Understanding Quadratic Functions and Their Graphs | 理解二次函数及其图像

📚 Understanding Quadratic Functions and Their Graphs | 理解二次函数及其图像

A quadratic function is one of the central topics in IGCSE Mathematics. It connects algebra, coordinate geometry and problem solving, and it appears in many real-world situations such as projectile motion, area optimisation and profit modelling. In this article, you will learn how to recognise a quadratic function, sketch its graph, find its vertex and intercepts, use completing the square, and apply transformations to sketch new curves.

二次函数是 IGCSE 数学的核心内容之一，它把代数、坐标几何与问题解决联系起来，并出现在许多现实情境中，例如抛体运动、面积最优化和利润建模。在本文中，你将学习如何识别二次函数、画出它的图像、求顶点与截距、使用配方法，以及运用变换来绘制新曲线。

1. What Is a Quadratic Function? | 什么是二次函数？

A quadratic function is a polynomial of degree 2. Its general form is:

f(x) = ax² + bx + c, a ≠ 0

Here a, b and c are real numbers. The coefficient a cannot be zero, because if a = 0, the function becomes linear. The graph of a quadratic function is a smooth curve called a parabola.

这里的 a、b 和 c 都是实数。系数 a 不能为零，因为如果 a = 0，函数就变成了一次函数。二次函数的图像是一条平滑的曲线，称为抛物线。

2. Shape and Direction of a Parabola | 抛物线的形状与开口方向

The sign of a determines whether the parabola opens upward or downward, and the size of a controls how wide or narrow the curve is.

a 的正负决定了抛物线开口向上还是向下，而 a 的大小控制着曲线的宽窄。

If a > 0, the parabola opens upward and it has a minimum point.
如果 a > 0，抛物线开口向上，并且有最小值点。
If a < 0, the parabola opens downward and it has a maximum point.
如果 a < 0，抛物线开口向下，并且有最大值点。
A larger value of |a| makes the curve narrower; a smaller value of |a| makes it wider.
|a| 的值越大，曲线越狭窄；|a| 的值越小，曲线越宽阔。

a > 0	Opens upward, like a cup, and has a minimum value.
a < 0	Opens downward, like a cap, and has a maximum value.

For the graph of f(x) = x², the curve is symmetric about the y-axis and its minimum point is at the origin (0, 0).

对于 f(x) = x² 的图像，曲线关于 y 轴对称，其最小值点在原点 (0, 0) 处。

3. Vertex and Axis of Symmetry | 顶点与对称轴

Every parabola has a vertical line of symmetry. For f(x) = ax² + bx + c, the axis of symmetry is given by:

x = −b/(2a)

The vertex is the turning point of the parabola. Its x-coordinate is the same as the axis of symmetry, and its y-coordinate is found by substituting this value into the function:

顶点是抛物线的转折点。它的 x 坐标与对称轴相同，y 坐标则通过把该值代入函数求得：

Vertex = (−b/(2a), f(−b/(2a)))

If a > 0, this vertex is the minimum point; if a < 0, it is the maximum point.

如果 a > 0，该顶点就是最低点；如果 a < 0，该顶点就是最高点。

4. Intersections with the Axes | 与坐标轴的交点

To find the y-intercept, set x = 0. Then f(0) = c, so the y-intercept is (0, c).

求 y 截距时，令 x = 0。此时 f(0) = c，所以 y 截距为 (0, c)。

To find the x-intercepts, set f(x) = 0 and solve the equation ax² + bx + c = 0. The number of real solutions depends on the discriminant:

求 x 截距时，令 f(x) = 0，并解方程 ax² + bx + c = 0。实数解的个数由判别式决定：

Δ = b² − 4ac

If Δ > 0, there are two distinct x-intercepts.
如果 Δ > 0，有两个不同的 x 截距。
If Δ = 0, there is one repeated x-intercept, so the vertex lies on the x-axis.
如果 Δ = 0，有一个重根 x 截距，因此顶点位于 x 轴上。
If Δ < 0, there are no real x-intercepts, and the curve does not cross the x-axis.
如果 Δ < 0，没有实数 x 截距，曲线不穿过 x 轴。

Δ > 0	Two real roots; the curve cuts the x-axis twice.
Δ = 0	One repeated root; the curve touches the x-axis.
Δ < 0	No real roots; the curve does not meet the x-axis.

5. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form:

配方法将二次函数改写为以下形式：

f(x) = a(x − h)² + k

Once the function is in this form, the vertex is simply (h, k), and the axis of symmetry is x = h.

一旦函数写成这种形式，顶点就是 (h, k)，对称轴就是 x = h。

For example, take f(x) = x² − 6x + 11. Half of −6 is −3, so:

例如，取 f(x) = x² − 6x + 11。−6 的一半是 −3，所以：

f(x) = (x − 3)² − 9 + 11 = (x − 3)² + 2

Therefore the minimum point is (3, 2) and the axis of symmetry is x = 3.

因此最低点是 (3, 2)，对称轴是 x = 3。

When a ≠ 1, factor out a from the first two terms before completing the square. This is a common IGCSE skill that helps you sketch graphs and solve maximum or minimum problems quickly.

当 a ≠ 1 时，应先把前两项的公因数 a 提出来，再配方。这是 IGCSE 常见技能，可以帮助你快速画图和解决最大值或最小值问题。

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

You can sketch new quadratic graphs by applying simple transformations to y = x². These transformations also apply to any function y = f(x).

