Investigation 2 – Graphing a(x-h)² + k | 探究2 – 绘制 a(x-h)² + k 图像

📚 Investigation 2 – Graphing a(x-h)² + k | 探究2 – 绘制 a(x-h)² + k 图像

Welcome to your investigation of the quadratic function in its vertex form y = a(x – h)² + k. This powerful representation reveals the graph’s key features instantly, making it a cornerstone of IB Mathematics. By changing the three parameters a, h and k, you will see how the basic parabola y = x² can be shifted, stretched, compressed and reflected. This guided exploration will help you master the transformation approach, interpret the vertex and axis of symmetry, and solve real-world optimisation problems with confidence.

欢迎进入对二次函数顶点式 y = a(x – h)² + k 的探究。这种强大的表达形式能够让你立即看到图形的关键特征，是 IB 数学的基石。通过改变三个参数 a、h 和 k，你将看到最基本的抛物线 y = x² 如何被平移、拉伸、压缩和反射。这一引导性探索将帮助你掌握变换方法，解读顶点和对称轴，并自信地解决现实中的最值优化问题。

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1. The Vertex Form and Its Origin | 顶点式的由来

The vertex form of a quadratic function is written as y = a(x – h)² + k, where a, h and k are real numbers and a ≠ 0. It is derived by completing the square for a standard quadratic y = ax² + bx + c. For example, y = 2x² + 8x + 5 can be rewritten as y = 2(x + 2)² – 3. In this form, the coordinates of the turning point – called the vertex – are immediately visible as (h, k).

二次函数的顶点式写作 y = a(x – h)² + k，其中 a、h、k 均为实数且 a ≠ 0。它是由一般式 y = ax² + bx + c 经配方法推导而来。例如 y = 2x² + 8x + 5 可改写为 y = 2(x + 2)² – 3。在这一形式下，转折点（即顶点）的坐标直接可见为 (h, k)。

Completing the square is an essential algebraic skill for IB students, as it links the expanded form to the graphical interpretation. The process involves factoring out the coefficient of x² from the x-terms, adding and subtracting the square of half the linear coefficient, and then simplifying. The result is a compact expression that makes transformations transparent.

配方是 IB 学生必须掌握的代数技能，因为它将展开式与图形解释紧密联系起来。配方的过程包括从含 x 的项中提取 x² 的系数，加上并减去一次项系数一半的平方，然后进行化简。最终得到一个紧凑的表达式，使变换关系一目了然。

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2. The Role of a: Direction and Width | 参数 a 的作用：开口方向与宽窄

The parameter a controls the direction in which the parabola opens and its vertical stretch or compression. If a > 0, the parabola opens upwards (smiley face ²); if a < 0, it opens downwards (sad face). The magnitude of a determines the "width" of the graph: larger values of |a| produce a narrower, steeper parabola, while values of |a| between 0 and 1 produce a wider, flatter shape.

参数 a 控制着抛物线的开口方向以及垂直拉伸或压缩的程度。若 a > 0，抛物线开口向上（像笑脸）；若 a < 0，则开口向下（像哭脸）。|a| 的大小决定了图形的“宽窄”：|a| 的值越大，抛物线越窄、越陡；而 |a| 在 0 到 1 之间时，图形则更宽、更扁平。

For example, compare y = 2(x – 1)² + 3 and y = ½(x – 1)² + 3. Both have the same vertex (1, 3), but the first is steeper because each y-value is doubled relative to the basic square, whereas the second is much wider because the squaring effect is halved. This vertical scaling can be understood as multiplying all y-coordinates of the basic parabola y = x² by the factor a, after shifting the vertex to the origin.

例如，比较 y = 2(x – 1)² + 3 和 y = ½(x – 1)² + 3。它们的顶点同样是 (1, 3)，但第一条更陡，因为每个 y 值相对于基本平方被放大了两倍；而第二条则更宽，因为平方的效果被减半了。这种垂直缩放可理解为将基本抛物线 y = x² 的所有 y 坐标乘以因子 a（在将顶点移至原点之后）。

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3. The Vertex (h, k) and Its Significance | 顶点 (h, k) 及其意义

The vertex (h, k) is the highest or lowest point on the graph, depending on the sign of a. For a > 0, the vertex is a minimum point; for a < 0, it is a maximum point. The x-coordinate h is the value that makes the squared term (x – h)² equal to zero, thus giving the extremum value y = k. This makes the vertex ideal for optimisation problems, where you seek the best possible outcome.

