完成 $y = 2x + \dfrac{20}{x} - 15$ 的数值表。

x	1	2.5	5	10	12.5	15	20
y	7		-1	7	11.6	16.3	26

在网格上画出当 $0 < x \le 20$ 时 $y = 2x + \dfrac{20}{x} - 15$ 的图象。 利用图象，写出使 $2x + \dfrac{20}{x} < 22$ 的 x 的取值范围。 答 ... 通过作切线，求曲线在 x = 4 处的斜率。 答 ... 通过在网格上画一条合适的直线，求方程 x^2 - 13x + 20 = 0 的解。 答 x = ... 或 ...

1. Substitute $x=2.5$ into the formula for $y$.

For $x=2.5$,
$y=2(2.5)+\frac{20}{2.5}-15$

2. Evaluate the numerical expression accurately.

$y=5+8-15$
$y=-2$

1. List the points to be plotted using the completed table.

From the table, the coordinates are
$(1,7),\ (2.5,-2),\ (5,-1),\ (10,7),\ (12.5,11.6),\ (15,16.3),\ (20,26).$

2. Plot the coordinates accurately on the given grid.

Plot the points
$(1,7),\ (2.5,-2),\ (5,-1),\ (10,7),\ (12.5,11.6),\ (15,16.3),\ (20,26)$
on the grid, using the given scales on both axes.

3. Draw the required graph as a smooth curve over the stated domain.

Join the plotted points with a smooth curve, not straight line segments. The graph is only for
$0<x\le 20,$
so it should be drawn only to the right of the $y$-axis up to $x=20$.

1. Relate the given inequality to the graph that has been drawn.

The graph drawn is
$y=2x+\frac{20}{x}-15.$
Given
$2x+\frac{20}{x}<22,$
subtract $15$ from both sides:
$2x+\frac{20}{x}-15<7.$
So we need the part of the graph where
$y<7.$

2. Read from the graph where the curve is below the line $y=7$.

From the graph, the curve meets $y=7$ at
$x=1 \quad \text{and} \quad x=10.$
The curve is below $y=7$ between these two values of $x$.

Since the inequality is strict, the endpoints are not included.

1. Locate the point on the curve where $x=4$ and draw the tangent there.

From the graph of $y=2x+\dfrac{20}{x}-15$, locate the point on the curve with $x=4$.

Draw a straight line that just touches the curve at this point and follows the direction of the curve there.

2. Use two points on the tangent to calculate its gradient.

Choose two well-spaced points on the tangent and calculate
$\text{gradient}=\frac{\text{vertical change}}{\text{horizontal change}}.$

For an accurately drawn tangent, suitable readings give approximately
$\text{gradient}\approx \frac{1-(-5)}{8-0}=\frac{6}{8}=0.75.$

1. Rewrite the quadratic equation in a form involving the curve already drawn.

Since $0<x\leq 20$, divide $x^2-13x+20=0$ by $x$:
$x-13+\frac{20}{x}=0.$
Using the curve
$y=2x+\frac{20}{x}-15,$
we want a straight line such that
$2x+\frac{20}{x}-15=x-2.$
This simplifies to
$x+\frac{20}{x}-13=0,$
which is equivalent to the required equation. So draw the straight line
$y=x-2.$

2. Read the solutions from the intersections of the line and the curve.

Draw the straight line $y=x-2$ on the same grid. The solutions are the $x$-coordinates of the points where this line intersects the curve.

From the graph,
$x\approx 1.8 \quad \text{or} \quad x\approx 11.2.$