Complete the table of values for $y = 10 - \frac{x^2}{2} - \frac{4}{x}$.

1. Find the missing $y$-value by substituting $x=5$ into the formula.

For $x=5$,
$y=10-\frac{5^2}{2}-\frac{4}{5}$
$=10-\frac{25}{2}-0.8$
$=10-12.5-0.8=-3.3$

1. Use the completed table to list the coordinates to be plotted.

From the table, including the completed value from part (a), plot:
$(0.5,1.9),\ (0.7,4.0),\ (1,5.5),\ (2,6),\ (3,4.2),\ (4,1),\ (5,-3.3),\ (6,-8.7).$

2. Plot the points accurately on the given grid.

Mark each coordinate point on the grid using the $x$-value horizontally and the $y$-value vertically. The graph is only required for
$0.5\le x\le 6.$

3. Join the plotted points with a smooth curve.

Draw a smooth curve through the plotted points, not a series of straight line segments. The curve rises from $(0.5,1.9)$ to a maximum near $(2,6)$, then falls through the remaining points.

1. Draw the tangent at the point $(2,6)$ on the curve.

On the completed graph, draw a straight line that just touches the curve at $(2,6)$ and follows the direction of the curve at that point.

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

Using two well-spaced points read from the tangent line, calculate
$\text{gradient}=\frac{\text{vertical change}}{\text{horizontal change}}.$
For an accurately drawn tangent, this gives approximately
$\text{gradient}\approx -1.$

1. Draw a straight line with gradient 3 as a guide.

A line with gradient $3$ rises $3$ units for every $1$ unit moved to the right. For example, draw a straight guide line through two convenient points such as $(0,0)$ and $(1,3)$.

2. Find the point on the curve where the tangent is parallel to this line.

Draw a line parallel to the gradient-$3$ guide line so that it just touches the curve. Reading the $x$-coordinate of the point of contact from the graph gives
$x \approx 1.0.$

1. Rewrite the cubic equation in terms of the expression used in the curve.

Given
$x^3-x^2-14x+8=0.$
Since $x>0$ in the range $0.5\le x\le 6$, divide by $2x$:
$\frac{x^2}{2}-\frac{x}{2}-7+\frac{4}{x}=0.$
So
$\frac{x^2}{2}+\frac{4}{x}=\frac{x}{2}+7.$

2. Express the equation as the intersection of the given curve and a straight line.

For the curve,
$y=10-\frac{x^2}{2}-\frac{4}{x}.$
Using
$\frac{x^2}{2}+\frac{4}{x}=\frac{x}{2}+7,$
we get
$y=10-\left(\frac{x}{2}+7\right)=3-\frac{x}{2}.$
Therefore, draw the straight line
$y=3-\frac{x}{2}$
on the same graph.

3. Read the $x$-coordinates of the intersections from the graph.

The solutions are the $x$-coordinates where the line
$y=3-\frac{x}{2}$
meets the curve
$y=10-\frac{x^2}{2}-\frac{4}{x}.$
From the graph, the intersections occur at approximately
$x\approx 0.56 \quad \text{and} \quad x\approx 4.0.$