顶点 (h, k) 是图形上的最高点或最低点，具体取决于 a 的符号。当 a > 0 时，顶点为最小值点；当 a < 0 时，顶点为最大值点。x 坐标 h 是使平方项 (x – h)² 等于零的值，从而给出极值 y = k。这使得顶点非常适用于优化问题，即寻找可能的最佳结果。

In the IB curriculum, you will often be asked to interpret the vertex in context. For instance, if a function models the height of a projectile over time, the vertex gives the maximum height reached and the time at which it occurs. The vertex form naturally highlights these quantities without the need to differentiate, linking algebra to real-world meaning.

在 IB 课程中，你经常需要结合情境解释顶点的意义。例如，如果一个函数描述抛射物的高度随时间的变化，顶点给出了最高高度以及达到该高度的时刻。顶点式自然地突出了这些量，无需通过求导，就将代数与现实意义联系了起来。

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4. The Axis of Symmetry | 对称轴

Every parabola has an axis of symmetry – a vertical line that passes through the vertex and divides the graph into two mirror-image halves. From the vertex form, the axis is simply the line x = h. This symmetry is a consequence of the square function: for any value d, the points with x-coordinates h + d and h – d yield the same y-value because (d)² = (–d)².

每条抛物线都有一个对称轴——一条经过顶点并将图形分成镜像对称两半的竖直线。从顶点式来看，对称轴就是直线 x = h。这种对称性是平方函数的结果：对于任意 d，x 坐标为 h + d 和 h – d 的点具有相同的 y 值，因为 (d)² = (–d)²。

Using the axis of symmetry simplifies graphing and solving equations. If you know one point on the right side of the parabola, you can instantly find its mirror point on the left side by reflecting across x = h. This saves time and reduces calculation errors, especially when sketching graphs by hand or finding the second intersection with a horizontal line.

利用对称轴可以简化作图和解方程。如果你知道抛物线右侧的一个点，通过关于 x = h 的反射就能立刻找到左侧的镜像点。这既节省时间又减少计算错误，尤其是在手绘草图或求与水平线的第二个交点时尤为有用。

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5. Vertical Translations from y = x² | 从 y = x² 的垂直平移

Starting with the parent function y = x², adding a constant k gives y = x² + k. This shifts the entire graph vertically by k units: upward if k > 0, downward if k < 0. The shape, width and orientation do not change; only the position of each point moves up or down. The vertex moves from (0, 0) to (0, k).

从母函数 y = x² 开始，加上一个常数 k 得到 y = x² + k。这会将整个图形垂直平移 k 个单位：若 k > 0 则向上平移，若 k < 0 则向下平移。形状、宽窄和开口方向均不改变；只是每个点的位置向上或向下移动。顶点从 (0, 0) 移动到 (0, k)。

For example, y = x² + 4 raises the standard parabola by 4 units, placing the vertex at (0, 4) and making the minimum value 4. Similarly, y = x² – 2 lowers it by 2 units, with a vertex at (0, –2). In the IB exam, recognising a vertical shift from a table of values or a written description is a fundamental skill that supports the understanding of more complex transformations.

例如，y = x² + 4 将标准抛物线向上平移了 4 个单位，顶点位于 (0, 4)，最小值为 4。类似地，y = x² – 2 将其向下平移 2 个单位，顶点在 (0, –2)。在 IB 考试中，根据数值表或文字描述识别出垂直平移是一项基本技能，它有助于理解更为复杂的变换。

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6. Horizontal Translations and the Sign Inside the Bracket | 水平平移与括号内的符号

Replacing x with (x – h) in y = x² yields y = (x – h)², which shifts the graph horizontally by h units. However, the direction is opposite to the sign inside the bracket: (x – h) moves the vertex to the right by h when h > 0, while (x + h) moves it to the left by h. This counter-intuitive shift is a common source of mistakes, so always verify by setting the bracket equal to zero: x – h = 0 gives x = h.

在 y = x² 中将 x 替换为 (x – h) 得到 y = (x – h)²，这会将图形水平平移 h 个单位。但是，平移方向与括号内的符号相反：(x – h) 在 h > 0 时将顶点向右移动 h，而 (x + h) 则将其向左移动 h。这种反直觉的平移是常见的错误来源，因此务必将括号设为零进行验证：由 x – h = 0 得 x = h。

For instance, y = (x – 3)² has vertex (3, 0) – the graph is shifted 3 units right relative to y = x². In contrast, y = (x + 5)² shifts the graph 5 units left to vertex (–5, 0). Understanding this horizontal transformation is crucial for placing the axis of symmetry correctly and for reading off the x-coordinate of the vertex directly from an equation.

