Using Graphs to Solve Quadratic Equations | 使用图像求解二次方程

📚 Using Graphs to Solve Quadratic Equations | 使用图像求解二次方程

Quadratic equations are a core topic in the Edexcel IGCSE Mathematics syllabus. While algebraic methods such as factorisation, completing the square, and the quadratic formula are powerful, the graphical method offers a visual and often quicker way to find approximate solutions. This article explains how to use graphs to solve quadratic equations of the form ax² + bx + c = 0, and how to solve more complex equations by drawing appropriate lines on the same axes.

二次方程是 Edexcel IGCSE 数学大纲中的核心内容。因式分解、配方法和二次公式等代数方法虽然强大，但图像法提供了一种直观且往往更快捷的方式来寻找近似解。本文讲解如何使用图像求解形如 ax² + bx + c = 0 的二次方程，以及如何通过在同一个坐标系中绘制合适的直线来解决更复杂的方程。

1. The Quadratic Graph: y = ax² + bx + c | 二次函数图像：y = ax² + bx + c

A quadratic function in x has the general form y = ax² + bx + c, where a ≠ 0. When plotted on a coordinate grid, its graph is always a smooth curve called a parabola. If a > 0, the parabola opens upwards like a ‘U’ shape; if a < 0, it opens downwards like an 'n' shape. The simplest quadratic graph is y = x², a U-shaped curve with its vertex (turning point) at the origin (0, 0).

二次函数的一般形式为 y = ax² + bx + c，其中 a ≠ 0。当在坐标网格上绘制时，其图像总是一条平滑的曲线，称为抛物线。如果 a > 0，抛物线开口朝上，呈字母 ‘U’ 形；如果 a < 0，抛物线开口朝下，呈字母 'n' 形。最简单的二次函数图像是 y = x²，这是一条以原点 (0, 0) 为顶点（转向点）的 U 形曲线。

Before solving equations graphically, you must be able to plot the graph accurately. You are usually given a table of x-values, and you calculate the corresponding y-values by substituting each x into the quadratic expression. Plot the points carefully and join them with a smooth, continuous curve — never use straight line segments between the points.

在通过图像解方程之前，你必须能够准确地绘制二次函数图像。通常会给你一个 x 值表格，你通过将每个 x 代入二次表达式中计算对应的 y 值。仔细描点，并用平滑连续的曲线连接各点——切勿用直线段连接相邻点。

2. Key Features of a Parabola | 抛物线的关键特征

Every parabola has three important features that help us solve equations. The roots (or zeros) are the x-coordinates where the curve crosses the x-axis; at these points, y = 0. The line of symmetry passes vertically through the vertex with equation x = -b/(2a); it divides the parabola into two mirror-image halves. The vertex is the turning point (maximum or minimum) of the parabola, and its x-coordinate is also -b/(2a); the y-coordinate is found by substituting this x back into the equation.

每条抛物线都有三个帮助我们解方程的重要特征。根（或零点）是曲线与 x 轴交点的 x 坐标；在这些点上，y = 0。对称轴垂直穿过顶点，其方程为 x = -b/(2a)；它将抛物线分为两个镜像对称的部分。顶点是抛物线的转向点（最大值或最小值），其 x 坐标同样是 -b/(2a)；将该 x 代回方程即可求出 y 坐标。

For example, for y = x² – 4x + 3, the line of symmetry is x = 2 and the vertex is at (2, -1). The roots are x = 1 and x = 3, because (x – 1)(x – 3) = 0. Observing these features on a sketched graph immediately tells you the solutions of x² – 4x + 3 = 0.

例如，对于 y = x² – 4x + 3，对称轴是 x = 2，顶点是 (2, -1)。根是 x = 1 和 x = 3，因为 (x – 1)(x – 3) = 0。在绘制的图像上观察这些特征，立刻就能知道 x² – 4x + 3 = 0 的解。

3. Solving ax² + bx + c = 0: Reading the x-Intercepts | 解 ax² + bx + c = 0：读取 x 轴交点

The most direct method for solving a quadratic equation graphically is to plot the graph of y = ax² + bx + c and read the x-coordinates where the curve crosses the x-axis (where y = 0). These x-values are exactly the solutions of the equation ax² + bx + c = 0.

