题目

Problem

A function $f$ is defined by

\begin{align*} f(x)=3+12x-2x^2,\qquad x\in\mathbb{R}. \end{align*}

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

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

where $a$ , $b$ and $c$ are integers to be found.

(3)

The curve with equation $y=f(x)-7$ crosses the $x$ -axis at the points $P$ and $Q$ and crosses the $y$ -axis at the point $R$ .

(b) Find the area of triangle $PQR$ , giving your answer in the form $m\sqrt{n}$ , where $m$ and $n$ are integers to be found.

(4)

解答

展开

先把 $-2$ 提出来，然后在括号里配方。注意目标形式是 $a-b(x+c)^2$ ，所以最后要把常数项整理到前面。

展开

\begin{align*} f(x)=&\,3+12x-2x^2\\ =&\,-2(x^2-6x)+3\\ =&\,-2\{(x-3)^2-9\}+3\\ =&\,21-2(x-3)^2. \end{align*}

展开

先写出新曲线 $y=f(x)-7$ 。点 $R$ 是 $y$ 轴截距，点 $P,Q$ 是两个 $x$ 轴截距。三角形的底边在 $x$ 轴上，高就是 $R$ 到 $x$ 轴的距离。

展开

Using part (a),

\begin{align*} y=&\,f(x)-7\\ =&\,21-2(x-3)^2-7\\ =&\,14-2(x-3)^2. \end{align*}

At the $y$ -axis, $x=0$ , so

\begin{align*} y=14-2(0-3)^2=-4. \end{align*}

Thus $R=(0,-4)$ .

For the $x$ -intercepts,

\begin{align*} 14-2(x-3)^2=&\,0\\ (x-3)^2=&\,7\\ x=&\,3\pm\sqrt7. \end{align*}

\begin{align*} PQ=&\,(3+\sqrt7)-(3-\sqrt7)\\ =&\,2\sqrt7. \end{align*}

The height of the triangle is $4$ , therefore

\begin{align*} \text{Area} =&\,\frac12(2\sqrt7)(4)\\ =&\,4\sqrt7. \end{align*}