The table shows some values for $y = 4 - 2x - \dfrac{5}{x}$ for $0.5 \le x \le 5.5$.

1. Evaluate $y$ when $x=3$.

For the column where $x=3$, $p$ is the corresponding value of $y$.

$p=4-2(3)-\frac{5}{3}$
$=4-6-1.666\ldots$
$=-3.666\ldots$

Correct to one decimal place,

$p=-3.7$

1. Use the completed table of values for plotting.

From part (a), $p=-3.7$. Hence the points to plot are
$(0.5,-7),\ (1,-3),\ (1.5,-2.3),\ (2,-2.5),\ (2.5,-3),\ (3,-3.7),\ (3.5,-4.4),\ (4,-5.3),\ (4.5,-6.1),\ (5,-7),\ (5.5,-7.9).$

2. Plot the points accurately on the grid.

Mark each ordered pair on the grid with the given $x$-coordinate and corresponding $y$-coordinate, using the scale shown on the axes.

3. Draw the curve of the function over the given interval.

Join the plotted points with a smooth curve from $x=0.5$ to $x=5.5$. The curve should rise from $(0.5,-7)$ to near $(1.5,-2.3)$, then fall smoothly through the remaining points.

1. Rewrite the equation so it can be related to the plotted curve.

The curve drawn is
$y=4-2x-\frac{5}{x}.$
Given
$2x+\frac{5}{x}=8,$
then
$-2x-\frac{5}{x}=-8.$
So
$y=4-8=-4.$

2. Use the graph to read the $x$-coordinates where the curve has $y=-4$.

Draw or use the horizontal line
$y=-4.$
Read the two points where this line meets the curve.

From the graph,
$x\approx 0.8 \quad \text{or} \quad x\approx 3.2.$

1. Construct a reference line with gradient 3 on the graph.

A line with gradient $3$ has
$\frac{\text{vertical change}}{\text{horizontal change}}=3.$
For example, use a rise of $3$ units for a run of $1$ unit to draw a straight reference line of gradient $3$.

2. Find where a tangent to the curve has this gradient and read the coordinates.

Draw a tangent to the curve parallel to the reference line of gradient $3$. The tangent touches the curve at approximately
$x=1,\qquad y=-3.$

1. Rewrite the quadratic equation so that it contains the expression from the plotted curve.

Starting with
$3x^2-14x+10=0.$
Since the graph is for $y=4-2x-\dfrac{5}{x}$, divide the equation by $x$:
$3x-14+\frac{10}{x}=0.$
Rearrange:
$3x=14-\frac{10}{x}.$
Divide by $2$:
$\frac{3}{2}x=7-\frac{5}{x}.$

2. Express the equation in terms of the curve equation $y=4-2x-\dfrac{5}{x}$.

From the graph,
$y=4-2x-\frac{5}{x}.$
Using
$\frac{3}{2}x=7-\frac{5}{x},$
subtract $2x$ from both sides:
$-\frac{1}{2}x=7-2x-\frac{5}{x}.$
But
$7-2x-\frac{5}{x}=\left(4-2x-\frac{5}{x}\right)+3=y+3.$
So
$-\frac{1}{2}x=y+3,$
therefore
$y=-\frac{1}{2}x-3.$

3. Draw the suitable straight line on the existing grid.

Draw the straight line
$y=-\frac{1}{2}x-3.$
For example, use two points on the line:
$(0,-3) \quad \text{and} \quad (4,-5).$
The solutions occur where this line intersects the curve $y=4-2x-\dfrac{5}{x}$.

4. Read the $x$-coordinates of the intersections from the graph.

From the graph, the line $y=-\dfrac{1}{2}x-3$ intersects the curve at approximately
$x\approx 0.9 \quad \text{and} \quad x\approx 3.7.$