图像法解二次方程最直接的方法是：绘制 y = ax² + bx + c 的图像，读取曲线与 x 轴交点的 x 坐标（此时 y = 0）。这些 x 值恰好就是方程 ax² + bx + c = 0 的解。

Consider the equation x² – 2x – 3 = 0. Plot y = x² – 2x – 3 using a table of values. Substituting x = -2 gives (-2)² – 2(-2) – 3 = 4 + 4 – 3 = 5; x = -1 gives 1 + 2 – 3 = 0; x = 0 gives -3; x = 1 gives -4; x = 2 gives -3; x = 3 gives 0; x = 4 gives 5. The curve crosses the x-axis at x = -1 and x = 3, so the solutions are x = -1 or x = 3. This agrees perfectly with the factorised form (x + 1)(x – 3) = 0.

考虑方程 x² – 2x – 3 = 0。用数值表绘制 y = x² – 2x – 3。将 x = -2 代入得 (-2)² – 2(-2) – 3 = 4 + 4 – 3 = 5；x = -1 得 1 + 2 – 3 = 0；x = 0 得 -3；x = 1 得 -4；x = 2 得 -3；x = 3 得 0；x = 4 得 5。曲线在 x = -1 和 x = 3 处穿过 x 轴，因此解为 x = -1 或 x = 3。这与因式分解形式 (x + 1)(x – 3) = 0 完全一致。

4. Three Possible Cases for the Roots | 根的三种可能情况

When you draw a quadratic graph, the number of x-intercept points tells you how many real roots the corresponding equation has. There are exactly three possibilities, and being able to predict them from the discriminant Δ = b² – 4ac is very useful for exam questions.

当你绘制二次函数图像时，x 轴交点的数目告诉你相应方程有多少个实数根。恰好有三种可能情况，能够从判别式 Δ = b² – 4ac 预测它们对考试题目非常有用。

Two distinct real roots: The parabola crosses the x-axis at two different points. This occurs when Δ > 0.

两个不同的实数根：抛物线与 x 轴在两个不同点相交。当 Δ > 0 时出现。
One repeated real root: The parabola just touches the x-axis at its vertex. This occurs when Δ = 0.

一个重根：抛物线仅在顶点处接触 x 轴。当 Δ = 0 时出现。
No real roots: The parabola does not intersect the x-axis at all; it lies entirely above (a > 0) or entirely below (a < 0) the axis. This occurs when Δ < 0.

没有实数根：抛物线完全不与 x 轴相交；它完全位于 x 轴上方（a > 0）或完全位于下方（a < 0）。当 Δ < 0 时出现。

For instance, y = x² – 6x + 9 = (x – 3)² touches the x-axis at x = 3 only; the equation x² – 6x + 9 = 0 has one repeated root. By contrast, y = x² + x + 1 has Δ = 1 – 4 = -3 < 0, so its graph never reaches the x-axis and the equation has no real solutions.

例如，y = x² – 6x + 9 = (x – 3)² 仅在 x = 3 处接触 x 轴；方程 x² – 6x + 9 = 0 有一个重根。相比之下，y = x² + x + 1 的判别式 Δ = 1 – 4 = -3 < 0，所以其图像永远达不到 x 轴，方程没有实数解。

5. Solving x² + bx + c = k: Adding a Horizontal Line | 解 x² + bx + c = k：添加水平直线

Sometimes the equation you need to solve is not in the standard ‘= 0’ form, such as x² – 2x – 3 = 2. You can still use the same plotted curve y = x² – 2x – 3. Simply draw the horizontal line y = k (here k = 2) on the same axes. The x-coordinates of the points where this line intersects the parabola are the solutions of the equation.

