The variables x and y are connected by the equation $y = x^3 - 3x + 1$. Some corresponding values of x and y are given in the table below.

x	-3	-2	-1	0	1	2	3
y	-17	k	3	1	-1	3	19

1. Evaluate $y$ when $x=-2$.

Since $k$ is the value of $y$ when $x=-2$,
$k=(-2)^3-3(-2)+1=-8+6+1=-1.$

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

From the table and part (a), the points to plot are
$(-3,-17),\ (-2,-1),\ (-1,3),\ (0,1),\ (1,-1),\ (2,3),\ (3,19).$

2. Plot the coordinates accurately and draw a smooth cubic curve through them.

On the given grid, plot each coordinate accurately. Join the points with a smooth curve, not straight line segments, from $x=-3$ to $x=3$.

1. Relate the equation to the graph.

For the graph $y=x^3-3x+1$, solving
$x^3-3x+1=0$
means finding where $y=0$, i.e. the points where the curve cuts the $x$-axis.

2. Read the three x-intercepts from the graph.

From the graph, the curve cuts the $x$-axis at approximately
$x=-1.88,\quad x=0.35,\quad x=1.53.$

1. Locate the point on the curve where $x=-1.5$ and draw the tangent there.

On the graph of $y=x^3-3x+1$, locate the point on the curve directly above $x=-1.5$. Draw a straight line that just touches the curve at this point and follows the direction of the curve there.

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

Choose two well-spaced points on the tangent and calculate

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

A correctly drawn tangent at $x=-1.5$ gives

$\text{gradient}\approx 3.75,$

so the gradient is about $3.8$.

1. Find two points on the line $y=3x-1$ and draw the straight line through them.

Choose convenient values of $x$.

When $x=0$,
$y=3(0)-1=-1,$
so one point is $(0,-1)$.

When $x=1$,
$y=3(1)-1=2,$
so another point is $(1,2)$.

Plot $(0,-1)$ and $(1,2)$ on the same grid and join them with a straight line.

1. Equate the curve and the line to form the equation solved by their intersections.

At the points where the curve and line meet,
$x^3-3x+1=3x-1.$

2. Rearrange the equation into the form $x^3=ax+b$ and identify $a$.

$x^3-3x+1=3x-1$
$x^3=6x-2$
Comparing with $x^3=ax+b$, $a=6$.

1. Use the intersections of the line and curve to form the equation being solved.

At the points of intersection,
$x^3-3x+1=3x-1.$

2. Rearrange into the form $x^3=ax+b$ and identify $b$.

$x^3-3x+1=3x-1$
$x^3=6x-2$
Comparing with $x^3=ax+b$, we get
$b=-2.$