📚 Using Graphs to Solve Other Equations | 利用图像解其他方程
Graphs are not just tools for visualising functions – imultaneous equations and inequalities. This article explains the key techniques, common question types, and how to extract accurate answers from sketches and plotted curves.
图像不仅是观察函数的工具——它们还可以用来求解那些用代数方法解起来困难或繁琐的方程。在 IGCSE 数学中，你经常需要绘制或者使用给定的图像，来找到诸如 f(x) = g(x)、f(x) = k 等方程的近似解，甚至用于解联立方程组与不等式。本文将讲解关键技巧、常见题型，以及如何从草图和精确曲线中提取准确的答案。
1. Why Use Graphs to Solve Equations? | 为什么用图像解方程？
arrange and solve using the quadratic formula, but the exam board wants you to practise the graphical method. More importantly, when the graph of a function is already provided, you can solve a whole family of related equations by simply adding straight lines or shifting the perspective – without doing any more algebra.
在许多 IGCSE 题目中，像 3x² − 2x − 5 = 2x + 1 这样的方程似乎很容易移项后用求根公式求解，但考试局希望考生练习图像法。更重要的是，当函数的图像已经给出时，你只需添上一条直线或转换视角，就能解出一系列相关方程，完全不必再做额外的代数运算。
The graphical approach also helps you build an intuitive understanding of roots, intersections, and inequalities, which is especially powerful when you move on to higher-level topics.
图形方法还能帮助你建立关于根、交点和不等式的直观理解，这对后续更高层次的学习非常有益。
2. The Core Idea: Intersection of Two Graphs | 核心思想：两个图像的交点
(x) = g(x). If you draw the graphs of y = f(x) and y = g(x) on the same set of axes, the x‑ ersect are the solutions to the equation. This is because at any intersection point, the y‑values are equal, and therefore f(x) = g(x).
每一个方程都可以改写成 f(x) = g(x) 的形式。若在同一坐标系中画出 y = f(x) 和 y = g(x) 的图像，两条曲线交点的 x 坐标就是方程的解。这是因为在任何一个交点上，y 值相等，所以 f(x) = g(x)。
lving x² + x − 6 = 0 is equivalent to finding where the parabola y = x² + x − 6 cuts the x‑ ersection of y = x² + x and y = 6, which changes the picture but gives the same solutions.
例如，解 x² + x − 6 = 0 等价于求抛物线 y = x² + x − 6 与 x 轴（即 y = 0）的交点。但你也可以将其看作 y = x² + x 与 y = 6 的交点，虽然图像变了，但解是相同的。
3. Solving f(x) = g(x) by Drawing Two Graphs | 通过绘制两个图像解 f(x) = g(x)
n graph paper, carefully labelling the axes. After sketching smooth curves, read off the x‑ e asked to draw the graph of y = x² − 2x − 3 and then use it to solve x² − 2x − 3 = x + 1.
一种直接的方法是在方格纸上画出 y = f(x) 和 y = g(x)，仔细标出坐标轴。画出平滑曲线后，读取所有交点的 x 坐标。考试中，可能会让你先画出 y = x² − 2x − 3 的图像，然后利用它来解方程 x² − 2x − 3 = x + 1。
e the approximate solutions to the original equation. Always check the scale and read as precisely as possible – answers are usually accepted within a small tolerance.
为此，你需要在同一网格上画出直线 y = x + 1。抛物线与直线会交于两点（例如 x ≈ −1.6 和 x ≈ 4.6），这两个值就是原方程的近似解。务必看清比例尺，尽量精确读取——通常答案允许有一个小的误差范围。
4. Solving f(x) = k with Horizontal Lines | 利用水平线解 f(x) = k
ou can solve any equation of the form f(x) = k by drawing the horizontal line y = k. The x‑ n, you can solve many different equations by simply moving the horizontal line up or down.
如果你已经有了 y = f(x) 的图像，只需画一条水平线 y = k，曲线与该直线的交点的 x 坐标就是方程 f(x) = k 的解。这种方法特别实用，因为一旦画好了曲线，你只需上下移动水平线，就能解出许多不同的方程。
of y = x² − 4x + 3, you can solve x² − 4x + 3 = 0 by looking where the curve crosses the x‑ 4x − 2 = 0, rearrange it as x² − 4x + 3 = 5, and again use the same horizontal line y = 5.
例如，已知 y = x² − 4x + 3 的图像，你可以通过观察曲线与 x 轴（y = 0）的交点来解 x² − 4x + 3 = 0；要解 x² − 4x + 3 = 5，就画一条 y = 5 的直线，找到两个交点；要解 x² − 4x − 2 = 0，可将其变形为 x² − 4x + 3 = 5，同样用水平线 y = 5 来解。
5. Using a Given Graph to Solve Related Equations | 利用已知图像解相关方程
IGCSE exam questions often give a ready‑ the given graph does not directly help. However, by cleverly rearranging the new equation to match the original function, you can turn it into a line‑intersection problem.
