📚 Quadratic Function Graphs: Opening, Vertex, and Intercepts | 二次函数图像：开口、顶点与交点
A quadratic function is one of the most important topics in A-Level mathematics. Its graph is a smooth curve called a parabola, and understanding its key features — opening direction, vertex, and intercepts — allows you to sketch it accurately and solve many real-world problems.
二次函数是 A-Level 数学中最重要的课题之一。它的图像是一条平滑的曲线，称为抛物线；理解其关键特征——开口方向、顶点和交点——可以帮助你准确地画出图像，并解决许多实际问题。
1. Standard Form and General Form | 标准形式与一般形式
A quadratic function can be written in two common ways. The general form is y = ax² + bx + c, where a, b, and c are constants and a ≠ 0. The completed square form is y = a(x − h)² + k, which directly reveals the vertex (h, k).
二次函数有两种常见的书写方式。一般形式是 y = ax² + bx + c，其中 a、b、c 是常数且 a ≠ 0。配方法形式是 y = a(x − h)² + k，它直接给出顶点 (h, k)。
The coefficient a controls both the opening direction and the steepness of the curve. If a > 0, the parabola opens upward; if a < 0, it opens downward. The larger |a| is, the narrower the parabola.
系数 a 同时控制开口方向和曲线的陡峭程度。若 a > 0，抛物线开口向上；若 a < 0，开口向下。|a| 越大，抛物线越窄。
2. Opening Direction: The Sign of a | 开口方向：a 的符号
For the quadratic y = ax² + bx + c, the sign of a determines whether the parabola has a minimum or maximum point. When a > 0, the arms rise to infinity and the vertex is a minimum. When a < 0, the arms fall to infinity and the vertex is a maximum.
对于二次函数 y = ax² + bx + c，a 的符号决定抛物线是有最小值点还是最大值点。当 a > 0 时，两臂向上延伸至无穷，顶点为最小值；当 a < 0 时，两臂向下延伸至无穷，顶点为最大值。
To determine the opening direction quickly, just look at the leading coefficient. For example, y = 2x² − 3x + 1 opens upward, while y = −x² + 4x − 5 opens downward.
要快速判断开口方向，只需看首项系数。例如，y = 2x² − 3x + 1 开口向上，而 y = −x² + 4x − 5 开口向下。
3. The Vertex: Complete the Square | 顶点：配方法
The vertex is the turning point of the parabola. It can be found by completing the square. Starting from y = ax² + bx + c, we write:
顶点是抛物线的转折点。可以通过配方法求得。从 y = ax² + bx + c 出发，我们写成：
y = a(x + b/2a)² + c − b²/4a
Therefore, the x-coordinate of the vertex is x = −b/2a, and the y-coordinate is y = c − b²/4a. In completed square form y = a(x − h)² + k, the vertex is simply (h, k).
因此，顶点的 x 坐标为 x = −b/2a，y 坐标为 y = c − b²/4a。在完成平方形式 y = a(x − h)² + k 中，顶点就是 (h, k)。
For example, take y = x² − 6x + 5. Completing the square gives y = (x − 3)² − 4, so the vertex is (3, −4).
例如，取 y = x² − 6x + 5。配方得到 y = (x − 3)² − 4，所以顶点是 (3, −4)。
4. The Axis of Symmetry | 对称轴
Every parabola is symmetric about a vertical line passing through its vertex. This line is called the axis of symmetry, with equation x = −b/2a. It divides the parabola into two mirror-image halves.
每条抛物线都关于经过其顶点的竖直线对称。这条直线称为对称轴，方程为 x = −b/2a。它将抛物线分成两个镜像对称的部分。
Knowing the axis of symmetry helps you plot points more efficiently: once you plot one point on one side, you automatically know its mirror point on the other side at the same height.
了解对称轴有助于更高效地作图：一旦你在一边绘制了一个点，就能自动知道另一边等高的对称点。
5. The y-Intercept | y 轴截距
The y-intercept is the point where the graph crosses the y-axis. This occurs when x = 0. Substituting x = 0 into y = ax² + bx + c gives y = c. Therefore, the y-intercept is always (0, c).
y 轴截距是图像与 y 轴的交点。这发生在 x = 0 时。将 x = 0 代入 y = ax² + bx + c 得到 y = c。因此，y 轴截距始终是 (0, c)。
This is the easiest point to find on the graph. Always plot it first, as it is a fixed reference point that does not depend on a or b.
这是图像上最容易找到的点。作图时请先标出它，因为它是一个固定参考点，与 a 或 b 无关。
6. The x-Intercepts: Roots of the Equation | x 轴截距：方程的根
The x-intercepts are the points where the graph crosses the x-axis, meaning y = 0. To find them, we solve the quadratic equation ax² + bx + c = 0. This can be done by factorisation, completing the square, or using the quadratic formula:
x 轴截距是图像与 x 轴相交的点，即 y = 0。要求出它们，我们需要解二次方程 ax² + bx + c = 0。这可以通过因式分解、配方法或求根公式完成：
x = (−b ± √(b² − 4ac)) / 2a
Each real solution corresponds to an x-intercept. If the equation has two distinct real roots, the parabola crosses the x-axis at two points. If it has one repeated root, the parabola touches the x-axis at exactly one point — the vertex itself.
每个实数解对应一个 x 轴截距。如果方程有两个不同的实根，抛物线与 x 轴有两个交点。如果有一个重根，抛物线在 x 轴上恰好相切于一点——即顶点本身。
7. The Discriminant: How Many Intercepts? | 判别式：有多少个交点？
The discriminant is defined as Δ = b² − 4ac. Its value tells us the number of x-intercepts without fully solving the equation.
