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CIE 9709 2022 June Paper 13 Q4

A Level / CIE / P1

CIE 9709 2022 June Paper 13 Paper · Question 4

题目

Problem

(a) The curve with equation y=x2+2x−5y=x^2+2x-5 is translated by (−13)\begin{pmatrix}-1\\3\end{pmatrix}.

Find the equation of the translated curve, giving your answer in the form y=ax2+bx+cy=ax^2+bx+c.

(3)

(b) The curve with equation y=x2+2x−5y=x^2+2x-5 is transformed to a curve with equation y=4x2+4x−5y=4x^2+4x-5.

Describe fully the single transformation that has been applied.

(2)
题目中文翻译

(a) 方程为 y=x2+2x−5y=x^2+2x-5 的曲线按向量 (−13)\begin{pmatrix}-1\\3\end{pmatrix} 平移。

求平移后曲线的方程,答案写成 y=ax2+bx+cy=ax^2+bx+c 的形式。

(3)

(b) 方程为 y=x2+2x−5y=x^2+2x-5 的曲线经过变换得到方程为 y=4x2+4x−5y=4x^2+4x-5 的曲线。

完整描述所实施的单一变换。

(2)
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

(a)

解法一

思路

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向左平移 11 时,把新横坐标加 11 得到旧横坐标。向上平移再给函数值加 33。

答题过程

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Writing f(x)=x2+2x−5f(x)=x^2+2x-5, the translated graph is

y=f(x+1)+3=(x+1)2+2(x+1)−5+3=x2+4x+1.y=f(x+1)+3=(x+1)^2+2(x+1)-5+3=x^2+4x+1.

(b)

解法一

思路

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比较两式,留意二次项和一次项的倍数不同,而常数不变。这对应把输入改成 2x2x。

答题过程

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The new expression is f(2x)=(2x)2+2(2x)−5f(2x)=(2x)^2+2(2x)-5. This is a stretch parallel to the xx-axis by scale factor 1/21/2.