IGCSE 考试经常给出一条函数图像，例如 y = x² − 3x + 1，然后要求你解方程 x² − 3x − 2 = 0。乍一看，已知图像帮不上忙。但通过巧妙地将新方程变形，使其与原有函数匹配，就能把问题转化为直线与曲线相交的问题。
the left side is exactly the function whose graph is provided. So the solutions are the x‑ rizontal line on the diagram, read off the intersection values, and you have your answers.
在 x² − 3x − 2 = 0 两边同时加 3，得到 x² − 3x + 1 = 3。此时左边恰好就是题目已给的函数。因此方程的解就是曲线与直线 y = 3 的交点横坐标。在图像上画出这条水平线，读取交点的值，就得到了答案。
6. Solving Simultaneous Equations Graphically | 图解法解联立方程组
y plotting both on the same axes. The solutions are the coordinates of the intersection points, and the x‑ sfy the combined equation x² + 2x − 1 = 2x + 3, which simplifies to x² = 4, giving x = ±2.
用图像法解联立方程组，本质上就是找两个函数图像的交点。例如，方程组 y = x² + 2x − 1 与 y = 2x + 3，可以在同一坐标系中绘制两条图像来求解。交点的坐标就是方程组的解，其中 x 值满足 x² + 2x − 1 = 2x + 3，化简后得 x² = 4，即 x = ±2。
In an exam, you might plot a parabola and a straight line; the two intersection points give two pairs of (x, y) solutions. If the line is tangent to the curve, there is exactly one solution (a repeated root). If they do not intersect, there are no real solutions.
考试中，你可能需要画一条抛物线和一条直线；两个交点给出两组 (x, y) 解。如果直线与曲线相切，则只有一个解（重根）；如果它们不相交，则没有实数解。
7. Solving Inequalities Using Graphs | 利用图像解不等式
A graph is not limited to equations – it can also help you solve inequalities. Consider the inequality f(x) > g(x). On a graph, this means the curve of f(x) is above the curve of g(x) for certain intervals of x. You simply shade the region where the inequality holds, and then write the ranges of x that belong to that region.
图像不仅能处理等式，还可以帮助你解不等式。比如不等式 f(x) > g(x)，在图像上表示在某些 x 区间内，f(x) 曲线在 g(x) 曲线的上方。你只需将满足不等式的区域涂上阴影，然后写出属于该区域的 x 范围。
graph of y = x² − 4x + 3 and the line y = 2, the inequality x² − 4x + 3 > 2 translates to “where is the parabola above y = 2?”. Find the intersection x‑ nd then the solution is x < 0.3 or x > 3.7, written in set notation or using inequalities.
例如，已知 y = x² − 4x + 3 的图像以及直线 y = 2，不等式 x² − 4x + 3 > 2 可转化为“抛物线在 y = 2 上方时对应的 x 值”。先求出交点的 x 坐标（约 0.3 和 3.7），那么解就是 x < 0.3 或 x > 3.7，以集合符号或不等式形式书写即可。
8. Worked Example: Quadratic and Linear Intersection | 实例讲解：二次函数与直线的交点
Let us work through a typical IGCSE question.
Given the graph of y = x² − 2x − 3, solve the equation x² − 2x − 3 = x + 1 graphically.
我们来看一道典型的 IGCSE 题目。
已知 y = x² − 2x − 3 的图像，用图解法解方程 x² − 2x − 3 = x + 1。
Step 1: Draw the straight line y = x + 1 on the same grid. This line has gradient 1 and y‑intercept (0, 1).
步骤 1：在同一坐标网格上画出直线 y = x + 1，该直线斜率为 1，y 截距为 (0, 1)。
1.4 and x = 4.4 (actual exact roots are −1 and 4, but graphical readings are approximate).
步骤 2：观察抛物线与直线的交点。从精确绘制的图像中，你会看到交点大约在 x = −1.4 和 x = 4.4 处（精确根是 −1 和 4，但图形读取的是近似值）。
swers correct to 1 decimal place, check that your readings are consistent and perhaps estimate halfway between grid lines.
步骤 3：写出解 x ≈ −1.4 和 x ≈ 4.4。如果题目要求答案精确到一位小数，要确保读数合理，必要时在网格线之间进行估计。
equation. (−1.4)² − 2(−1.4) − 3 ≈ 1.76 + 2.8 − 3 = 1.56, and (−1.4)+1 = −0.4; 1.56 ≠ −0.4 – mber graphical solutions are approximate, and more precise drawing gives better agreement.
