AI辅助创作

📚 Quadratic Functions and Graphs | 二次函数与图像

Quadratic functions are one of the core topics in IGCSE Mathematics. This article explains how to solve quadratic equations, sketch their graphs, and apply key techniques to exam-style problems.

二次函数是IGCSE数学的核心内容之一。本文将讲解如何求解二次方程、绘制其图像,并运用关键技巧解答考试型题目。


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

A quadratic function is written as f ants and a ≠ 0. The graph of such a function is a parabola. If a > 0, the parabola opens upwards; if a < 0, it opens downwards.

二次函数写作 f(x) = ax² + bx + c,其中 a、b、c 为常数且 a ≠ 0。此类函数的图像是一条抛物线。若 a > 0,抛物线开口向上;若 a < 0,开口向下。

f(x) = ax² + bx + c , a ≠ 0


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

or example, solve x² − 5x + 6 = 0. We factor to (x − 2)(x − 3) = 0, giving x = 2 or x = 3.

当二次方程可写成两个一次因式相乘时,令每个因式等于零。例如解 x² − 5x + 6 = 0,因式分解为 (x − 2)(x − 3) = 0,得 x = 2 或 x = 3。

x² − 5x + 6 = (x − 2)(x − 3) = 0 → x = 2, x = 3


3. Completing the Square | 配方法

+ 6x + 10, half of 6 is 3, so (x + 3)² − 9 + 10 = (x + 3)² + 1. The vertex is at (−3, 1).

配方法将二次式改写为 a(x + p)² + q。对于 y = x² + 6x + 10,6 的一半是 3,所以 (x + 3)² − 9 + 10 = (x + 3)² + 1。顶点为 (−3, 1)。

x² + 6x + 10 = (x + 3)² + 1


4. The Quadratic Formula | 二次公式

For ax² + bx + c = 0, the solutions are given by x = (−b ± √b² − 4ac) / (2a). This formula works for all quadratic equations, including those that cannot be factored.

对于 ax² + bx + c = 0,解为 x = (−b ± √b² − 4ac) / (2a)。该公式适用于所有二次方程,包括无法因式分解的情况。

x = (−b ± √b² − 4ac) / (2a)

For example, 2x² + 3x − 2 = 0 gives x = (−3 ± √9 + 16) / 4 = (−3 ± 5) / 4, so x = 0.5 or x = −2.

例如,2x² + 3x − 2 = 0 得 x = (−3 ± √25) / 4 = (−3 ± 5) / 4,故 x = 0.5 或 x = −2。


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

The expression b² − 4ac is called the discriminant, written as Δ. Its sign tells us how many real roots the equation has.

表达式 b² − 4ac 称为判别式,记作 Δ。它的符号决定方程有多少实根。

Δ = b² − 4acNumber of real roots实根个数
Δ > 0Two distinct roots两个不同的实根
Δ = 0One repeated root一个重根
Δ < 0No real roots没有实根

If Δ is a perfect square and a, b, c are rational, the equation can also be solved by factorisation.

若 Δ 是一个完全平方数且 a、b、c 为有理数,则方程也可用因式分解法求解。


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6. Sketching the Graph of a Quadratic | 绘制二次函数图像

ing ax² + bx + c = 0, and the turning point by completing the square or using x = −b/(2a).

绘制 y = ax² + bx + c 的草图,需要找出 y 轴截距 (0, c),通过解 ax² + bx + c = 0 找 x 轴截距,并通过配方或用 x = −b/(2a) 找顶点。

Turning point x-coordinate: x = −b/(2a)


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

s shifted horizontally by h and vertically by k. The sign of a determines the orientation.

从配方式 y = a(x − h)² + k 可以看出,y = x² 的图像水平平移 h,垂直平移 k。a 的符号决定开口方向。


8. Solving Quadratic Inequalities | 解二次不等式

To solve ax² + bx + c > 0 or < 0, first find the roots, then sketch the parabola to decide the intervals where the inequality holds.

解 ax² + bx + c > 0 或 < 0,先求根,再画抛物线草图判断满足不等式的区间。

For example, x² − x − 2 < 0 factors as (x − 2)(x + 1) < 0. The roots are 2 and −1. Since the parabola opens upwards, the solution is −1 < x < 2.

例如,x² − x − 2 < 0 因式分解为 (x − 2)(x + 1) < 0。根为 2 和 −1。因为抛物线开口向上,所以解为 −1 < x < 2。

(x − 2)(x + 1) < 0 → −1 < x < 2


9. Quadratic Functions in Real-World Problems | 二次函数在实际问题中的应用

Quadratic functions model projectile motion, area optimization, and profit maximisation. In such problems, the vertex often represents the maximum or minimum value.

二次函数常用来模拟抛体运动、面积优化和利润最大化。在这类问题中,顶点通常代表最大值或最小值。

he area is A = x(50 − x) = 50x − x². The maximum area occurs at x = 25, giving A = 625 m².

例如:一块矩形场地周长为 100 m。若一边长为 x m,则面积 A = x(50 − x) = 50x − x²。最大面积出现在 x = 25,此时 A = 625 m²。


10. Common Exam Mistakes | 常见考试错误


11. Practice Questions | 练习题目

1. Solve x² − 4x − 5 = 0 by factorisation.

1. 用因式分解法解 x² − 4x − 5 = 0。

2. Find the coordinates of the turning point of y = 2x² − 8x + 7.

2. 求 y = 2x² − 8x + 7 的顶点坐标。

3. Determine the number of real roots of 3x² + 2x + 5 = 0.

3. 判断 3x² + 2x + 5 = 0 的实根个数。


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