判别式定义为 Δ = b² − 4ac。它的值告诉我们 x 轴截距的数量，而无需完全求解方程。
| Δ value | Number of x-intercepts | Graph behaviour |
| Δ > 0 | Two distinct | Crosses the x-axis twice |
| Δ = 0 | One repeated | Touches the x-axis at the vertex |
| Δ < 0 | None | Does not touch the x-axis |
When Δ < 0, the quadratic has no real roots. The parabola lies entirely above the x-axis if a > 0, or entirely below if a < 0.
当 Δ < 0 时，二次函数没有实根。若 a > 0，抛物线完全位于 x 轴上方；若 a < 0，则完全位于 x 轴下方。
8. Sketching the Graph: A Step-by-Step Method | 绘制图像：分步方法
To sketch a quadratic graph accurately, follow these steps. First, determine the opening direction using the sign of a. Second, find the vertex using x = −b/2a and substitute to find the y-coordinate, or complete the square. Third, find the y-intercept at (0, c). Fourth, find the x-intercepts by solving ax² + bx + c = 0. Finally, plot these points and draw a smooth symmetric curve.
为了准确绘制二次函数图像，请按以下步骤操作。首先，用 a 的符号确定开口方向。其次，用 x = −b/2a 并代入求出 y 坐标来找到顶点，或采用配方法。第三，找到 y 轴截距 (0, c)。第四，通过解 ax² + bx + c = 0 找到 x 轴截距。最后，标出这些点并画出平滑对称的曲线。
Always check whether the parabola has two, one, or zero x-intercepts before sketching. This prevents common errors such as drawing a curve that crosses the x-axis when it should not.
作图前务必检查抛物线与 x 轴有两个、一个还是零个交点。这可以避免常见错误，比如画出了本不该与 x 轴相交的曲线。
9. Worked Example 1: y = x² − 4x + 3 | 示例 1：y = x² − 4x + 3
Take y = x² − 4x + 3. Here a = 1 > 0, so the parabola opens upward. The vertex is at x = −(−4)/(2×1) = 2, and y = 2² − 4×2 + 3 = −1, so the vertex is (2, −1).
取 y = x² − 4x + 3。这里 a = 1 > 0，所以开口向上。顶点在 x = −(−4)/(2×1) = 2，y = 2² − 4×2 + 3 = −1，因此顶点是 (2, −1)。
The y-intercept is (0, 3). Solving x² − 4x + 3 = 0 gives (x − 1)(x − 3) = 0, so the x-intercepts are x = 1 and x = 3. Plotting these four points, we draw a symmetric parabola with vertex at the lowest point.
y 轴截距是 (0, 3)。解 x² − 4x + 3 = 0 得到 (x − 1)(x − 3) = 0，所以 x 轴截距是 x = 1 和 x = 3。标出这四个点后，我们画出以顶点为最低点的对称抛物线。
10. Worked Example 2: y = −2x² + 8x − 5 | 示例 2：y = −2x² + 8x − 5
For y = −2x² + 8x − 5, a = −2 < 0, so the parabola opens downward. The vertex is at x = −8/(2×(−2)) = 2, and y = −2×4 + 16 − 5 = 3, so the vertex is (2, 3).
对于 y = −2x² + 8x − 5，a = −2 < 0，所以开口向下。顶点在 x = −8/(2×(−2)) = 2，y = −2×4 + 16 − 5 = 3，因此顶点是 (2, 3)。
The y-intercept is (0, −5). The discriminant is Δ = 8² − 4×(−2)×(−5) = 64 − 40 = 24 > 0, so there are two x-intercepts. Using the quadratic formula gives x = (−8 ± √24)/(−4), which simplifies to x = 2 ± √6/2. We plot these to complete the sketch.
y 轴截距是 (0, −5)。判别式为 Δ = 8² − 4×(−2)×(−5) = 64 − 40 = 24 > 0，所以有两个 x 轴截距。使用求根公式得到 x = (−8 ± √24)/(−4)，化简为 x = 2 ± √6/2。我们标出这些点以完成图像。
11. Common Mistakes and Exam Tips | 常见错误与考试技巧
A frequent error is forgetting that a > 0 means a minimum but a < 0 means a maximum. Another common mistake is misidentifying the vertex when using the completed square form: y = a(x − h)² + k has vertex (h, k), not (−h, k), because the sign inside the bracket is already subtracted.
一个常见错误是忘记 a > 0 表示最小值而 a < 0 表示最大值。另一个常见错误是在使用配方法形式时弄错顶点：y = a(x − h)² + k 的顶点是 (h, k)，而不是 (−h, k)，因为括号内的符号已经是减号。
In exams, always show the completed square form or state the formula you use. When sketching, do not forget to label the vertex coordinates, the intercepts, and the axis of symmetry. Check whether the graph should be ‘U’ shaped or ‘n’ shaped before drawing.
在考试中，务必展示配方法形式或说明所使用的公式。作图时，不要忘记标注顶点坐标、交点坐标和对称轴。在动笔之前，先确认图形是“U”形还是“n”形。
12. Connection to Translations and Transformations | 与平移和变换的联系
The completed square form y = a(x − h)² + k reveals that any parabola is a transformation of the basic graph y = x². The value h shifts the graph horizontally, k shifts it vertically, and a stretches or reflects it.
配方法形式 y = a(x − h)² + k 揭示了任何抛物线都是基础图像 y = x² 的变换。h 使图像水平平移，k 使图像垂直平移，a 则拉伸或反射图像。
Understanding this connection helps you answer questions that ask for the equation of a parabola given its vertex and another point: substitute the vertex into the form y = a(x − h)² + k, then use the extra point to solve for a.
理解这种联系有助于你解答这样的问题：已知顶点和另一个点，求抛物线方程。将顶点代入形式 y = a(x − h)² + k，再用另一个点求解 a 即可。
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