完成 $y = x - \frac{4}{x^2}$ 的值表。在适当的地方，数值保留到一位小数。

(a)
完成 $y = x - \dfrac{4}{x^2}$ 的值表。在适当的地方，数值保留到一位小数。
(b)
在第 11 页的网格上，画出 $-15 \le x \le 15$ 时 $y = x - \dfrac{4}{x^2}$ 的图像。

该图像在 $x = 0$ 处有一条渐近线（图像靠近但不穿过 y 轴）。
(c)
1. (i)
  在同一个网格上，画出 $-15 \le x \le 15$ 时 $2y - 2x = -7$ 的图像。
2. (ii)
  写出直线与曲线相交交点的 x 坐标。
  
  答案：x = ... 或 ...
3. (iii)
  第 (ii) 部分中的这些 x 值是方程
  
  $Ax^2 + B = 0$
  
  的解。求 A 的值和 B 的值。
  
  答案：A = ... B = ...

幂函数图像一次函数图像列方程解应用题数系与四则运算近似与估算直角坐标系有序数对与图像直线方程公式变形用字母表示数 English

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

7(a)

1. Find the missing value of $y$ when $x=-10$.

For $x=-10$ ,
$y=x-\frac{4}{x^2}$
$y=-10-\frac{4}{(-10)^2}$
$y=-10-\frac{4}{100}$
$y=-10-0.04=-10.04$
To one decimal place,
$y=-10.0$

答案

Missing table value: $\boxed{-10.0}$

7(b)

1. Use the completed table values as coordinates to be plotted.

From part (a), the coordinates are
$(-15,-15.0)$ , $(-10,-10.0)$ , $(-5,-5.2)$ , $(-1,-5)$ , $(1,-3)$ , $(5,4.8)$ , $(10,10.0)$ , $(15,15.0)$ .

2. Plot the points accurately on the grid.

Plot the points:
$(-15,-15.0),\ (-10,-10.0),\ (-5,-5.2),\ (-1,-5)$
on the left of the $y$ -axis, and
$(1,-3),\ (5,4.8),\ (10,10.0),\ (15,15.0)$
on the right of the $y$ -axis.

3. Draw the curve with the correct shape and asymptote behaviour.

Join the plotted points with a smooth curve in two separate branches. Since $x=0$ is an asymptote, the curve must go near but must not touch or cross the $y$ -axis.

答案

Graph of $y=x-\dfrac{4}{x^2}$ drawn for $-15\leq x\leq 15$ , passing through the plotted points and approaching but not touching the $y$ -axis.

7(c)(i)

1. Rewrite the equation in the form $y=mx+c$.

$2y-2x=-7$

$2y=2x-7$

$y=x-3.5$

2. Choose two convenient points and draw the line through them for the given range.

For $y=x-3.5$ ,

when $x=-15$ , $y=-15-3.5=-18.5$ .

when $x=15$ , $y=15-3.5=11.5$ .

Plot $(-15,-18.5)$ and $(15,11.5)$ , then join them with a straight line.

答案

Straight line $y=x-3.5$ drawn on the same grid for $-15\leq x\leq 15$ .

7(c)(ii)

1. Identify where the straight line crosses the curve and read the corresponding $x$-coordinates.

From the completed graph, the line $y=x-3.5$ intersects the curve $y=x-\dfrac{4}{x^2}$ at two points.

Reading the $x$ -coordinates of these two intersection points gives approximately
$x\approx -1.1 \quad \text{and} \quad x\approx 1.1.$

答案

$x=\boxed{-1.1}$ or $\boxed{1.1}$

7(c)(iii)

1. Form the equation satisfied by the intersection points and compare coefficients.

At the points of intersection, $y=x-\dfrac{4}{x^2}$ also satisfies

$2y-2x=-7.$

Substitute $y=x-\dfrac{4}{x^2}$ :

$2\left(x-\frac{4}{x^2}\right)-2x=-7$

$2x-\frac{8}{x^2}-2x=-7$

$-\frac{8}{x^2}=-7$

Multiply by $x^2$ :

$-8=-7x^2$

$7x^2-8=0.$

Comparing with $Ax^2+B=0$ ,

$A=7,\quad B=-8.$

答案

$A=\boxed{7},\quad B=\boxed{-8}$

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完成 y=x−4x2y = x - \frac{4}{x^2}y=x−x24​ 的值表。 在适当的地方，数值保留到一位小数。