变量 x 和 y 由方程 $y = 2x + \dfrac{6}{x} - 7$ 联结。下表给出了 x 和 y 的一些对应值。

(a)
计算 a 的值。
(b)
在网格中，画出 $0.5 \le x \le 7$ 范围内 $y = 2x + \dfrac{6}{x} - 7$ 的图像。
(c)
点 A 的坐标为 (3,,-1)。可以作该曲线的一条切线，使该切线经过点 A。
1. (i)
  在同一个网格上画出这条切线。
2. (ii)
  求该切线的方程。
(d)
通过在你的图上画一条合适的直线，解释为什么方程 $3x^2 - 8x + 6 = 0$ 在 $0.5 \le x \le 7$ 范围内没有解。

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完整解法

11(a)

1. Identify the x-value for which the y-value is a and substitute it into the equation.

From the table, $a$ is the value of $y$ when $x=2$ .

$a=2(2)+\frac{6}{2}-7$
$=4+3-7$
$=0$

答案

$a=0$

11(b)

1. Use the completed table of values for plotting.

From part (a), $a=0$ , so the point for $x=2$ is $(2,0)$ .

The points to plot are:
$(0.5,6),\ (1,1),\ (1.5,0),\ (2,0),\ (3,1),\ (4,2.5),\ (5,4.2),\ (6,6),\ (7,7.9).$

2. Plot the points accurately and draw the curve over the required interval.

Plot all the points on the given grid.

Join the points with a smooth curve from $x=0.5$ to $x=7$ , without using straight line segments between successive points.

答案

A smooth curve drawn through the plotted points $(0.5,6)$ , $(1,1)$ , $(1.5,0)$ , $(2,0)$ , $(3,1)$ , $(4,2.5)$ , $(5,4.2)$ , $(6,6)$ and $(7,7.9)$ for $0.5 \le x \le 7$ .

11(c)(i)

1. Draw the tangent through the given external point so that it just touches the curve.

On the completed graph, mark $A(3,-1)$ . Place a ruler through $A$ and adjust it until the straight line just touches the curve at one point. Draw this line.

The point of contact is approximately $(1.5,0)$ , so the tangent should pass through $A(3,-1)$ and touch the curve near $(1.5,0)$ .

答案

A tangent line drawn through $A(3,-1)$ , touching the curve at approximately $(1.5,0)$ .

11(c)(ii)

1. Find the gradient of the tangent using two points on the tangent.

Using the points $(1.5,0)$ and $A(3,-1)$ ,
$m=\frac{-1-0}{3-1.5}=\frac{-1}{1.5}=-\frac{2}{3}.$

2. Use the gradient and a point on the tangent to find the equation of the line.

Use $y=mx+c$ and substitute $m=-\dfrac{2}{3}$ and $(3,-1)$ :
$-1=-\frac{2}{3}(3)+c$
$-1=-2+c$
$c=1.$
Hence the tangent is
$y=-\frac{2}{3}x+1.$

答案

$y=-\dfrac{2}{3}x+1$

11(d)

1. Rewrite the given equation in a form involving the plotted curve.

Since $0.5\le x\le 7$ , $x\ne 0$ , so we may divide by $x$ .

$3x^2-8x+6=0$
$3x-8+\frac{6}{x}=0$
$3x+\frac{6}{x}-8=0$

2. Connect the equation to the curve and identify the straight line to draw.

The curve already drawn is

$y=2x+\frac{6}{x}-7.$

For the quadratic equation,

$3x+\frac{6}{x}-8=0$

can be written as

$2x+\frac{6}{x}-7=1-x.$

So the suitable straight line to draw is

$y=1-x.$

3. Use the graph to justify that there is no solution in the interval.

Draw the straight line $y=1-x$ on the same grid.

From the graph, the line $y=1-x$ does not meet the curve $y=2x+\dfrac{6}{x}-7$ for $0.5\le x\le 7$ .

Therefore there is no value of $x$ in this interval satisfying

$3x^2-8x+6=0.$

答案

Draw the straight line $y=1-x$ . Since this line does not intersect the curve $y=2x+\dfrac{6}{x}-7$ for $0.5\le x\le 7$ , the equation $3x^2-8x+6=0$ has no solution in this interval.

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变量 x 和 y 由方程 y=2x+6x−7y = 2x + \dfrac{6}{x} - 7y=2x+x6​−7 联结。 下表给出了 x 和 y 的一些对应值。