步骤 4：代回原方程进行验证。(−1.4)² − 2(−1.4) − 3 ≈ 1.76 + 2.8 − 3 = 1.56，而 (−1.4)+1 = −0.4，1.56 ≠ −0.4——但要记住图像解是近似的，更精确的绘图会得到更好的一致性。
9. Accuracy and Common Pitfalls | 精度与常见误区
The accuracy of a graphical solution depends heavily on the scale and neatness of your drawing. Always use a sharp pencil, label axes with equal spacing, and plot points as accurately as possible. A common mistake is to misread the x‑coordinate of an intersection, especially when it does not fall exactly on a grid line. Use a ruler to align the point with the x‑axis and read carefully.
图像解的精度很大程度上取决于作图的准确度和整洁度。务必使用削尖的铅笔，均匀标出坐标轴刻度，并尽可能精确地描点。常见的错误是误读交点的 x 坐标，特别是当交点不正好落在网格线上时。用直尺将点对准 x 轴，然后仔细读数。
Another pitfall is forgetting that an equation may have multiple solutions. If you only draw a small portion of the curve, you might miss an intersection. Always extend the graph to cover the x‑range specified in the question, and look beyond the immediate window if necessary.
另一个误区是忘记一个方程可能有多个解。如果你只画了曲线的一小部分，很可能会漏掉某个交点。务必让图像覆盖题目指定的 x 范围，必要时还要观察更远的地方。
Finally, when rearranging an equation to use a given graph, be careful with signs. A simple algebraic mistake can lead to drawing the wrong line and obtaining incorrect solutions. Always double‑check your rearrangement by testing with a known point.
最后，在变形方程以便利用已知图像时，要注意符号。一个简单的代数错误会导致你画出错误的直线，得到错误的解。始终通过代入已知点来复查你变形后的式子。
10. Practice Question with Solution | 练习题与解答
Question: The graph of y = x² + 2x − 5 is drawn on the grid below.
(a) Use the graph to find the roots of the equation x² + 2x − 5 = 0.
(b) By drawing a suitable line, solve x² + 2x − 5 = 4.
) Rearrange x² + 2x − 1 = 0 into the form x² + 2x − 5 = k, and hence solve it graphically.
题目：函数 y = x² + 2x − 5 的图像已画在下面的网格中。
(a) 利用图像求方程 x² + 2x − 5 = 0 的根。
(b) 通过画一条适当的直线，解方程 x² + 2x − 5 = 4。
(c) 将方程 x² + 2x − 1 = 0 变形成 x² + 2x − 5 = k 的形式，并据此用图解法求解。
Solution:
(a) The roots are where the curve cuts the x‑ ccur at approximately x ≈ −3.45 and x ≈ 1.45. So the solutions are x ≈ −3.45 and x ≈ 1.45.
line meets the curve at x ≈ −3.7 and x ≈ 1.7. Thus the solutions are x ≈ −3.7 and x ≈ 1.7.
this horizontal line, and read off x ≈ −2.2 and x ≈ 0.2. These are the required solutions.
解答：
(a) 方程的根即曲线与 x 轴（y = 0）的交点。从图像中可知，交点大约在 x ≈ −3.45 和 x ≈ 1.45。此即方程的解。
(b) 画出水平线 y = 4。该直线与曲线交于 x ≈ −3.7 和 x ≈ 1.7，因此解为 x ≈ −3.7 与 x ≈ 1.7。
(c) 将 x² + 2x − 1 = 0 两边同时减去 4，得 x² + 2x − 5 = −4。因此需要找出已知曲线与直线 y = −4 的交点。画出这条水平线，读取 x ≈ −2.2 与 x ≈ 0.2，即为所求的解。
11. Summary and Key Takeaways | 总结与要点
Graphical methods bring equations to life and offer a powerful alternative to algebra, particularly in IGCSE exams where you must interpret given graphs or produce your own. Remember: any equation can be viewed as finding the intersections of two functions. Master the art of rearranging an unfamiliar equation so that one side matches a graph you already have, and the other side becomes a simple line. Always work neatly, read scales carefully, and state answers with the required precision.
图像法让方程变得直观可见，是代数方法的有力补充，尤其是在 IGCSE 考试中，你必须能够解读给定图像或自行画图。请记住：任何方程都可以看作求两个函数图像的交点。熟练掌握将陌生方程变形，使其一边与你已有的图像匹配、另一边变成简单直线的技巧。始终规整作图，仔细读取刻度，并按照要求精度写出答案。
g(x), f(x) = k, simultaneous equations, or inequalities, the same principles apply. A well‑ blem into a visual search for crossing points, giving you both answers and deeper insight.
无论是解 f(x) = g(x)、f(x) = k、联立方程组还是不等式，同样的原则都适用。一张精心绘制的图像能将复杂问题转化为寻找交点的视觉过程，既给出答案，又带来更深的理解。
Published by TutorHao | Mathematics 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