有时你需要解的方程不是标准的 ‘= 0’ 形式，例如 x² – 2x – 3 = 2。你仍然可以使用已经绘制的曲线 y = x² – 2x – 3。只需在同一坐标轴上画水平直线 y = k（此处 k = 2）。该直线与抛物线交点处的 x 坐标就是方程的解。

Using the curve y = x² – 2x – 3 from Section 3, draw the line y = 2. Reading the intersections gives approximately x ≈ -1.8 and x ≈ 3.8. Let us verify algebraically: x² – 2x – 3 = 2 → x² – 2x – 5 = 0 → x = [2 ± √(4 + 20)]/2 = 1 ± √6 ≈ -1.45 and 3.45. The graph gives close approximations; the accuracy depends on the scale of your axes.

使用第 3 节中的曲线 y = x² – 2x – 3，画直线 y = 2。读取交点可得 x ≈ -1.8 和 x ≈ 3.8。让我们用代数验证：x² – 2x – 3 = 2 → x² – 2x – 5 = 0 → x = [2 ± √(4 + 20)]/2 = 1 ± √6 ≈ -1.45 和 3.45。图像给出接近的近似值；精度取决于坐标轴的比例。

6. Solving ax² + bx + c = mx + n: Intersection of Curve and Line | 解 ax² + bx + c = mx + n：曲线与直线的交点

To solve an equation of the form ax² + bx + c = mx + n, where the right-hand side is a linear expression, you can plot both y = ax² + bx + c and y = mx + n on the same axes. The x-coordinates of their intersection points satisfy both equations simultaneously, hence they are the solutions of the original quadratic equation.

要解形如 ax² + bx + c = mx + n 的方程（右边是一次表达式），你可以在同一坐标轴上绘制 y = ax² + bx + c 和 y = mx + n。它们交点处的 x 坐标同时满足两个方程，因此就是原二次方程的解。

Worked example: Use the curve y = x² – 2x – 3 to solve x² – 2x – 3 = x – 1.

实例：利用曲线 y = x² – 2x – 3 求解 x² – 2x – 3 = x – 1。

Draw the straight line y = x – 1 on the same grid as the parabola. The line has slope 1 and y-intercept -1. The two graphs intersect at two points; reading the x-coordinates from the graph gives x ≈ -0.6 and x ≈ 3.6. To check: rearranging gives x² – 3x – 2 = 0, so x = (3 ± √17)/2 ≈ 3.56 and -0.56, confirming the graphical readings.

在抛物线的同一坐标网格中绘制直线 y = x – 1。该直线斜率为 1，y 截距为 -1。两条图有两个交点；从图像读取 x 坐标得 x ≈ -0.6 和 x ≈ 3.6。验证：移项得 x² – 3x – 2 = 0，所以 x = (3 ± √17)/2 ≈ 3.56 和 -0.56，与图像读数一致。

7. Rearranging Before Drawing | 先重新整理方程再绘制图像

In many exam questions, you will be given a pre-drawn parabola (such as y = x² – 4x + 3) and asked to solve a different quadratic equation, like x² – 4x + 1 = 0. You must rearrange the new equation so that one side matches the equation of the given curve, then draw the appropriate line.

在许多考试题中，你会被给出一条已绘制好的抛物线（如 y = x² – 4x + 3），并要求解一个不同的二次方程，如 x² – 4x + 1 = 0。你必须重新整理新方程，使一边与给定曲线的方程匹配，然后画出相应的直线。

For x² – 4x + 1 = 0, rewrite it as x² – 4x + 3 = 2. The left-hand side is exactly the given curve y = x² – 4x + 3, and the right-hand side is k = 2. Therefore, draw the horizontal line y = 2 on the given graph; its intersections with the parabola give the solutions. Alternatively, rearrange as x² – 4x + 3 = 2x – 2, and draw the line y = 2x – 2 instead.

对于 x² – 4x + 1 = 0，将其改写为 x² – 4x + 3 = 2。左边正是给定曲线 y = x² – 4x + 3，右边是 k = 2。因此，在给定的图像上画水平线 y = 2；它与抛物线的交点即为解。或者，将其整理为 x² – 4x + 3 = 2x – 2，然后画直线 y = 2x – 2。

To find the line to draw, follow this rule: write the target equation, then subtract or adjust terms so that the quadratic part exactly equals f(x) of the given curve y = f(x). The remaining non-zero expression on the other side of the equality is the equation of the line you must draw.

