📚 Exercise 5C: Mastering Function Transformations | 练习5C：掌握函数变换
Welcome to a comprehensive walkthrough of Exercise 5C, focusing on one of the most vital topics in IB Mathematics: function transformations. Mastering how shifts, stretches, and reflections alter the graph of a parent function is essential for success in both Analysis & Approaches and Applications & Interpretation courses. This article unpacks key concepts, provides step-by-step examples, and highlights common pitfalls to ensure you tackle Exercise 5C with confidence.
欢迎来到练习5C的全面解析，聚焦IB数学中最关键的主题之一：函数变换。掌握平移、伸缩和反射如何改变父函数的图像，对于在分析与方法和应用与解释课程中取得成功至关重要。本文将拆解关键概念，提供分步例题，并强调常见陷阱，确保你自信应对练习5C。
1. Understanding Parent Functions | 理解父函数
nsformations, it is crucial to recognise common parent functions such as f(x) = x², f(x) = √x, f(x) = 1/x, f(x) = eˣ, f(x) = ln x, f(x) = |x| , and tan x. Their basic shapes serve as the foundation for all shifted and scaled graphs.
在应用变换之前，识别常见的父函数至关重要，例如 f(x) = x²、f(x) = √x、f(x) = 1/x、f(x) = eˣ、f(x) = ln x、f(x) = |x| 以及三角函数 sin x、cos x 和 tan x。它们的基本形状是所有平移和缩放图像的基石。
IB exam questions often start by asking you to sketch a transformed function without a calculator. Internalising the key points of each parent graph—such as the vertex of a parabola or the asymptotes of a reciprocal function—means you can quickly apply changes.
IB考题经常要求你在不使用计算器的情况下绘制变换后的函数图像。内化每个父函数图像的关键点——例如抛物线的顶点或倒数函数的渐近线——意味着你能快速应用变化。
- Quadratic: vertex (0,0), axis of symmetry x=0.
- Square root: starts at (0,0), domain x ≥ 0.
- Exponential: y-intercept (0,1), horizontal asymptote y=0.
- 二次函数：顶点 (0,0)，对称轴 x=0。
- 平方根函数：起点 (0,0)，定义域 x ≥ 0。
- 指数函数：y轴截距 (0,1)，水平渐近线 y=0。
2. Vertical Translation: f(x) + k | 垂直平移：f(x) + k
Adding a constant k to a function shifts the graph vertically. If k > 0, the graph moves up; if k < 0, it moves down. The shape remains unchanged.
给函数加上常数 k 会使图像垂直移动。若 k > 0，图像上移；若 k < 0，图像下移。形状保持不变。
ndard parabola and lifts it 3 units upward. Every y-coordinate increases by 3, so the vertex becomes (0,3).
例如，g(x) = x² + 3 将标准抛物线上移 3 个单位。每个 y 坐标增加 3，因此顶点变为 (0,3)。
Similarly, h(x) = x² − 2 moves the parabola down by 2, placing the vertex at (0,−2).
同样，h(x) = x² − 2 将抛物线下移 2，顶点落在 (0,−2)。
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