The variables x and y are connected by the equation $y = \frac{3}{2}x + \frac{5}{x} - 7$. Some corresponding values of x and y are given in the table below.

x	0.5	1	1.5	2	3	4	5	6	7
y	3.75	-0.5	-1.42	p	-0.83	0.25	1.5	2.83	4.21

1. Identify the x-value corresponding to p and substitute it into the formula.

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

$p=\frac{3}{2}(2)+\frac{5}{2}-7$

2. Evaluate the expression accurately.

$p=3+2.5-7$
$p=-1.5$

1. Use the table values, including the value found in part (a), to prepare the coordinates to be plotted.

From the table and part (a), the points are:

$(0.5,3.75),\ (1,-0.5),\ (1.5,-1.42),\ (2,-1.5),\ (3,-0.83),\ (4,0.25),\ (5,1.5),\ (6,2.83),\ (7,4.21).$

2. Set up the axes using the required scale.

Use the scale $2\text{ cm}$ to represent $1$ unit on both axes.

The horizontal axis should cover at least $0.5\le x\le 7$, and the vertical axis should cover the given $y$-values, from about $-1.5$ to $4.21$.

3. Plot the points accurately and join them with a smooth curve.

Plot all the coordinates accurately on the grid.

Join the points with a smooth curve, not straight line segments, for $0.5\le x\le 7$. The curve decreases from $(0.5,3.75)$ to a minimum near $x=2$, then increases through the remaining points.

1. Locate the point on the curve at $x=2.5$ and draw a tangent there.

On the graph drawn in part (b), locate the point on the curve where $x=2.5$. 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 clear points on the tangent line and calculate
$\text{gradient}=\frac{\text{vertical change}}{\text{horizontal change}}.$
From a correctly drawn tangent, this gives approximately
$\text{gradient}\approx 0.7.$

1. Rearrange the line equation into a form suitable for plotting.

$2y+x=2$
$2y=2-x$
$y=1-\frac{x}{2}$

2. Find two points on the line for the required interval and draw the line.

For $0\le x\le 7$,

when $x=0$,
$y=1-\frac{0}{2}=1,$
so one point is $(0,1)$.

When $x=7$,
$y=1-\frac{7}{2}=-2.5,$
so another point is $(7,-2.5)$.

Plot $(0,1)$ and $(7,-2.5)$, then join them with a straight line.

1. Identify the points where the straight line meets the curve on the completed graph.

From the graph, locate the two intersection points of the curve $y=\dfrac{3}{2}x+\dfrac{5}{x}-7$ and the line $2y+x=2$.

2. Read off the corresponding x-coordinates.

Reading the $x$-coordinates of these two points gives approximately
$x=0.8 \quad \text{and} \quad x=3.2.$

1. Use the fact that, at an intersection, the line and curve have the same y-value.

The curve is
$y=\frac{3}{2}x+\frac{5}{x}-7.$
The line is
$2y+x=2,$
so
$y=1-\frac{x}{2}.$
At a point of intersection,
$\frac{3}{2}x+\frac{5}{x}-7=1-\frac{x}{2}.$

2. Simplify the equation to obtain the given quadratic equation.

$\frac{3}{2}x+\frac{x}{2}+\frac{5}{x}-8=0$
$2x+\frac{5}{x}-8=0$
Multiplying throughout by $x$,
$2x^{2}+5-8x=0$
$2x^{2}-8x+5=0.$