完成 $y=\dfrac{2}{x^{2}}-2x$ 的数值表。在适当的地方，数值保留到小数点后一位。在对面的网格上，画出 $-3\le x\le 3$ 范围内的 $y=\dfrac{2}{x^{2}}-2x$ 图像。在同一个网格上，画出 $-3\le x\le 3$ 范围内的 $y+x=6$ 图像。写出直线与曲线交点的 $x$ 坐标。答案 $x$ = ... 和 ... 这些 $x$ 的值是方程 $x^{3}+Ax^{2}+B=0$ 的解。求 $A$ 和 $B$ 的值。答案 $A$ = ... $B$ = ...

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它不会直接告诉你答案，而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

5(a)

1. Evaluate the formula for each given value of $x$.

Using $y=\dfrac{2}{x^2}-2x$ :

\begin{aligned} x=-3: \quad y&=\frac{2}{(-3)^2}-2(-3)=\frac{2}{9}+6=6.222\ldots\approx 6.2,\\ x=-2: \quad y&=\frac{2}{(-2)^2}-2(-2)=\frac{2}{4}+4=4.5,\\ x=-1: \quad y&=\frac{2}{(-1)^2}-2(-1)=2+2=4.0,\\ x=-0.5: \quad y&=\frac{2}{(-0.5)^2}-2(-0.5)=8+1=9.0,\\ x=-0.4: \quad y&=\frac{2}{(-0.4)^2}-2(-0.4)=12.5+0.8=13.3,\\ x=0.5: \quad y&=\frac{2}{(0.5)^2}-2(0.5)=8-1=7.0,\\ x=1: \quad y&=\frac{2}{1^2}-2(1)=2-2=0.0,\\ x=2: \quad y&=\frac{2}{2^2}-2(2)=0.5-4=-3.5,\\ x=3: \quad y&=\frac{2}{3^2}-2(3)=\frac{2}{9}-6=-5.777\ldots\approx -5.8. \end{aligned}

答案

\begin{array}{c|rrrrrrrrr} x&-3&-2&-1&-0.5&-0.4&0.5&1&2&3\\ \hline y&6.2&4.5&4.0&9.0&13.3&7.0&0.0&-3.5&-5.8 \end{array}

5(b)

1. Use the completed table to identify the points to be plotted.

From the table in part (a), plot the points
$(-3,6.2),\ (-2,4.5),\ (-1,4.0),\ (-0.5,9.0),\ (-0.4,13.3),\ (0.5,7.0),\ (1,0.0),\ (2,-3.5),\ (3,-5.8).$

2. Plot the points accurately on the grid.

On the given axes, mark each ordered pair at its correct position. For example, $(-3,6.2)$ is 3 units left of the origin and slightly above $y=6$ , while $(3,-5.8)$ is 3 units right of the origin and slightly above $y=-6$ .

3. Join the plotted points with the correct curve shape.

Join the plotted points with a smooth curve. Since $y=\dfrac{2}{x^{2}}-2x$ is undefined at $x=0$ , draw two separate smooth branches: one for $-3\le x<0$ and one for $0<x\le3$ . Do not join the curve across the $y$ -axis.

答案

A smooth curve of $y=\dfrac{2}{x^{2}}-2x$ drawn for $-3\le x\le 3$ , passing through the plotted points $(-3,6.2)$ , $(-2,4.5)$ , $(-1,4.0)$ , $(-0.5,9.0)$ , $(-0.4,13.3)$ , $(0.5,7.0)$ , $(1,0.0)$ , $(2,-3.5)$ and $(3,-5.8)$ , with separate branches on either side of $x=0$ .

5(c)(i)

1. Rewrite the line equation in graphing form.

$y+x=6$
$y=6-x$

2. Find two points on the line within the required range.

For $x=-3$ ,
$y=6-(-3)=9.$
For $x=3$ ,
$y=6-3=3.$
So two points are $(-3,9)$ and $(3,3)$ .

3. Draw the straight line on the given grid for the specified interval.

Plot $(-3,9)$ and $(3,3)$ on the same grid, then join them with a straight line segment from $x=-3$ to $x=3$ .

答案

Draw the straight line $y=6-x$ for $-3\le x\le 3$ , for example through the points $(-3,9)$ and $(3,3)$ .

5(c)(ii)

1. Locate the intersection points of the line and the curve on the same grid.

On the completed graph, find where the straight line $y=6-x$ intersects the curve $y=\dfrac{2}{x^2}-2x$ .

There are two intersection points within $-3\leq x\leq 3$ .

2. Read off the two corresponding $x$-coordinates.

Reading the $x$ -coordinates from the grid gives approximately
$x\approx -0.6$
and
$x\approx 0.55.$

答案

$x \approx -0.6$ and $x \approx 0.55$

5(c)(iii)

1. Equate the curve and the line at their points of intersection.

At an intersection,
$y=\frac{2}{x^2}-2x$
and
$y+x=6 \quad \Rightarrow \quad y=6-x.$
So
$\frac{2}{x^2}-2x=6-x.$

2. Clear the denominator and rearrange into cubic form.

Multiply throughout by $x^2$ :
$2-2x^3=6x^2-x^3.$
Rearrange:
$2-x^3-6x^2=0$
$x^3+6x^2-2=0.$

3. Compare coefficients with the given equation.

Compare
$x^3+6x^2-2=0$
with
$x^3+Ax^2+B=0.$
Hence
$A=6, \qquad B=-2.$

答案

$A=6$

$B=-2$

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