Complete the table of values for $y = 2x + \dfrac{1}{x^{2}} - 4$. Values are given to two decimal places where appropriate. On the grid opposite, draw the graph of $y = 2x + \dfrac{1}{x^{2}} - 4$ for $-2 \le x \le 3$. $y = b$ cuts the graph of $y = 2x + \dfrac{1}{x^{2}} - 4$ at one point for $-2 \le x \le 3$, state the range of values of $b$. Answer ... On the same grid, draw the graph of $3y - 5x = 4$ for $-2 \le x \le 3$. Write down the x-coordinates of the points where the graph of $3y - 5x = 4$ intersects the curve for $-2 \le x \le 3$. Answer x = ... and ... These values of x are solutions of the equation $x^{3} + Ax^{2} + B = 0$. Find the value of A and the value of B. Answer A = ... B = ...

1. Evaluate $y$ when $x=0.5$ to complete the table.

For $x=0.5$,
$y=2(0.5)+\frac{1}{(0.5)^2}-4$
$=1+\frac{1}{0.25}-4$
$=1+4-4=1$

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

From the table, plot the points

$(-2,-7.75),\ (-1,-5),\ (-0.5,-1),\ (-0.3,6.51),\ (0.3,7.71),\ (0.5,1),\ (1,-1),\ (2,0.25),\ (3,2.1).$

The value at $x=0.5$ is $1$ from part (a).

2. Plot all points accurately on the grid using the given scale.

On the grid, mark each point at its corresponding $x$-coordinate and $y$-coordinate. For example, $(-2,-7.75)$ is close to $-8$ on the $y$-axis, and $(-0.3,6.51)$ is slightly above $6.5$.

3. Draw the required graph as a smooth curve over the stated domain.

Join the plotted points with smooth curves. Do not join across $x=0$, because

$\frac{1}{x^2}$

is undefined when $x=0$.

So the graph has one branch for $-2\le x<0$ and another branch for $0<x\le 3$.

1. Identify the horizontal levels that give exactly one intersection with the curve.

From the graph, the left branch increases from $y=-7.75$ at $x=-2$ to very large positive values near $x=0$.

The right branch has its lowest point at $(1,-1)$.

So a horizontal line $y=b$ cuts the curve at exactly one point only when it cuts the left branch but is below the lowest point of the right branch:

$-7.75 \leqslant b < -1$

1. Rewrite the equation in a form suitable for plotting.

$3y-5x=4$

$3y=5x+4$

$y=\dfrac{5x+4}{3}$

2. Find two or more points on the line within the given range of x-values.

Choose convenient values of $x$ in $-2\leq x\leq 3$.

When $x=-2$,

$y=\dfrac{5(-2)+4}{3}=\dfrac{-6}{3}=-2$.

When $x=1$,

$y=\dfrac{5(1)+4}{3}=\dfrac{9}{3}=3$.

So two points are $(-2,-2)$ and $(1,3)$.

3. Draw the required straight line on the given grid.

Plot $(-2,-2)$ and $(1,3)$ on the same grid, then join them with a straight line.

Extend the line only over the given interval $-2\leq x\leq 3$. At $x=3$,

$y=\dfrac{5(3)+4}{3}=\dfrac{19}{3}\approx 6.33$,

so the drawn line segment ends at approximately $(3,6.33)$.

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

On the completed graph, find the two points where the straight line $3y-5x=4$ intersects the curve $y=2x+\dfrac{1}{x^2}-4$.

2. Read off the corresponding $x$-coordinates from the grid.

Reading the $x$-coordinates of these two intersection points from the graph gives approximately

$x=-0.43 \quad \text{and} \quad x=0.43.$

1. Write the straight line equation as an expression for $y$.

From $3y-5x=4$,

$3y=5x+4$

$y=\frac{5x+4}{3}$

2. Equate the line and curve equations and clear denominators.

At the points of intersection,

$2x+\frac{1}{x^2}-4=\frac{5x+4}{3}$

Multiply throughout by $3x^2$:

$6x^3+3-12x^2=5x^3+4x^2$

3. Rearrange into the required cubic form and compare coefficients.

$6x^3+3-12x^2=5x^3+4x^2$

$x^3-16x^2+3=0$

Comparing with

$x^3+Ax^2+B=0,$

$A=-16,\qquad B=3$