The table shows some values for $y = \frac{3}{10}x^3 - x + 1$ for $-3 \le x \le 3$.

x	-3	-2	-1	0	1	2	3
y	p	0.6	1.7	1	0.3	1.4	6.1

1. Substitute $x=-3$ into the formula and evaluate $y$.

Since $p$ is the value of $y$ when $x=-3$,
$p=\frac{3}{10}(-3)^3-(-3)+1$
$=\frac{3}{10}(-27)+3+1$
$=-8.1+4=-4.1$

1. Draw and label the axes using the given scale and ranges.

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

Draw the horizontal axis for
$-3 \le x \le 3$
and the vertical axis for
$-4.5 \le y \le 6.5.$

Label both axes clearly and mark the scale accurately.

2. Plot all points from the completed table accurately.

Using $p=-4.1$ from part (a), plot the points
$(-3,-4.1),\;(-2,0.6),\;(-1,1.7),\;(0,1),\;(1,0.3),\;(2,1.4),\;(3,6.1).$

3. Join the plotted points with an appropriate smooth curve.

Join the plotted points with a smooth curve, not straight line segments, following the general shape of the cubic graph over $-3 \le x \le 3$.

1. Draw the tangent to the curve at $x=2$.

On the graph, locate the point on the curve where $x=2$, which is approximately $(2,1.4)$. 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 estimate its gradient.

Choose two convenient points on the tangent and calculate

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

For the curve $y=\frac{3}{10}x^3-x+1$, the tangent at $x=2$ has gradient about

$\frac{\Delta y}{\Delta x}\approx 2.6.$

1. Relate the given equation to the graph already drawn and identify the integer values of $k$ giving exactly one intersection.

Since the curve drawn is
$y=\dfrac{3}{10}x^3-x+1$,
the equation
$\dfrac{3}{10}x^3-x=k-1$
can be rewritten as
$\dfrac{3}{10}x^3-x+1=k,$
so its solutions are the intersections of the curve with the horizontal line $y=k$.

From the curve, a horizontal line cuts the curve exactly once when it is above the local maximum, about $1.7$, and not above the endpoint value $6.1$.

Since $k$ is a positive integer,
$k=2,3,4,5,6.$
Hence
$2\leqslant k\leqslant 6.$

1. Rewrite the target equation using the expression for the existing curve.

The curve already drawn is
$y=\dfrac{3}{10}x^3-x+1$.

Given
$\dfrac{3}{10}x^3-1=\dfrac{3}{2}x$.

Add $-x+2$ to both sides:
$\dfrac{3}{10}x^3-x+1=\dfrac{3}{2}x-x+2.$

2. Simplify the right-hand side and compare with $y=ax+b$.

$\dfrac{3}{2}x-x+2=\dfrac{1}{2}x+2$

So the line to draw is
$y=\dfrac{1}{2}x+2.$

Comparing with $y=ax+b$,
$a=\dfrac{1}{2},\qquad b=2.$

1. Use the line found in part (e)(i) and choose two convenient points to draw it.

From part (e)(i),
$y=ax+b=\frac12x+2.$
For example,
$x=-3:\quad y=\frac12(-3)+2=0.5,$
$x=3:\quad y=\frac12(3)+2=3.5.$
Plot $(-3,0.5)$ and $(3,3.5)$ and join them with a straight line for $-3\leq x\leq 3$.

2. Identify where the straight line intersects the curve already drawn.

The required solutions occur where
$\frac{3}{10}x^3-x+1=\frac12x+2,$
because these intersection points represent the same $y$-value on the curve and on the line.

3. Read off the three $x$-coordinates of the intersection points from the graph.

Reading from the graph, the three intersections have approximate $x$-coordinates
$x\approx -1.8,\quad x\approx -0.8,\quad x\approx 2.5.$