Let $y = 9x^3 + 5x^2 - 3x + 1$.

1. Evaluate $y$ when $x=-0.25$ and round to 1 decimal place.

For the missing entry, $x=-0.25$.

$y=9(-0.25)^3+5(-0.25)^2-3(-0.25)+1$

$y=9(-0.015625)+5(0.0625)+0.75+1$

$y=-0.140625+0.3125+0.75+1=1.921875$

Correct to 1 decimal place,

$y=1.9$

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

From the table, the points to plot are
$(-1,0),\ (-0.75,2.3),\ (-0.5,2.6),\ (-0.25,1.9),\ (0,1),\ (0.25,0.7),\ (0.5,1.9),\ (0.75,5.4),\ (1,12).$

2. Plot the points accurately on the supplied grid.

Plot each coordinate pair on the grid for $-1\le x\le 1$, using the given scales on the $x$-axis and $y$-axis.

3. Draw the curve through the plotted points.

Join the plotted points with a smooth curve, not straight line segments, over the interval $-1\le x\le 1$. The curve should pass close to all the plotted points and stay within the given domain.

1. Relate the equation to the graph drawn in part (b).

The graph drawn is
$y=9x^3+5x^2-3x+1$.

Given
$9x^3+5x^2=3x+5$,

rearrange:
$9x^3+5x^2-3x+1=6$.

So the required solution is where the curve has $y=6$.

2. Use the graph to read the x-coordinate where the curve meets the line y = 6.

Draw or imagine the horizontal line $y=6$ on the graph.

It intersects the curve at about
$x=0.78$.

Hence, $x\approx0.78$.

1. Identify the direction of a line with gradient $-3$.

A gradient of $-3$ means
$\frac{\text{vertical change}}{\text{horizontal change}}=-3.$
So a suitable guide line has a fall of $3$ units for every $1$ unit moved to the right.

2. Use the graph to locate tangents with this gradient.

On the graph of $y=9x^3+5x^2-3x+1$, draw tangents parallel to the guide line of gradient $-3$. There are two such points on the curve.

3. Read the required $x$-coordinates from the graph.

Reading the $x$-coordinates of these two tangent points from the graph gives approximately
$x\approx -0.37 \quad \text{and} \quad x\approx 0.$

1. Rearrange the equation into gradient-intercept form.

$8x=5y-15$
$5y=8x+15$
$y=\frac{8}{5}x+3$
$y=1.6x+3$

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

When $x=-1$,
$y=1.6(-1)+3=1.4.$
When $x=1$,
$y=1.6(1)+3=4.6.$
So two points on the line are $(-1,1.4)$ and $(1,4.6)$.

3. Draw the line segment on the existing grid.

Plot $(-1,1.4)$ and $(1,4.6)$ on the grid, then join them with a straight line segment for $-1\le x\le 1$.

1. Write the line equation in terms of $y$.

From part (e)(i),
$8x=5y-15$
$5y=8x+15$
$y=\frac{8x+15}{5}.$

2. Equate the curve and the line to form the cubic equation for their intersections.

At the points of intersection,
$9x^3+5x^2-3x+1=\frac{8x+15}{5}.$
Multiply by $5$:
$45x^3+25x^2-15x+5=8x+15.$
Bring all terms to one side:
$45x^3+25x^2-23x-10=0.$

3. Compare the cubic equation with the given form to find $A$ and $B$.

Compare
$45x^3+25x^2-23x-10=0$
with
$45x^3+25x^2-Ax-2B=0.$
So
$A=23$
and
$2B=10 \Rightarrow B=5.$