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P1 2022 Jan Q2

A Level / Edexcel / P1

IAL 2022 Jan Paper · Question 2

题目

Problem

The function ff is defined by

f(x)=11−4x−2x2.\begin{align*} f(x)=11-4x-2x^2. \end{align*}

(a) Express f(x)f(x) in the form

a+b(x+c)2\begin{align*} a+b(x+c)^2 \end{align*}

where aa, bb and cc are integers to be found.

(3)

(b) Sketch the graph with equation y=f(x)y=f(x), showing the coordinates of the point at which the graph crosses the yy-axis.

(2)

(c) Write down the equation of the line of symmetry of the graph with equation y=f(x)y=f(x).

(1)

解答

(a)

解法一

思路

展开

先把 x2x^2 和 xx 项提出共同系数 −2-2，再配方。配方时注意括号外的 −2-2 会改变常数项。

答题过程

展开 f(x)= 11−4x−2x2= −2x2−4x+11= −2(x2+2x)+11= −2((x+1)2−1)+11= −2(x+1)2+2+11= 13−2(x+1)2.\begin{align*} f(x) =&\,11-4x-2x^2\\ =&\,-2x^2-4x+11\\ =&\,-2(x^2+2x)+11\\ =&\,-2\bigl((x+1)^2-1\bigr)+11\\ =&\,-2(x+1)^2+2+11\\ =&\,13-2(x+1)^2. \end{align*}

So

f(x)=13−2(x+1)2.\begin{align*} f(x)=13-2(x+1)^2. \end{align*}

(b)

解法一

思路

展开

由 (a) 可知顶点是 (−1,13)(-1,13)，且二次项系数为负，所以图像开口向下。求 yy 轴截距时令 x=0x=0。

答题过程

展开

From part (a),

y=13−2(x+1)2.\begin{align*} y=13-2(x+1)^2. \end{align*}

The graph is a downward-opening parabola with vertex (−1,13)(-1,13).

At the yy-axis, x=0x=0, so

y= 11−4(0)−2(0)2= 11.\begin{align*} y=&\,11-4(0)-2(0)^2\\ =&\,11. \end{align*}

Therefore the graph crosses the yy-axis at

(0,11).\begin{align*} (0,11). \end{align*}

(c)

解法一

思路

展开

配方形式 13−2(x+1)213-2(x+1)^2 说明顶点的 xx 坐标是 −1-1。二次函数的对称轴就是经过顶点的竖直直线。

答题过程

展开

The line of symmetry is

x=−1.\begin{align*} x=-1. \end{align*}