$y = \dfrac{x^3}{4} - 2x + 3$ 。

(a)
完成 $y = \dfrac{x^3}{4} - 2x + 3$ 的数值表。
(b)
在对面的网格中，画出 $-4 \le x \le 4$ 时 $y = \dfrac{x^3}{4} - 2x + 3$ 的图像。
(c)
解释你的图像是如何说明方程 $\dfrac{x^3}{4} - 2x + 3 = 0$ 只有唯一解的。

...
(d)
方程 $x^3 - 10x + 4 = 0$ 可以通过寻找直线 $y = ax + b$ 与曲线 $y = \dfrac{x^3}{4} - 2x + 3$ 的交点来求解。
1. (i)
  求 $a$ 的值和 $b$ 的值。
  
  答案 $a =$ ...
  
  $b =$ ...
2. (ii)
  通过画出直线 $y = ax + b$ ，求解方程 $x^3 - 10x + 4 = 0$ 。
  
  答案 $x =$ ... 或 ... 或 ...

幂函数图像一次函数与二次函数有序数对与图像直角坐标系一次式的加减求公式中的未知量数系与四则运算直线方程一次函数图像统计图表分析与解读 English

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完整解法

3(a)

1. Find the missing table value by substituting $x=-4$ into the formula.

For $x=-4$ ,
$y=\frac{(-4)^3}{4}-2(-4)+3$
$=\frac{-64}{4}+8+3$
$=-16+8+3=-5$

答案

The missing value is $-5$ .

Completed entry: when $x=-4$ , $y=-5$ .

3(b)

1. Use the completed table values as coordinates for plotting.

From the table,

\begin{array}{c|ccccccccc} x&-4&-3&-2&-1&0&1&2&3&4\\ \hline y&-5&2.25&5&4.75&3&1.25&1&3.75&11 \end{array}

So the points to plot are
$(-4,-5),\ (-3,2.25),\ (-2,5),\ (-1,4.75),\ (0,3),\ (1,1.25),\ (2,1),\ (3,3.75),\ (4,11).$

2. Plot all the points accurately on the given grid.

Plot each coordinate pair on the grid, using the given scales on the $x$ -axis and $y$ -axis. For example, $(-4,-5)$ is 4 units left of the origin and 5 units below the origin, while $(4,11)$ is 4 units right of the origin and 11 units above the origin.

3. Join the plotted points with a smooth curve over the stated domain.

Join the plotted points with a smooth curve, not with straight line segments, from $x=-4$ to $x=4$ . The curve should pass through all plotted points and have the general cubic shape implied by
$y=\frac{x^3}{4}-2x+3.$

答案

A smooth cubic curve drawn through the points
$(-4,-5),\ (-3,2.25),\ (-2,5),\ (-1,4.75),\ (0,3),\ (1,1.25),\ (2,1),\ (3,3.75),\ (4,11).$

3(c)

1. Explain the graphical meaning of the solution and count the x-intercepts.

The solutions of $\dfrac{x^3}{4}-2x+3=0$ are the points where the graph of $y=\dfrac{x^3}{4}-2x+3$ has $y=0$ , i.e. where it cuts the $x$ -axis. From the graph, the curve cuts the $x$ -axis only once.

答案

The curve crosses the $x$ -axis only once, so $\dfrac{x^3}{4}-2x+3=0$ has only one solution.

3(d)(i)

1. Write the equation formed when the line intersects the curve.

At the points of intersection,
$\frac{x^3}{4}-2x+3=ax+b.$
Rearranging,
$\frac{x^3}{4}-(2+a)x+(3-b)=0.$

2. Match coefficients with the given cubic equation to find $a$ and $b$.

The given equation is
$x^3-10x+4=0.$
Divide by $4$ so that the coefficient of $x^3$ matches:
$\frac{x^3}{4}-\frac{5}{2}x+1=0.$
Compare with
$\frac{x^3}{4}-(2+a)x+(3-b)=0.$
So
$2+a=\frac{5}{2} \quad\Rightarrow\quad a=\frac{1}{2},$
and
$3-b=1 \quad\Rightarrow\quad b=2.$

答案

$a=\dfrac{1}{2}$

$b=2$

3(d)(ii)

1. Use the values of $a$ and $b$ from part (d)(i) to identify the straight line to draw.

From part (d)(i),
$a=\frac12,\qquad b=2.$
So the required line is
$y=\frac12x+2.$

2. Draw the straight line accurately on the same axes as the curve.

Choose two convenient points on
$y=\frac12x+2.$
For example,
$x=0 \Rightarrow y=2,$
$x=4 \Rightarrow y=4.$
Draw the straight line through $(0,2)$ and $(4,4)$ on the same graph as the curve.

3. Read off the $x$-coordinates of the intersection points.

The solutions of
$x^3-10x+4=0$
are the $x$ -coordinates where the line
$y=\frac12x+2$
intersects the curve
$y=\frac{x^3}{4}-2x+3.$
Reading from the graph gives approximately
$x=-3.35,\quad x=0.41,\quad x=2.94.$

答案

$x \approx -3.35$ or $0.41$ or $2.94$

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y=x34−2x+3y = \dfrac{x^3}{4} - 2x + 3y=4x3​−2x+3。