Answer (b), (c) and (d)

1. Find the missing value of $y$ when $x=-1$.

Substitute $x=-1$ into $y=\dfrac{x^3}{5}-x^2+2$:

$y=\dfrac{(-1)^3}{5}-(-1)^2+2$

$y=-\dfrac{1}{5}-1+2=0.8$

2. Find the missing value of $y$ when $x=5$.

Substitute $x=5$ into $y=\dfrac{x^3}{5}-x^2+2$:

$y=\dfrac{5^3}{5}-5^2+2$

$y=\dfrac{125}{5}-25+2=25-25+2=2$

1. Use the completed table of values to identify all points to be plotted.

From the table in part (a), the coordinates are

$(-2,-3.6),\ (-1,0.8),\ (0,2),\ (1,1.2),\ (2,-0.4),\ (3,-1.6),\ (4,-1.2),\ (5,2).$

2. Plot the points accurately on the given grid.

Plot the points on the Cartesian grid using the given scale:

$(-2,-3.6),\ (-1,0.8),\ (0,2),\ (1,1.2),\ (2,-0.4),\ (3,-1.6),\ (4,-1.2),\ (5,2).$

Each point should be placed according to its $x$-coordinate and corresponding $y$-coordinate.

3. Draw the graph as a smooth curve over the required domain.

Join the plotted points with a smooth curve for the interval

$-2\le x\le 5.$

The curve should not be drawn as separate straight line segments. It should pass smoothly through or very close to all the plotted points.

1. Locate the point on the curve where $x=4$ and draw the tangent there.

From the graph/table, when $x=4$, $y=-1.2$. Locate the point $(4,-1.2)$ on the curve and 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

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

Reading from a correctly drawn tangent gives

$\text{gradient}\approx 1.6.$

1. Relate the given equation to the curve already drawn.

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

Given
$\dfrac{x^3}{5}-x^2+1=0$,
so
$\dfrac{x^3}{5}-x^2+2=1$.

Hence draw the straight line
$y=1$.

2. Draw the suitable straight line on the grid.

On the same grid as the curve, draw the horizontal line $y=1$.

3. Read off the $x$-coordinates of the intersections.

The solutions are the $x$-coordinates where the curve intersects the line $y=1$.

From the graph, the intersections occur at approximately
$x=-0.9$, $x=1.1$ and $x=4.8$.

1. Choose a negative horizontal line that does not give exactly three intersections.

From the graph, the minimum point of the curve is about $y=-1.7$. A horizontal line such as $y=-2$ lies below this minimum, so it will not cut the curve three times. Hence a possible value is

$k=-2.$