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

1. Evaluate $y$ when $x=2$.

Since $k$ is the value of $y$ when $x=2$,
$k=\frac{2^3}{2}-5(2)-2=\frac{8}{2}-10-2=4-10-2=-8.$

1. List the coordinates to be plotted from the table, using the value of $k$ found earlier.

From the table and $k=-8$, plot the points
$(-3,-0.5),\ (-2,4),\ (-1,2.5),\ (0,-2),\ (1,-6.5),\ (2,-8),\ (3,-3.5),\ (4,10).$

2. Draw the curve accurately through the plotted points for the given range of $x$.

Join the plotted points with a smooth curve, not straight line segments, for
$-3\le x\le 4.$

1. Rewrite the equation in terms of the graph already drawn.

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

Given
$\dfrac{x^3}{2}-5x=7$,
subtract $2$ from both sides:
$\dfrac{x^3}{2}-5x-2=5.$
So this corresponds to points on the curve where $y=5$.

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

On the graph, the horizontal line $y=5$ cuts the curve $y=\dfrac{x^3}{2}-5x-2$ at only one point.

Therefore, there is only one value of $x$ satisfying the equation.

1. Draw a tangent to the curve at $(1,-6.5)$.

On the graph, draw a straight line that just touches the curve at $(1,-6.5)$ and follows the direction of the curve at that point.

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

Choose two well-spaced points on the tangent. For example, the tangent should pass approximately through $(0,-3)$ and $(2,-10)$.

$\text{Gradient}=\frac{\text{change in }y}{\text{change in }x}$

$=\frac{-10-(-3)}{2-0}=\frac{-7}{2}=-3.5$

1. Find suitable points on the line within the given range of $x$.

For $y=4-x$:

When $x=-1$,
$y=4-(-1)=5.$

When $x=4$,
$y=4-4=0.$

So two points on the line are $(-1,5)$ and $(4,0)$.

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

Plot $(-1,5)$ and $(4,0)$ on the existing grid.

Join these two points with a straight line segment for $-1\le x\le 4$.

1. Identify the intersection of the curve and the line on the graph and read its x-coordinate.

From the graph, the line $y=4-x$ intersects the curve $y=\dfrac{x^3}{2}-5x-2$ at about $x=3.4$.

1. Equate the curve and line equations at their intersection.

At the intersection,
$\frac{x^3}{2}-5x-2=4-x.$

2. Rearrange the equation into the form $x^3+Ax+B=0$ and compare coefficients.

$\frac{x^3}{2}-5x-2=4-x$
$\frac{x^3}{2}-4x-6=0$
Multiplying by $2$,
$x^3-8x-12=0.$
Comparing with $x^3+Ax+B=0$,
$A=-8,\quad B=-12.$