Complete the table of values for $y=\dfrac{2}{x^{2}}-2x. Values are given to one decimal place where appropriate. On the grid opposite, draw the graph of $y=\dfrac{2}{x^{2}}-2x$ for $-3\le x\le 3$. On the same grid, draw the graph of $y+x=6$ for $-3\le x\le 3$. Write down the x-coordinates of the points where the line intersects the curve. Answer x = ... and ... These values of x are solutions of the equation $x^{3}+Ax^{2}+B=0$. Find the value of A and of B. Answer A = ... B = ...

1. Evaluate the formula for each given value of $x$.

Using $y=\dfrac{2}{x^2}-2x$:

1. Use the completed table to identify the points to be plotted.

From the table in part (a), plot the points
$(-3,6.2),\ (-2,4.5),\ (-1,4.0),\ (-0.5,9.0),\ (-0.4,13.3),\ (0.5,7.0),\ (1,0.0),\ (2,-3.5),\ (3,-5.8).$

2. Plot the points accurately on the grid.

On the given axes, mark each ordered pair at its correct position. For example, $(-3,6.2)$ is 3 units left of the origin and slightly above $y=6$, while $(3,-5.8)$ is 3 units right of the origin and slightly above $y=-6$.

3. Join the plotted points with the correct curve shape.

Join the plotted points with a smooth curve. Since $y=\dfrac{2}{x^{2}}-2x$ is undefined at $x=0$, draw two separate smooth branches: one for $-3\le x<0$ and one for $0<x\le3$. Do not join the curve across the $y$-axis.

1. Rewrite the line equation in graphing form.

$y+x=6$
$y=6-x$

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

For $x=-3$,
$y=6-(-3)=9.$
For $x=3$,
$y=6-3=3.$
So two points are $(-3,9)$ and $(3,3)$.

3. Draw the straight line on the given grid for the specified interval.

Plot $(-3,9)$ and $(3,3)$ on the same grid, then join them with a straight line segment from $x=-3$ to $x=3$.

1. Locate the intersection points of the line and the curve on the same grid.

On the completed graph, find where the straight line $y=6-x$ intersects the curve $y=\dfrac{2}{x^2}-2x$.

There are two intersection points within $-3\leq x\leq 3$.

2. Read off the two corresponding $x$-coordinates.

Reading the $x$-coordinates from the grid gives approximately
$x\approx -0.6$
and
$x\approx 0.55.$

1. Equate the curve and the line at their points of intersection.

At an intersection,
$y=\frac{2}{x^2}-2x$
and
$y+x=6 \quad \Rightarrow \quad y=6-x.$
So
$\frac{2}{x^2}-2x=6-x.$

2. Clear the denominator and rearrange into cubic form.

Multiply throughout by $x^2$:
$2-2x^3=6x^2-x^3.$
Rearrange:
$2-x^3-6x^2=0$
$x^3+6x^2-2=0.$

3. Compare coefficients with the given equation.

Compare
$x^3+6x^2-2=0$
with
$x^3+Ax^2+B=0.$
Hence
$A=6, \qquad B=-2.$