你可以通过对 y = x² 进行简单变换来画出新的二次函数图像。这些变换也适用于任意函数 y = f(x)。

y = f(x) + k: translate the graph upward by k units.

y = f(x) + k：将图像向上平移 k 个单位。
y = f(x) − k: translate the graph downward by k units.

y = f(x) − k：将图像向下平移 k 个单位。
y = f(x + k): translate the graph to the left by k units.

y = f(x + k)：将图像向左平移 k 个单位。
y = f(x − k): translate the graph to the right by k units.

y = f(x − k)：将图像向右平移 k 个单位。
y = −f(x): reflect the graph in the x-axis.

y = −f(x)：将图像关于 x 轴反射。
y = f(−x): reflect the graph in the y-axis.

y = f(−x)：将图像关于 y 轴反射。

Notice that y = f(x + k) moves the graph left, not right. This sign rule is a common source of errors in examinations.

注意 y = f(x + k) 使图像向左移动，而不是向右移动。这个符号规则是考试中常见的错误来源。

7. Finding the Equation of a Quadratic from Its Graph | 由图像求二次函数的表达式

If you know the vertex of a parabola and one other point, you can find the equation in vertex form.

如果知道抛物线的顶点和另一个点的坐标，就可以用顶点式求出函数表达式。

Suppose the vertex is (2, −3) and the curve passes through (0, 5). Since the vertex is (h, k), we write:

假设顶点为 (2, −3)，并且曲线经过点 (0, 5)。因为顶点是 (h, k)，我们写成：

y = a(x − 2)² − 3

Substitute x = 0 and y = 5 into the equation:

把 x = 0 和 y = 5 代入方程：

5 = a(0 − 2)² − 3

5 = 4a − 3

4a = 8, so a = 2

Therefore the equation is y = 2(x − 2)² − 3. Expanding this gives the standard form y = 2x² − 8x + 5.

因此函数表达式为 y = 2(x − 2)² − 3。展开后得到标准形式 y = 2x² − 8x + 5。

8. Worked Examination Example | 考试例题解析

The function f is defined by f(x) = 2x² − 8x + 5. Write f(x) in the form a(x − h)² + k, state the minimum value, and find the y-intercept.

已知函数 f(x) = 2x² − 8x + 5。请将 f(x) 写成 a(x − h)² + k 的形式，说明最小值，并求 y 截距。

First, factor out 2 from the first two terms:

首先，把 2 从前两项中提出来：

f(x) = 2(x² − 4x) + 5

Complete the square inside the brackets. Half of −4 is −2, so:

在括号内配方。−4 的一半是 −2，所以：

x² − 4x = (x − 2)² − 4

Then continue:

然后继续计算：

f(x) = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3

Now the vertex form is f(x) = 2(x − 2)² − 3. Since a = 2 > 0, the graph opens upward and has a minimum value of −3 when x = 2.

现在顶点式为 f(x) = 2(x − 2)² − 3。因为 a = 2 > 0，图像开口向上，当 x = 2 时取得最小值 −3。

The y-intercept is found by setting x = 0:

求 y 截距时令 x = 0：

f(0) = 0 − 0 + 5 = 5

So the curve crosses the y-axis at (0, 5).

所以曲线与 y 轴交于点 (0, 5)。

9. Common Mistakes and Tips | 常见错误与提示

Many students lose marks on quadratic questions because of small sign errors or missing steps. Here are the most common pitfalls.

许多学生在二次函数问题上失分，原因往往是小符号错误或漏写步骤。以下是最常见的陷阱。

Using the wrong sign in the axis of symmetry formula. Remember that the axis is x = −b/(2a), not x = b/(2a).

对称轴公式的符号用错。记住对称轴是 x = −b/(2a)，而不是 x = b/(2a)。
Forgetting to factor out a before completing the square. This leads to an incorrect vertex.

配方法前忘记提出 a。这会导致顶点错误。
Confusing horizontal translations. y = f(x + k) moves the graph left, not right.

混淆水平平移。y = f(x + k) 使图像向左移动，而不是向右移动。
Assuming every quadratic has two x-intercepts. Always check the discriminant Δ = b² − 4ac.

假设每个二次函数都有两个 x 截距。一定要检查判别式 Δ = b² − 4ac。
When using the quadratic formula, write the equation in the form ax² + bx + c = 0 before reading a, b and c.

使用求根公式前，先把方程写成 ax² + bx + c = 0 的形式，再确定 a、b、c。

10. Summary and Practice | 总结与练习

The quadratic function f(x) = ax² + bx + c is one of the most heavily tested topics in IGCSE Mathematics.

二次函数 f(x) = ax² + bx + c 是 IGCSE 数学中考查最频繁的内容之一。

The graph is a parabola. It opens upward when a > 0 and downward when a < 0.

图像是抛物线。当 a > 0 时开口向上，当 a < 0 时开口向下。
The axis of symmetry is x = −b/(2a), and the vertex lies on this line.

对称轴是 x = −b/(2a)，顶点位于这条直线上。
Completing the square gives the minimum or maximum point directly.

配方法可以直接给出最小值点或最大值点。
The discriminant Δ = b² − 4ac tells you how many x-intercepts the graph has.

判别式 Δ = b² − 4ac 可以告诉你图像与 x 轴有多少个交点。

Try these quick practice questions. First, for f(x) = x² − 4x + 3, find the vertex and the x-intercepts. Second, describe the transformation from y = x² to y = (x + 1)² − 2. The vertex is (2, −1), and the x-intercepts are 1 and 3; the graph is translated 1 unit left and 2 units down.

请快速完成以下练习。第一题：对 f(x) = x² − 4x + 3，求顶点和 x 截距。第二题：描述从 y = x² 到 y = (x + 1)² − 2 的变换。第一题答案是顶点为 (2, −1)，x 截距为 1 和 3；第二题答案是图像向左平移 1 个单位，再向下平移 2 个单位。

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Understanding Quadratic Functions and Their Graphs | 理解二次函数及其图像