The variables x and y are connected by the equation $y = \frac{x^3}{5} - 2x + 1$. Some corresponding values of x and y are given in the table below.

1. Evaluate the formula at $x=-2$ to find the missing table value $p$.

When $x=-2$,
$p=\frac{(-2)^3}{5}-2(-2)+1$
$=\frac{-8}{5}+4+1$
$=-1.6+5=3.4$

1. Use the completed table values as coordinates to be plotted.

From the table and part (a), the points to plot are

$(-4,-3.8),\ (-3,1.6),\ (-2,3.4),\ (-1,2.8),\ (0,1),\ (1,-0.8),\ (2,-1.4),\ (3,0.4),\ (4,5.8)$.

2. Plot the points accurately on the given graph paper.

On the given axes, mark each coordinate pair accurately. For example, plot $(-2,3.4)$ two units left of the origin and $3.4$ units above the $x$-axis, and plot $(2,-1.4)$ two units right of the origin and $1.4$ units below the $x$-axis.

3. Draw the curve over the required interval.

Join the plotted points with a smooth curve, not straight line segments, starting at $x=-4$ and ending at $x=4$. The graph should pass close to all the plotted points and show the cubic shape for $-4\le x\le 4$.

1. Identify the required line graphically.

The line $y=mx-2$ has $y$-intercept $-2$, so it passes through $(0,-2)$.

On the graph, draw the line through $(0,-2)$ which just touches the curve near $(2,-1.4)$ and intersects the curve once more. This gives exactly 2 points of intersection.

2. Find the gradient of this line to obtain $m$.

Using the two points $(0,-2)$ and approximately $(2,-1.4)$ on the line,

$m=\frac{-1.4-(-2)}{2-0}$

$m=\frac{0.6}{2}=0.3$

1. Rewrite the equation so that it matches the plotted curve.

The curve drawn is
$y=\dfrac{x^3}{5}-2x+1$.

For
$\dfrac{x^3}{5}-2x-3=0$,
add $4$ to both sides:
$\dfrac{x^3}{5}-2x+1=4.$
So this equation is equivalent to finding where the curve has $y=4$.

2. Use the graph to count the intersections with $y=4$.

Draw or consider the horizontal line $y=4$ on the graph.

From the graph, this horizontal line intersects the curve only once. Therefore there is only one corresponding $x$-value.

1. Rewrite the equation in a form suitable for drawing the line.

$2y+x=4$

$2y=4-x$

$y=2-\dfrac{x}{2}$

2. Choose two points on the line and draw it on the given axes.

When $x=0$, $y=2$, so one point is $(0,2)$.

When $y=0$, $x=4$, so another point is $(4,0)$.

Plot $(0,2)$ and $(4,0)$, then join them with a straight line.

1. Write the line equation in terms of $y$.

From part (e), the line is

$2y+x=4$

$2y=4-x$

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

2. Use the fact that points of intersection satisfy both the curve and the line.

The curve is

$y=\frac{x^3}{5}-2x+1$

At a point of intersection,

$\frac{x^3}{5}-2x+1=2-\frac{x}{2}$

3. Simplify the equation to obtain the required cubic equation.

Multiply throughout by $10$:

$10\left(\frac{x^3}{5}-2x+1\right)=10\left(2-\frac{x}{2}\right)$

$2x^3-20x+10=20-5x$

$2x^3-15x-10=0$

Therefore the $x$-coordinates of the intersections are the solutions of

$2x^3-15x-10=0.$