在坐标纸上回答这道大题的全部内容。 变量 x 和 y 通过方程 $y = x^3 - 3x^2 + 1$ 相关联。 下表给出了一些对应的 x 和 y 值。

x	-2	-1	0	1	2	3	4
y	-19	-3	1	-1	p	1	17

1. Evaluate the formula when $x=2$, since $p$ is the corresponding $y$-value in the table.

For $x=2$,
$p=2^3-3(2^2)+1=8-12+1=-3.$

1. Set up the axes using the scales and ranges given in the question.

Draw the horizontal $x$-axis for $-2\leq x\leq 4$ using the scale

$2\text{ cm}:1\text{ unit}$.

Draw the vertical $y$-axis for $-20\leq y\leq 20$ using the scale

$1\text{ cm}:2\text{ units}$.

Label the axes clearly as $x$ and $y$.

2. Use the table, including the value of $p$ found earlier, to identify all points to be plotted.

From part (a), $p=-3$.

The points to plot are therefore

$(-2,-19),\ (-1,-3),\ (0,1),\ (1,-1),\ (2,-3),\ (3,1),\ (4,17).$

3. Plot the points accurately and join them with a smooth curve.

Plot each ordered pair accurately on the axes.

Join the plotted points with a smooth curve, not with straight line segments, since the relationship is

$y=x^3-3x^2+1.$

1. Recognise that solving $x^3-3x^2+1=0$ means finding where the curve has $y=0$.

For the curve $y=x^3-3x^2+1$, the equation
$x^3-3x^2+1=0$
means
$y=0.$
So we look for the intersections of the curve with the $x$-axis.

2. Read the $x$-coordinates of all intersections with the $x$-axis from the graph.

From the graph, the curve cuts the $x$-axis at approximately
$x=-0.5,\quad x=0.7,\quad x=2.9.$

1. Recognise the extra line needed so that intersections solve the equation.

The curve already drawn is
$y=x^3-3x^2+1$.

For
$x^3-3x^2+1=-4x$,
draw the straight line
$y=-4x$.

The solution is the $x$-coordinate where this line meets the curve.

2. Read the intersection point from the graph.

From the graph, the line $y=-4x$ meets the curve at about
$x=-0.2$.

1. Identify the point on the curve where the tangent is required.

From the table/graph, when $x=3$, $y=1$.

So draw a tangent to the curve at the point $(3,1)$.

2. Use two well-spaced points on the tangent to find its gradient.

Choose two clear points on the tangent line and calculate

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

For example, a correctly drawn tangent at $(3,1)$ has gradient about

$\frac{10-(-8)}{4-2}=\frac{18}{2}=9.$

1. Identify a horizontal line that will meet the curve at two points.

From the plotted curve/table, the points $(0,1)$ and $(3,1)$ both lie on the curve.

So the horizontal line

$y=1$

meets the curve at two distinct points.

2. State the corresponding value of k.

Since the line is $y=k$ and we have chosen the line $y=1$,

$k=1.$