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

x	1	1.5	2	2.5	3	4	5	6
y	3	0.7	0.3	0.3	0.6	1.3	p	3.1

1. Identify that p is the y-value when x = 5 and evaluate the formula.

For $x=5$,
$p=5+\frac{5}{5^2}-3$
$=5+\frac{5}{25}-3$
$=5+0.2-3=2.2$

1. Use the table values to identify the coordinates to be plotted.

From the table and part (a), the points are
$(1,3),\ (1.5,0.7),\ (2,0.3),\ (2.5,0.3),\ (3,0.6),\ (4,1.3),\ (5,2.2),\ (6,3.1).$

2. Plot all the coordinates accurately on the given axes.

Plot each coordinate on the grid at its given $x$-value and $y$-value, using the given axes for $1\le x\le 6$. In particular, plot $(5,2.2)$ using $p=2.2$.

3. Join the plotted points with a smooth curve over the given interval.

Draw a smooth curve through the plotted points from $x=1$ to $x=6$. The curve falls steeply from $(1,3)$ to about $(2,0.3)$, stays near its minimum around $x=2$ to $2.5$, then rises through $(3,0.6)$, $(4,1.3)$, $(5,2.2)$ and $(6,3.1)$.

1. Relate the equation to the graph and identify the graphical condition for a solution.

The equation $x+\dfrac{5}{x^2}-3=0$ means $y=0$ on the graph of $y=x+\dfrac{5}{x^2}-3$.

From the graph, the curve is always above the x-axis and does not cut or touch the x-axis.

1. Identify the point on the curve where the tangent is needed and draw the tangent.

At $x=1.5$, the curve has $y\approx 0.7$ from the table, so the point is approximately $(1.5,0.7)$. Draw a tangent to the curve at this point.

2. Calculate the gradient of the tangent using two points on the tangent.

Choose two well-spaced points on the tangent and calculate

$\text{gradient}=\frac{\text{vertical change}}{\text{horizontal change}}.$

From the tangent, this gives approximately

$\text{gradient}\approx -2.0.$

1. Use the gradient to find another point on the line.

Gradient $=2=\dfrac{\text{change in }y}{\text{change in }x}$.

Starting from $(0.5,2)$, increase $x$ by $1$:

$y\text{ increases by }2\times 1=2$

So another point on the line is

$(0.5+1,\ 2+2)=(1.5,4).$

2. Draw the required straight line on the same axes.

Plot $(0.5,2)$ and $(1.5,4)$ on the same axes, then join them with a straight line and extend it across the grid.

1. Find the equation of the line using its gradient and a point on it.

For a straight line, $y=mx+c$.

Given gradient $m=2$, so
$y=2x+c$.

Since the line passes through $(0.5,2)$,
$2=2(0.5)+c$
$2=1+c$
$c=1$

Hence the equation is
$y=2x+1.$

1. Write the condition for a point of intersection between the curve and the straight line.

The curve is
$y=x+\frac{5}{x^2}-3$
and the straight line from part (e)(ii) is
$y=2x+1.$
At a point of intersection, the two $y$-values are equal, so
$x+\frac{5}{x^2}-3=2x+1.$

2. Rearrange the intersection equation to match the given cubic equation.

$x+\frac{5}{x^2}-3=2x+1$
$\frac{5}{x^2}=x+4$
Multiplying by $x^2$,
$5=x^3+4x^2$
$5-4x^2-x^3=0.$

1. Read the x-coordinate of the intersection of the line and the curve from the graph.

From the graph, the straight line $y=2x+1$ intersects the curve $y=x+\dfrac{5}{x^2}-3$ at $x=1$.

Therefore the solution of
$5-4x^2-x^3=0$
is
$x=1.$