找出所需绘制的直线，遵循以下规则：写出目标方程，然后通过加减项使二次部分恰好等于给定曲线 y = f(x) 的 f(x)。等式另一边剩余的非零表达式就是你必须绘制的直线方程。

8. Worked Example: Solving x² – 5x + 4 = 0 Graphically | 实例：图像法求解 x² – 5x + 4 = 0

Let us go through a complete example step by step. Plot the graph of y = x² – 5x + 4 for -1 ≤ x ≤ 5, then solve x² – 5x + 4 = 0.

让我们逐步完成一个完整实例。在 -1 ≤ x ≤ 5 范围内绘制 y = x² – 5x + 4 的图像，然后解 x² – 5x + 4 = 0。

x	-1	0	1	2	3	4	5
y	10	4	0	-2	-2	0	4

Plot the points (-1, 10), (0, 4), (1, 0), (2, -2), (3, -2), (4, 0) and (5, 4), and join them with a smooth U-shaped curve. The graph crosses the x-axis at x = 1 and x = 4, so the solutions of x² – 5x + 4 = 0 are x = 1 or x = 4. Indeed, (x – 1)(x – 4) = 0 confirms this.

描出点 (-1, 10)、(0, 4)、(1, 0)、(2, -2)、(3, -2)、(4, 0) 和 (5, 4)，并用平滑的 U 形曲线连接。图像在 x = 1 和 x = 4 处穿过 x 轴，因此 x² – 5x + 4 = 0 的解为 x = 1 或 x = 4。事实上，(x – 1)(x – 4) = 0 也验证了这一点。

Now, using the same curve, solve x² – 5x + 4 = 2. Draw the line y = 2 on the same axes. The line intersects the parabola at approximately x = 0.4 and x = 4.6. Algebraically, x² – 5x + 2 = 0 gives x = (5 ± √17)/2 ≈ 4.56 and 0.44, matching the graphical estimate.

现在，使用同一条曲线解 x² – 5x + 4 = 2。在同一坐标轴上画直线 y = 2。该直线与抛物线相交于大约 x = 0.4 和 x = 4.6 处。代数上，x² – 5x + 2 = 0 给出 x = (5 ± √17)/2 ≈ 4.56 和 0.44，与图像估算一致。

9. The Discriminant and Graphical Interpretation | 判别式与图像解释

The discriminant Δ = b² – 4ac is not just an algebraic tool; it directly predicts what the graph looks like relative to the x-axis. This connection is frequently tested in Edexcel IGCSE papers, both in algebra and graph questions.

判别式 Δ = b² – 4ac 不仅是代数工具；它直接预测图像相对于 x 轴的位置关系。这种联系在 Edexcel IGCSE 考试中经常被考查，无论代数题还是图像题。

Δ > 0: the parabola cuts the x-axis at two distinct points → two real roots.

Δ > 0：抛物线与 x 轴相交于两个不同点 → 两个实数根。
Δ = 0: the parabola touches the x-axis at one point → one repeated root.

Δ = 0：抛物线与 x 轴相切于一点 → 一个重根。
Δ < 0: the parabola does not touch or cross the x-axis → no real roots.

Δ < 0：抛物线不接触也不穿过 x 轴 → 没有实数根。

For example, y = 2x² – 4x + 3 has a = 2, b = -4, c = 3, so Δ = 16 – 24 = -8 < 0. Since a > 0, this parabola opens upwards and sits entirely above the x-axis; the equation 2x² – 4x + 3 = 0 has no real solutions, and the graph never crosses the x-axis.

例如，y = 2x² – 4x + 3 中 a = 2, b = -4, c = 3，所以 Δ = 16 – 24 = -8 < 0。由于 a > 0，这条抛物线开口朝上并且完全位于 x 轴上方；方程 2x² – 4x + 3 = 0 没有实数解，图像永远不会穿过 x 轴。

10. Estimating Solutions from Graphs | 从图像估算解

Graphical solutions are by nature approximate, unless the roots happen to be integers that align exactly with grid lines. When reading solutions from a graph, always write your answers to the degree of accuracy the graph allows — usually 1 decimal place if the grid is in 1-unit intervals. Use a ruler to read the x-coordinate vertically down from an intersection point to the x-axis.

