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CIE 9709 2026 June Paper 11 Q3

A Level / CIE / P1

CIE 9709 2026 June Paper 11 Paper · Question 3

题目

Problem

The curve with equation y=x3y=x^3 is stretched with scale factor 44 in the xx-direction and then translated by (5−2)\begin{pmatrix}5\\-2\end{pmatrix}.

Find the equation of the transformed curve. You do not need to simplify your answer.

(3)
题目中文翻译

方程为 y=x3y=x^3 的曲线先沿 xx 轴方向作伸缩,伸缩因子为 44,再按向量 (5−2)\begin{pmatrix}5\\-2\end{pmatrix} 平移。

求变换后曲线的方程。答案不必化简。

(3)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

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横向伸缩时用 x/4x/4 替换原来的 xx。随后右移 55、下移 22。

答题过程

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The stretch gives y=(x/4)3y=(x/4)^3. The translation then replaces xx by x−5x-5 and subtracts 22 from the height. Thus the transformed equation is

y=(x−54)3−2.y=\Big(\frac{x-5}{4}\Big)^3-2.