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CIE 9709 2025 Nov Paper 13 Q6

A Level / CIE / P1

CIE 9709 2025 Nov Paper 13 Paper · Question 6

题目

Problem

The diagram shows the graphs of y=x3y=x^3 and y=f(x)y=f(x). The graph of y=x3y=x^3 is transformed to the graph of y=f(x)y=f(x) by a sequence of transformations.

(a) Describe fully a suitable sequence of transformations. Make clear the order in which the transformations are applied.

[5]

(b) You are given that f(x)=a(x+b)3+cf(x)=a(x+b)^3+c.

State the values of the constants aa, bb and cc.

[3]
题目中文翻译

图中显示的是 y=x3y=x^3 和 y=f(x)y=f(x) 的图像。y=x3y=x^3 的图像通过一系列变换得到 y=f(x)y=f(x) 的图像。

(a) 完整描述一种合适的变换顺序。需要清楚说明这些变换应用的先后顺序。

(b) 已知 f(x)=a(x+b)3+cf(x)=a(x+b)^3+c。

写出常数 aa、bb 和 cc 的值。

解答

(a)

解法一

思路

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原曲线的水平拐点是 (0,0)(0,0),新曲线的水平拐点是 (3,−4)(3,-4)。比较两条曲线上各自向右一格的点,可以看出竖直伸缩比例。

答题过程

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On y=x3y=x^3, the point one unit to the right of the stationary inflection has height 11 above it. On y=f(x)y=f(x), the corresponding point (4,−2)(4,-2) is two units above (3,−4)(3,-4). A suitable sequence is:

  1. Stretch by scale factor 22 in the yy-direction.
  2. Translate by (3−4)\begin{pmatrix}3\\-4\end{pmatrix}.

In this order the image has equation y=2(x−3)3−4y=2(x-3)^3-4.

解法二

思路

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若先平移再伸缩,平移造成的纵向位移也会乘以 22。因此先下移 22,伸缩后就会下移 44。

答题过程

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First translate by (3−2)\begin{pmatrix}3\\-2\end{pmatrix}, giving y=(x−3)3−2y=(x-3)^3-2. Then stretch by scale factor 22 in the yy-direction to obtain y=2(x−3)3−4y=2(x-3)^3-4. The stationary inflection maps to (3,−4)(3,-4) and (1,1)(1,1) maps to (4,−2)(4,-2).

(b)

解法一

思路

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把变换后的方程与题目给出的形式逐项比较。注意 x−3x-3 对应 b=−3b=-3。

答题过程

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Comparing f(x)=2(x−3)3−4f(x)=2(x-3)^3-4 with a(x+b)3+ca(x+b)^3+c gives a=2a=2, b=−3b=-3 and c=−4c=-4.