图像解本质上是近似值，除非根恰好是与网格线对齐的整数。从图像读取解时，始终以图像所能达到的精度写出答案——如果网格以 1 个单位为间隔，通常取 1 位小数。用直尺从交点垂直向下读取 x 轴上的 x 坐标。

In Edexcel mark schemes, a range of acceptable answers is normally given (for example, accept 0.3 to 0.5 and 4.4 to 4.7). This acknowledges that different students may draw slightly different curves or read positions with small variations. Always use suitable scales on both axes so that the parabola is large enough to give reliable readings.

在 Edexcel 评分标准中，通常会给出一个可接受答案的范围（例如，接受 0.3 至 0.5 以及 4.4 至 4.7）。这考虑到不同学生可能画出略微不同的曲线或读数时有微小差异。始终在两个坐标轴上使用合适的比例，使抛物线足够大以保证读数可靠。

11. Common Mistakes and Exam Tips | 常见错误与考试技巧

Several errors frequently cost students marks in graphical quadratic questions. Being aware of them will help you avoid them.

学生在二次函数图像题中经常因一些错误而丢分。了解这些错误有助于你避免它们。

Mistake 1: Drawing straight lines between plotted points. Parabolas must be smooth curves. Use a sharp pencil and draw the curve free-hand in one continuous motion through all points.

错误 1：描点后画直线段连接。抛物线必须是平滑曲线。使用削尖的铅笔，用一次连贯的动作穿过所有点画出曲线。
Mistake 2: Forgetting to rearrange the target equation. If asked to solve x² – 4x + 1 = 0 using the graph of y = x² – 4x + 3, you must first rearrange to determine which line to draw.

错误 2：忘记重新整理目标方程。如果要求利用 y = x² – 4x + 3 的图像解 x² – 4x + 1 = 0，你必须先重新整理以确定要画哪条直线。
Mistake 3: Reading y-coordinates instead of x-coordinates. The solutions of the equation are the x-coordinates of the intersection points, not the y-coordinates.

错误 3：读取的是 y 坐标而不是 x 坐标。方程的解是交点的 x 坐标，而不是 y 坐标。
Mistake 4: Using too small a scale. A small graph leads to inaccurate readings. Choose a scale that makes the parabola fill at least half the grid.

错误 4：比例尺太小。图像太小会导致读数不准确。选择使抛物线至少占网格一半的比例尺。

Before the exam, practise plotting at least three different quadratic functions and solving associated equations by drawing lines. Also, always check whether your graphical solutions make sense by substituting them back into the original equation mentally.

考试前，练习至少绘制三个不同的二次函数并通过画线求解相关方程。此外，始终通过将图像解代回原方程来检查其是否合理。

12. Summary and Practice Questions | 总结与练习

To solve a quadratic equation graphically, plot the parabola y = ax² + bx + c and read the x-intercepts for standard form. For equations like ax² + bx + c = k, draw the horizontal line y = k. For mixed equations like ax² + bx + c = mx + n, draw the straight line y = mx + n and read the x-coordinates of the intersections. Always rearrange a new equation so the quadratic part matches the given curve, then identify the line required.

要通过图像解二次方程，绘制抛物线 y = ax² + bx + c 并读取标准形式下的 x 轴交点。对于形如 ax² + bx + c = k 的方程，画水平线 y = k。对于混合方程如 ax² + bx + c = mx + n，画直线 y = mx + n 并读取交点的 x 坐标。始终重新整理新方程，使二次部分与给定曲线匹配，然后确定所需绘制的直线。

Try these exercises. (1) Plot y = x² – 3x – 10 for -3 ≤ x ≤ 5 and use it to solve x² – 3x – 10 = 0. (2) Using the same graph, solve x² – 3x – 10 = -6. (3) Using the same graph, solve x² – 3x – 10 = 2x – 5 by drawing the appropriate line. (4) State the discriminant of x² + 2x + 5 and explain what

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Using Graphs to Solve Quadratic Equations | 使用图像求解二次方程