变量 x 与 y 之间满足方程 $y = \frac{1}{3}x(11 - x^2). 下表给出了一些对应的 x 和 y 的值（精确到小数点后一位）。

1. Use the value of $x$ corresponding to $p$ and evaluate the formula.

From the table, $p$ is the value of $y$ when $x=2$.

$p=\frac{1}{3}(2)\left(11-2^2\right)$
$=\frac{2}{3}(11-4)$
$=\frac{14}{3}=4.666\ldots$

Correct to 1 decimal place, $p=4.7.$

1. Use the completed table values as coordinates to be plotted.

From the table and part (a), the points to plot are
$(-4,6.7),\ (-3,-2),\ (-2,-4.7),\ (-1,-3.3),\ (0,0),\ (1,3.3),\ (2,4.7),\ (3,2).$

2. Plot the points accurately on the given grid.

On the given axes, mark each coordinate accurately, using the scale shown on the grid. In particular, the decimal values $6.7$, $-4.7$, $-3.3$, $3.3$ and $4.7$ should be plotted between the labelled integer/even grid lines as appropriate.

3. Draw the smooth curve through the plotted points over the required interval.

Join the plotted points with a single smooth curve, not straight line segments, from $x=-4$ to $x=3$. The curve should follow the shape indicated by the points and should not extend beyond the required range $-4\leq x\leq 3$.

1. Relate the equation to the graph already drawn.

The equation
$\frac13x(11-x^2)=3$
means the points on the curve where $y=3$. Draw the horizontal line $y=3$ on the same axes.

2. Read the three intersection points from the graph.

From the graph, the line $y=3$ cuts the curve at approximately
$x=-3.7,\quad x=0.9,\quad x=2.8.$

1. Draw a tangent at the given point on the curve.

On the graph, draw a straight line that just touches the curve at $(-1,-3.3)$ 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 and calculate
$\text{gradient}=\frac{\text{vertical change}}{\text{horizontal change}}.$
For a correctly drawn tangent at $(-1,-3.3)$, the estimate should be about
$\text{gradient}\approx -2.7.$

1. Rewrite the equation of the line in the form $y=mx+c$.

$5y+2x=7$
$5y=7-2x$
$y=\frac{7-2x}{5}=1.4-0.4x$

2. Find two points on the line within the required interval.

Using $y=1.4-0.4x$:

When $x=-4$,
$y=1.4-0.4(-4)=3.$

When $x=3$,
$y=1.4-0.4(3)=0.2.$

So two points are $(-4,3)$ and $(3,0.2)$.

3. Draw the straight line on the same axes over the stated range.

Plot $(-4,3)$ and $(3,0.2)$ on the existing grid and join them with a straight line segment for $-4\leq x\leq 3$.

1. Use the fact that at an intersection point, the curve and the line have the same coordinates.

Curve: $y=\dfrac13x(11-x^2)$

Line: $5y+2x=7$

At points of intersection, substitute $y=\dfrac13x(11-x^2)$ into the line equation:

$5\left(\dfrac13x(11-x^2)\right)+2x=7$

2. Simplify and rearrange to show the required cubic equation.

$\frac53x(11-x^2)+2x=7$

Multiply throughout by $3$:

$5x(11-x^2)+6x=21$

Expand:

$55x-5x^3+6x=21$

$-5x^3+61x-21=0$

Multiplying by $-1$:

$5x^3-61x+21=0$

1. Read the $x$-coordinates of the intersections of the line and the curve.

From the graph, the line $5y+2x=7$ intersects the curve $y=\frac13x(11-x^2)$ at about

$x\approx -3.7 \quad \text{and} \quad x\approx 0.35.$