$y = \dfrac{x^3}{4} - 2x + 3$.

1. Find the missing table value by substituting $x=-4$ into the formula.

For $x=-4$,
$y=\frac{(-4)^3}{4}-2(-4)+3$
$=\frac{-64}{4}+8+3$
$=-16+8+3=-5$

1. Use the completed table values as coordinates for plotting.

From the table,

So the points to plot are
$(-4,-5),\ (-3,2.25),\ (-2,5),\ (-1,4.75),\ (0,3),\ (1,1.25),\ (2,1),\ (3,3.75),\ (4,11).$

2. Plot all the points accurately on the given grid.

Plot each coordinate pair on the grid, using the given scales on the $x$-axis and $y$-axis. For example, $(-4,-5)$ is 4 units left of the origin and 5 units below the origin, while $(4,11)$ is 4 units right of the origin and 11 units above the origin.

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

Join the plotted points with a smooth curve, not with straight line segments, from $x=-4$ to $x=4$. The curve should pass through all plotted points and have the general cubic shape implied by
$y=\frac{x^3}{4}-2x+3.$

1. Explain the graphical meaning of the solution and count the x-intercepts.

The solutions of $\dfrac{x^3}{4}-2x+3=0$ are the points where the graph of $y=\dfrac{x^3}{4}-2x+3$ has $y=0$, i.e. where it cuts the $x$-axis. From the graph, the curve cuts the $x$-axis only once.

1. Write the equation formed when the line intersects the curve.

At the points of intersection,
$\frac{x^3}{4}-2x+3=ax+b.$
Rearranging,
$\frac{x^3}{4}-(2+a)x+(3-b)=0.$

2. Match coefficients with the given cubic equation to find $a$ and $b$.

The given equation is
$x^3-10x+4=0.$
Divide by $4$ so that the coefficient of $x^3$ matches:
$\frac{x^3}{4}-\frac{5}{2}x+1=0.$
Compare with
$\frac{x^3}{4}-(2+a)x+(3-b)=0.$
So
$2+a=\frac{5}{2} \quad\Rightarrow\quad a=\frac{1}{2},$
and
$3-b=1 \quad\Rightarrow\quad b=2.$

1. Use the values of $a$ and $b$ from part (d)(i) to identify the straight line to draw.

From part (d)(i),
$a=\frac12,\qquad b=2.$
So the required line is
$y=\frac12x+2.$

2. Draw the straight line accurately on the same axes as the curve.

Choose two convenient points on
$y=\frac12x+2.$
For example,
$x=0 \Rightarrow y=2,$
$x=4 \Rightarrow y=4.$
Draw the straight line through $(0,2)$ and $(4,4)$ on the same graph as the curve.

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

The solutions of
$x^3-10x+4=0$
are the $x$-coordinates where the line
$y=\frac12x+2$
intersects the curve
$y=\frac{x^3}{4}-2x+3.$
Reading from the graph gives approximately
$x=-3.35,\quad x=0.41,\quad x=2.94.$