Complete the table of values for $y = x - \frac{4}{x^2}$. Values are given to one decimal place where appropriate.

1. Find the missing value of $y$ when $x=-10$.

For $x=-10$,
$y=x-\frac{4}{x^2}$
$y=-10-\frac{4}{(-10)^2}$
$y=-10-\frac{4}{100}$
$y=-10-0.04=-10.04$
To one decimal place,
$y=-10.0$

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

From part (a), the coordinates are
$(-15,-15.0)$, $(-10,-10.0)$, $(-5,-5.2)$, $(-1,-5)$, $(1,-3)$, $(5,4.8)$, $(10,10.0)$, $(15,15.0)$.

2. Plot the points accurately on the grid.

Plot the points:
$(-15,-15.0),\ (-10,-10.0),\ (-5,-5.2),\ (-1,-5)$
on the left of the $y$-axis, and
$(1,-3),\ (5,4.8),\ (10,10.0),\ (15,15.0)$
on the right of the $y$-axis.

3. Draw the curve with the correct shape and asymptote behaviour.

Join the plotted points with a smooth curve in two separate branches. Since $x=0$ is an asymptote, the curve must go near but must not touch or cross the $y$-axis.

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

$2y-2x=-7$

$2y=2x-7$

$y=x-3.5$

2. Choose two convenient points and draw the line through them for the given range.

For $y=x-3.5$,

when $x=-15$, $y=-15-3.5=-18.5$.

when $x=15$, $y=15-3.5=11.5$.

Plot $(-15,-18.5)$ and $(15,11.5)$, then join them with a straight line.

1. Identify where the straight line crosses the curve and read the corresponding $x$-coordinates.

From the completed graph, the line $y=x-3.5$ intersects the curve $y=x-\dfrac{4}{x^2}$ at two points.

Reading the $x$-coordinates of these two intersection points gives approximately
$x\approx -1.1 \quad \text{and} \quad x\approx 1.1.$

1. Form the equation satisfied by the intersection points and compare coefficients.

At the points of intersection, $y=x-\dfrac{4}{x^2}$ also satisfies

$2y-2x=-7.$

Substitute $y=x-\dfrac{4}{x^2}$:

$2\left(x-\frac{4}{x^2}\right)-2x=-7$

$2x-\frac{8}{x^2}-2x=-7$

$-\frac{8}{x^2}=-7$

Multiply by $x^2$:

$-8=-7x^2$

$7x^2-8=0.$

Comparing with $Ax^2+B=0$,

$A=7,\quad B=-8.$