例如，y = (x – 3)² 的顶点为 (3, 0)——图像相对于 y = x² 向右平移了 3 个单位。相反，y = (x + 5)² 将图像向左平移 5 个单位，顶点为 (–5, 0)。正确理解这种水平变换对于准确放置对称轴以及直接从方程中读出顶点的 x 坐标至关重要。

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7. Combined Shifts: Moving the Vertex to Any Location | 组合平移：将顶点移至任意位置

When both h and k are present in y = (x – h)² + k, the graph undergoes both horizontal and vertical shifts simultaneously. The vertex ends up at (h, k) and the axis of symmetry is x = h. The order in which you apply the shifts does not matter – the final position is the same. This combination allows you to place the vertex anywhere in the coordinate plane.

当 y = (x – h)² + k 中同时出现 h 和 k 时，图像会同时进行水平和垂直平移。最终顶点位于 (h, k)，对称轴为 x = h。平移的顺序并不影响最终位置。这一组合让你能将顶点放置在坐标平面上的任意位置。

Consider y = (x – 2)² – 1. Start with y = x²; shift right by 2 units, then down by 1 unit. The vertex is at (2, –1) and the parabola opens upwards. To sketch it quickly, plot the vertex, draw the axis x = 2, find a couple of points on one side (e.g., x = 3 gives y = 0), and reflect them across the axis. This method is far more efficient than making a table of many values.

考虑 y = (x – 2)² – 1。从 y = x² 开始；向右平移 2 个单位，再向下平移 1 个单位。顶点位于 (2, –1)，开口向上。要快速画出草图，先标出顶点，画出对称轴 x = 2，在一侧找几个点（例如 x = 3 时 y = 0），然后将它们关于对称轴反射。这种方法远比列出许多点的数值表来得高效。

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8. Reflections and Vertical Stretch/Compression Combined | 反射与垂直拉伸/压缩的组合

The full vertex form y = a(x – h)² + k incorporates the parameter a, which can cause a reflection across the x-axis when a is negative, as well as a vertical stretch if |a| > 1 or a compression if 0 < |a| < 1. These transformations are applied after the shifts have positioned the vertex. In other words, the location (h, k) remains fixed while the rest of the graph is scaled and possibly flipped.

完整的顶点式 y = a(x – h)² + k 包含了参数 a。当 a 为负数时，它可以引起关于 x 轴的反射；同时，若 |a| > 1 则产生垂直拉伸，若 0 < |a| < 1 则产生垂直压缩。这些变换是在平移确定顶点位置之后应用的。换言之，顶点 (h, k) 保持不动，而图形的其余部分被缩放并可能翻转。

For example, y = –2(x – 1)² + 4 has a = –2, h = 1, k = 4. The negative sign reflects the parabola downwards, making the vertex (1, 4) a maximum point. The factor 2 makes the graph twice as steep as the basic parabola. To graph it, plot the vertex, then from the vertex, go 1 unit right and 2 units down (instead of 1 down) to reflect the steeper slope. Repeat on the left side using symmetry.

例如，y = –2(x – 1)² + 4 中 a = –2、h = 1、k = 4。负号使抛物线向下反射，顶点 (1, 4) 成为最大值点。系数 2 使得图形比基本抛物线陡两倍。作图时，先标出顶点，然后从顶点出发，向右 1 个单位、向下 2 个单位（而非向下 1 个单位），以体现更陡的斜率。在左侧利用对称性重复这一步骤。

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9. Step-by-Step Graphing Strategy | 逐步作图策略

To graph any quadratic given in vertex form y = a(x – h)² + k, follow this systematic procedure:

要绘制任意一个以顶点式 y = a(x – h)² + k 给出的二次函数图像，请按照以下系统步骤操作：

• Identify and plot the vertex (h, k).
• 标出并绘制顶点 (h, k)。
• Draw the axis of symmetry x = h as a dashed vertical line through the vertex.
  画出对称轴 x = h，这是一条经过顶点的竖直虚线。
• Determine the direction of opening from the sign of a: upward if a > 0, downward if a < 0.
  根据 a 的符号确定开口方向：若 a > 0 则向上，若 a < 0 则向下。
• Choose one or two x-values to the right of h and compute their y-values using the equation.
  在 h 的右侧选择一到两个 x 值，并利用方程计算对应的 y 值。
• Plot these points and reflect them across the axis of symmetry to obtain points on the left.
  描出这些点，并将它们关于对称轴反射，以得到左侧的点。
• Connect the points with a smooth U-shaped (or inverted U) curve.
  用光滑的 U 形（或倒 U 形）曲线连接所有点。

This method works regardless of whether the vertex is a minimum or maximum. It is particularly valuable in non-calculator papers, where sketching must be accurate but not necessarily perfectly scaled. The step-by-step logic reinforces the link between the algebraic form and the geometric picture.

