下表是 $y = x + \dfrac{5}{x} - 7$ 的函数值表。

x	0.5	1	1.5	2	3	4	5	6	7
y	3.5	-1.0	-2.2	-2.5	-2.3	-1.8	k	-0.2	0.7

1. Evaluate the y-value when x = 5

When $x=5$,
$k = 5 + \frac{5}{5} - 7 = 5 + 1 - 7 = -1.$

1. Use the completed table of coordinates for the graph

From the table and part (a), the points to plot are
$(0.5,3.5),\ (1,-1.0),\ (1.5,-2.2),\ (2,-2.5),\ (3,-2.3),\ (4,-1.8),\ (5,-1),\ (6,-0.2),\ (7,0.7).$

2. Plot the points accurately on the Cartesian grid

Plot each coordinate at its correct position on the grid, using the $x$-value as the horizontal coordinate and the $y$-value as the vertical coordinate.

3. Draw the graph as a smooth curve over the required interval

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

1. Relate the required equation to the plotted graph

The graph drawn is
$y=x+\dfrac{5}{x}-7$.

Given
$x+\dfrac{5}{x}=5$,
subtract $7$ from both sides:
$x+\dfrac{5}{x}-7=5-7=-2.$
So the required solutions are the $x$-coordinates where the graph has $y=-2$.

2. Draw the line corresponding to the required y-value

On the graph, draw the horizontal line
$y=-2.$

3. Read off the two intersection x-coordinates

The line $y=-2$ cuts the curve at approximately
$x=1.4 \quad \text{and} \quad x=3.6.$

1. Draw the tangent at the given point on the curve

At $(4,-1.8)$, draw a straight line which just touches the curve and has the same direction as the curve at that point.

2. Estimate the gradient of the tangent

Using two points on the tangent, estimate
$\text{gradient}=\frac{\text{vertical change}}{\text{horizontal change}}.$
From the tangent drawn on the graph,
$\text{gradient}\approx 0.7.$

1. Form an equation for the intersection points

At the intersections,

$x + \frac{5}{x} - 7 = -x + 3$

2. Rearrange the equation into quadratic form

$x + \frac{5}{x} - 7 = -x + 3$

$2x + \frac{5}{x} - 10 = 0$

Multiplying by $x$,

$2x^2 - 10x + 5 = 0$

3. Compare with the given quadratic to identify $P$ and $Q$

Compare

$2x^2 - 10x + 5 = 0$

with

$Px^2 + Qx + 5 = 0.$

Hence,

$P = 2, \quad Q = -10.$