📚 E - Finding a Quadratic from its Graph | 由图像求二次函数解析式
In IB Mathematics, being able to derive the equation of a quadratic function from its graph is a fundamental skill. Whether you are given the vertex, the intercepts, or simply a set of points, there are several reliable methods to reconstruct the function. This article walks through each approach, illustrates the reasoning with worked examples, and highlights the mistakes that cost marks.
在 IB 数学中，由图像推导二次函数解析式是一项基本技能。无论题目给出的是顶点、截距，还是仅仅几个点，都有多种可靠的方法来还原函数。本文将逐一介绍这些方法，结合实例展示推理过程，并指出会丢分的常见错误。
1. Introduction to Quadratic Graphs | 二次函数图像导论
≠ 0. Its graph is a parabola, opening upwards when a > 0 and downwards when a < 0. The curve is symmetric about a vertical line called the axis of symmetry, which passes through the turning point of the parabola.
二次函数的一般形式为 y = ax² + bx + c，其中 a ≠ 0。它的图像是一条抛物线，若 a > 0 则开口向上，若 a < 0 则开口向下。图像关于一条称为对称轴的竖直线对称，该直线穿过抛物线的转折点。
Because three parameters (a, b and c) define the curve, three independent pieces of information are enough to pin the function down completely. A graph usually supplies far more than three clues, so your task is to select the most convenient ones.
由于曲线由三个参数（a、b、c）决定，三个独立信息就足以完全确定这个函数。而一张图像通常提供的线索远多于三个，因此你的任务是挑选最方便使用的那几组信息。
2. Key Features of a Parabola | 抛物线的关键特征
the axis of symmetry, the vertical line x = h through the vertex; (3) the x-intercepts (roots), where the graph crosses the x-axis; (4) the y-intercept, where x = 0; and (5) the direction of opening.
观察二次函数图像时，请先识别以下特征：(1) 顶点，即最高点或最低点；(2) 对称轴，即过顶点的竖直线 x = h；(3) x 轴截距（根），即图像与 x 轴的交点；(4) y 轴截距，即 x = 0 处的位置；(5) 开口方向。
| Feature 特征 | On the graph 图像上如何体现 | Algebraic meaning 代数含义 |
| Vertex 顶点 | Turning point 曲线的转折点 | (h, k) in y = a(x − h)² + k |
| Axis of symmetry 对称轴 | Vertical line through vertex 过顶点的竖直线 | x = h = −b ÷ (2a) |
| x-intercepts x 轴截距 | Crossings of the x-axis 与 x 轴的交点 | x = p and x = q (the roots 两根) |
| y-intercept y 轴截距 | Crossing of the y-axis 与 y 轴的交点 | y = c when x = 0 |
| Direction 开口方向 | Upwards or downwards 向上或向下 | Sign of a 系数 a 的符号 |
3. The Three Forms of a Quadratic | 二次函数的三种形式
Every quadratic can be written in three equivalent ways, and each form makes a different feature immediately readable. Choosing the right form from the start saves you a great deal of algebra.
任何一个二次函数都可以写成三种等价形式，而每种形式都会让某一特征可以直接读出。一开始就选对形式，能替你省下大量代数运算。
y = ax² + bx + c
This is the general (expanded) form. It reveals the y-intercept c directly, but the vertex and the roots are hidden inside the coefficients.
这是一般式（展开式）。它能直接显示 y 轴截距 c，但顶点和根都隐藏在系数之中。
y = a(x − h)² + k
s of symmetry x = h at a glance. Use it whenever the turning point is marked on the graph.
这是顶点式（配方式），一眼就能看出顶点 (h, k) 与对称轴 x = h。只要图上标出了转折点，就应使用这一形式。
y = a(x − p)(x − q)
p and x = q. Use it whenever the graph clearly crosses the x-axis at two labelled points.
这是交点式（因式分解式），能直接读出 x 轴截距 x = p 和 x = q。只要图像清晰地与 x 轴交于两个已标注的点，就应使用这一形式。
4. Method 1: Vertex Plus One Extra Point | 方法一：顶点加一个额外点
nly a is still unknown. Pick any other clearly marked point on the curve, substitute its coordinates, and solve the resulting linear equation for a.
从图上读出顶点 (h, k)，代入 y = a(x − h)² + k。此时只有 a 还是未知数。再取曲线上任意一个清晰标注的点，代入其坐标，解出这个关于 a 的一元一次方程即可。
, so 4a = 8 and a = 2. The answer is y = 2(x − 2)² − 3, which expands to y = 2x² − 8x + 5.
例题：图像顶点为 (2, −3)，且经过点 (0, 5)。先写出 y = a(x − 2)² − 3。代入 (0, 5) 得 5 = a(0 − 2)² − 3 = 4a − 3，故 4a = 8，a = 2。答案为 y = 2(x − 2)² − 3，展开后为 y = 2x² − 8x + 5。
Always check that the sign of a agrees with the graph. If the parabola opens downwards, a must come out negative; if it does not, you have almost certainly misread the vertex coordinates.
务必检查 a 的符号是否与图像相符。若抛物线开口向下，a 必须为负；若不是，那几乎可以肯定你读错了顶点坐标。
5. Method 2: x-intercepts Plus One Extra Point | 方法二：x 轴截距加一个额外点
tute any third known point to find a. This route is usually shorter than the vertex route because the algebra is simpler.
