题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

A curve $C$ has equation $y=f(x)$ where

\begin{align*} f(x)=2(x+1)(x-3)^2. \end{align*}

(a) Sketch a graph of $C$ .

Show on your graph the coordinates of the points where $C$ cuts or meets the coordinate axes.

(3)

(b) Write $f(x)$ in the form $ax^3+bx^2+cx+d$ , where $a$ , $b$ , $c$ and $d$ are constants to be found.

(3)

(4)

解答

(a)

解法一

思路

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因式形式直接给出 $x$ 轴交点： $x=-1$ 是单根，图像穿过； $x=3$ 是重根，图像接触后折回。 $y$ 轴截距令 $x=0$ 。

答题过程

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The $x$ -intercepts are found from

\begin{align*} 2(x+1)(x-3)^2=0. \end{align*}

\begin{align*} x=-1\quad\text{or}\quad x=3. \end{align*}

At $x=0$ ,

\begin{align*} y=&\,2(0+1)(0-3)^2\\ =&\,18. \end{align*}

The sketch is a positive cubic. It crosses the $x$ -axis at $(-1,0)$ , touches the $x$ -axis at $(3,0)$ , and crosses the $y$ -axis at $(0,18)$ .

(b)

解法一

思路

展开

先展开平方项，再乘以 $2(x+1)$ 。

答题过程

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\begin{align*} f(x) =&\,2(x+1)(x-3)^2\\ =&\,2(x+1)(x^2-6x+9)\\ =&\,2(x^3-6x^2+9x+x^2-6x+9)\\ =&\,2(x^3-5x^2+3x+9)\\ =&\,2x^3-10x^2+6x+18. \end{align*}

(c)

解法一

思路

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由 (b) 求导，再代入 $x=\frac13$ 找切线斜率。同时还要代入原函数找切点的 $y$ 坐标。

答题过程

展开

From part (b),

\begin{align*} f(x)=2x^3-10x^2+6x+18. \end{align*}

\begin{align*} f'(x)=6x^2-20x+6. \end{align*}

At $x=\frac13$ ,

\begin{align*} f'\left(\frac13\right) =&\,6\left(\frac13\right)^2-20\left(\frac13\right)+6\\ =&\,\frac23-\frac{20}{3}+6\\ =&\,0. \end{align*}

The tangent is horizontal. The $y$ coordinate is

\begin{align*} f\left(\frac13\right) =&\,2\left(\frac13+1\right)\left(\frac13-3\right)^2\\ =&\,2\cdot\frac43\cdot\left(-\frac83\right)^2\\ =&\,\frac83\cdot\frac{64}{9}\\ =&\,\frac{512}{27}. \end{align*}

Therefore the tangent is

\begin{align*} y=\frac{512}{27}. \end{align*}

IAL 2021 Oct Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程