这一方法对于顶点是最小值还是最大值都适用。它在不可使用计算器的试卷中尤其宝贵，因为在那类试卷中草图要求准确，但不需要完全按比例。逐步的逻辑强化了代数形式与几何图像之间的联系。

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10. Finding Intercepts from Vertex Form | 从顶点式求截距

Even though the vertex form highlights the turning point, you may also need the x- and y-intercepts. To find the y-intercept, set x = 0 and evaluate y = a(0 – h)² + k = ah² + k. This is the point (0, ah² + k). For x-intercepts, set y = 0 and solve the equation a(x – h)² + k = 0. Rearranging gives (x – h)² = –k/a. There are two, one or no real solutions depending on the sign of –k/a.

尽管顶点式突出了转折点，你可能还需要求出 x 截距和 y 截距。要求 y 截距，设 x = 0 并计算 y = a(0 – h)² + k = ah² + k，得到点 (0, ah² + k)。对于 x 截距，设 y = 0 并解方程 a(x – h)² + k = 0。整理后得 (x – h)² = –k/a。根据 –k/a 的符号，方程可以有两个、一个或无实数解。

If –k/a > 0, the parabola crosses the x-axis at two points: x = h ± √(–k/a). If –k/a = 0, the vertex lies on the x-axis, giving a single intercept (a repeated root). If –k/a < 0, the graph does not intersect the x-axis, meaning the quadratic has no real roots. These conditions align with the discriminant of the standard form, but the vertex form provides a geometric explanation.

若 –k/a > 0，抛物线与 x 轴相交于两点：x = h ± √(–k/a)。若 –k/a = 0，顶点落在 x 轴上，给出一个截距（重根）。若 –k/a < 0，图像不与 x 轴相交，意味着二次函数没有实根。这些条件与一般式的判别式相对应，但顶点式提供了几何上的解释。

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11. Real-World Application: Maximising Area and Profit | 现实应用：面积与利润最大化

The vertex form shines in optimisation problems. Suppose you have 40 metres of fencing and want to enclose a rectangular area against a wall. If x is the length of the sides perpendicular to the wall, the area A = x(40 – 2x) = –2x² + 40x. Completing the square gives A = –2(x – 10)² + 200. The vertex (10, 200) tells you the maximum area is 200 m², achieved when x = 10 m.

顶点式在优化问题中表现出色。假设你有 40 米长的围栏，并希望靠墙围出一块矩形区域。设垂直于墙的两边长度为 x，则面积 A = x(40 – 2x) = –2x² + 40x。通过配方得到 A = –2(x – 10)² + 200。顶点 (10, 200) 告诉你，最大面积为 200 平方米，此时 x = 10 米。

Similarly, a business profit model P = –5(x – 30)² + 4500 indicates that the maximum profit is 4500 currency units when the price is set at 30 units. The negative a ensures a downward-opening parabola, confirming a maximum. Being able to interpret such expressions without calculus is a highly assessed skill in IB Mathematics: applications and interpretation.

同样，企业利润模型 P = –5(x – 30)² + 4500 表明，当价格设定为 30 单位时，最大利润为 4500 货币单位。负的 a 值保证了抛物线开口向下，确认是一个最大值。能够不借助微积分就解读这类表达式，是 IB 数学“应用与解释”课程中考查的重点技能。

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12. Summary and Common Pitfalls | 总结与常见误区

Mastering y = a(x – h)² + k means understanding that every quadratic graph is a transformation of y = x². The vertex (h, k) is your anchor; the sign and size of a dictate the shape and direction; the axis x = h gives symmetry. Always be alert to the sign inside the bracket: (x – h) shifts right, not left. And remember that a negative k does not automatically mean the graph is entirely below the x-axis – its position relative to the vertex determines the intercepts.

掌握 y = a(x – h)² + k 意味着理解每一个二次函数图像都是 y = x² 经过变换而成的。顶点 (h, k) 是你的锚点；a 的符号和大小决定了形状和方向；对称轴 x = h 提供了对称性。始终注意括号内的符号：(x – h) 是向右平移而非向左。同时记住，负的 k 并不自动意味着图像完全在 x 轴下方——图形相对于顶点的位置决定了截距的情况。

When approaching an investigation or exam question, write the function in vertex form if it isn’t already. Sketch the vertex immediately, note whether it is a max or min, and then use the axis and a couple of plotted points to complete the curve. This strategy will serve you well across all topics involving quadratic models, conic sections and beyond.

在处理探究或考试题目时，若函数尚未化为顶点式，请先将其写出。立刻画出顶点，注意它是最大值还是最小值，然后利用对称轴和几个描点完成曲线的绘制。这一策略将在所有涉及二次模型、圆锥曲线乃至更广的主题中为你提供良好的帮助。

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