当图像与 x 轴交于 x = p 和 x = q 时，写出 y = a(x − p)(x − q)，再代入任意第三个已知点求 a。由于代数运算更简单，这条路通常比顶点法更短。
g (0, −6): −6 = a(0 + 1)(0 − 3) = −3a, so a = 2. Hence y = 2(x + 1)(x − 3) = 2x² − 4x − 6.
例题：图像与 x 轴交于 (−1, 0) 和 (3, 0)，并经过 (0, −6)。写出 y = a(x + 1)(x − 3)。代入 (0, −6)：−6 = a(0 + 1)(0 − 3) = −3a，故 a = 2。因此 y = 2(x + 1)(x − 3) = 2x² − 4x − 6。
esponds to the factor (x + 1), not (x − 1). Subtracting the root is the rule, so always keep the bracket as x minus the intercept value.
注意符号陷阱：根为 x = −1 时，对应因式是 (x + 1)，而不是 (x − 1)。规则是"x 减去根的值"，所以括号内始终写成 x 减去截距值。
6. Method 3: Three Points on the Curve | 方法三：曲线上的三个点
x + c. You obtain three simultaneous linear equations in a, b and c, which you solve by elimination or substitution.
如果图上只标出三个普通点、没有任何特殊特征，就把每个点分别代入 y = ax² + bx + c，得到关于 a、b、c 的三个联立一次方程，再用消元法或代入法求解。
ng (2, 3): 4a + 2b + 1 = 3, so 2a + b = 1. Subtracting gives a = 2 and therefore b = −3. The equation is y = 2x² − 3x + 1.
例题：曲线经过 (0, 1)、(1, 0) 和 (2, 3)。由 (0, 1) 立刻得 c = 1。代入 (1, 0)：a + b + 1 = 0，即 a + b = −1。代入 (2, 3)：4a + 2b + 1 = 3，即 2a + b = 1。两式相减得 a = 2，进而 b = −3。方程为 y = 2x² − 3x + 1。
Whenever one of the given points lies on the y-axis, begin with it: it hands you c for free and reduces the system to two unknowns.
只要已知点中有一个落在 y 轴上，就先从它入手：它免费送给你 c，并把方程组化简为两个未知数。
7. Using the y-intercept and Symmetry | 利用 y 轴截距与对称性
÷ 2. This single fact links the intercept form and the vertex form without any expansion.
对称轴永远恰好位于两个 x 轴截距的正中间，因此 h = (p + q) ÷ 2。这一个事实就把交点式与顶点式联系了起来，无需任何展开运算。
x = 3, it must also pass through (5, 6). This trick lets you manufacture an extra point whenever the question seems to give too few.
一个有用的对称性质：图像在 x = h − d 与 x = h + d 处的 y 值相同。所以若曲线经过 (1, 6) 且已知对称轴为 x = 3，它必然也经过 (5, 6)。当题目看起来给的条件太少时，这个技巧可以帮你"造"出一个额外的点。
= (c − k) ÷ h². Memorising this shortcut is optional, but deriving it takes only one line.
若顶点为 (h, k) 且图像经过 (0, c)，代入 y = a(x − h)² + k 得 c = ah² + k，故 a = (c − k) ÷ h²。记住这个快捷公式并非必要，因为推出它只需一行。
8. Reading a from the Shape | 从形状读出 a
The magnitude of a controls how wide or narrow the parabola is. A large |a| produces a steep, narrow curve; a small |a| ide, flat one. Comparing the sketch with y = x² gives a quick sanity check on your answer.
a 的大小控制抛物线的宽窄。|a| 越大，曲线越陡越窄；|a| 越小，曲线越平缓越宽。把草图与 y = x² 作比较，可以快速检验答案是否合理。
The sign of a is fixed by the direction of opening: a > 0 opens upwards, a < 0 opens downwards. This is the cheapest check available, and it catches the single most common sign error in this topic.
a 的符号由开口方向决定：a > 0 开口向上，a < 0 开口向下。这是成本最低的检查手段，也能抓出本主题中最常见的符号错误。
Also note that the y-intercept of the general form equals c, but the y-intercept of the vertex form equals ah² + k. These two expressions must agree, which gives you a second independent verification after you finish.
还要注意，一般式的 y 轴截距等于 c，而顶点式的 y 轴截距等于 ah² + k。这两个表达式必须一致，这为你完成解答后提供了第二个独立的验证途径。
9. A Full Worked Example | 完整例题演示
Problem: a parabola opens downwards, has its maximum at (−1, 8), and passes through (1, 0). Find the equation in general form, then state the x-intercepts and the y-intercept.
题目：一条抛物线开口向下，最高点为 (−1, 8)，且经过点 (1, 0)。求其一般式，并写出 x 轴截距与 y 轴截距。
with the stated downward opening. Step 3: expand: y = −2(x² + 2x + 1) + 8 = −2x² − 4x + 6.
第一步：由于给出顶点，从 y = a(x + 1)² + 8 开始。第二步：代入 (1, 0)：0 = a(1 + 1)² + 8 = 4a + 8，得 a = −2。负值与题设的开口向下一致。第三步：展开得 y = −2(x² + 2x + 1) + 8 = −2x² − 4x + 6。
Step 4: roots. Set −2(x + 1)² + 8 = 0, so (x + 1)² = 4 and x + 1 = ±2, giving x =
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