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

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

For $x=-3$, $y=p$.

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

$p=\frac{-27}{5}+6+2$

$p=-5.4+8=2.6$

1. Use the table values, including the value of $p$ found in part (a), as coordinates to be plotted.

The points to plot are

$(-3,2.6),\ (-2,4.4),\ (-1,3.8),\ (0,2),\ (1,0.2),\ (2,-0.4),\ (3,1.4),\ (4,6.8).$

2. Plot all the ordered pairs accurately on the given Cartesian grid.

On the grid, mark each point at its correct horizontal position $x$ and vertical position $y$:

$(-3,2.6),\ (-2,4.4),\ (-1,3.8),\ (0,2),\ (1,0.2),\ (2,-0.4),\ (3,1.4),\ (4,6.8).$

3. Join the plotted points with a smooth curve over the required domain.

Draw a smooth continuous curve passing through the plotted points from $x=-3$ to $x=4$.

Do not join the points with straight line segments.

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

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

Given
$\dfrac{x^3}{5}-2x=3$,
add $2$ to both sides:
$\dfrac{x^3}{5}-2x+2=5.$
So the solution occurs where the curve has $y=5$.

2. Use the graph to explain the number of solutions.

Draw or imagine the horizontal line $y=5$ on the graph.

From the graph, the line $y=5$ cuts the curve $y=\dfrac{x^3}{5}-2x+2$ at exactly one point. Therefore, there is only one value of $x$ satisfying the equation.

1. Find two points on the line within the required domain.

For $y=-2x+5$:

When $x=-1$, $y=-2(-1)+5=7$, giving $(-1,7)$.

When $x=3$, $y=-2(3)+5=-1$, giving $(3,-1)$.

2. Plot the two points and draw the required line segment.

Plot $(-1,7)$ and $(3,-1)$ on the same grid.

Join them with a straight line for $-1\le x\le 3$.

1. Identify the point where the straight line meets the curve and read its x-coordinate from the graph.

From the graph, the line $y=-2x+5$ intersects the curve $y=\dfrac{x^3}{5}-2x+2$ at about $x=2.5$.

1. Use the fact that the intersection point lies on both the curve and the line.

At the point of intersection,
$\frac{x^3}{5}-2x+2=-2x+5.$

2. Simplify the equation and rearrange it into cubic form.

$\frac{x^3}{5}-2x+2=-2x+5$
Cancel $-2x$ from both sides:
$\frac{x^3}{5}+2=5$
$\frac{x^3}{5}=3$
$x^3=15$
So
$x^3-15=0.$

3. Compare with the required form $x^3+Ax+B=0$.

Compare
$x^3-15=0$
with
$x^3+Ax+B=0.$
There is no $x$ term, so $A=0$.
The constant term is $-15$, so $B=-15$.