Complete the table of values for $y = 2x + \dfrac{9}{x} - 11$. Values are given to 1 decimal place where appropriate.

x	0.5	1	1.5	2	2.5	3	4	5
y	8	0	-2	-2.5	-2.4	-2	-0.8

On the grid, draw the graph of $y = 2x + \dfrac{9}{x} - 11$ for $0 \le x \le 5$. By drawing a tangent, find the gradient of the curve at (1.5, -2). On the same axes, draw the graph of the straight line $2y = 12 - 3x$. Write down the x-coordinates of the points where the graphs meet. Answer $x$ = ... or ... These values of $x$ are the solutions of the equation $7x^2 + Ax + B = 0$. Find the value of $A$ and value of $B$. Answer $A$ = ... $B$ = ...

1. Evaluate the formula for the given $x$-values.

Using $y=2x+\dfrac{9}{x}-11$:

For example,
$x=2.5:\quad y=2(2.5)+\frac{9}{2.5}-11=5+3.6-11=-2.4.$

Similarly,

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

From part (a), the points to plot are
$(0.5,8),\ (1,0),\ (1.5,-2),\ (2,-2.5),\ (2.5,-2.4),\ (3,-2),\ (4,-0.8),\ (5,0.8).$

2. Plot the coordinates accurately on the grid.

Plot each point at its corresponding $x$- and $y$-coordinate on the given grid, using the scale of $1$ unit per major square on both axes.

3. Join the plotted points with an appropriate smooth curve.

Draw a smooth curve through the plotted points for the given range $0\le x\le 5$.

Since the formula contains $\dfrac{9}{x}$, the curve is not defined at $x=0$, so the drawn curve should start from the first plotted point near $x=0.5$ and continue smoothly to $x=5$.

1. Draw a tangent to the curve at the given point.

At $(1.5,-2)$, draw a straight line that just touches the curve and has the same direction as the curve at that point.

2. Choose two convenient points on the tangent and calculate its gradient.

Using two well-separated points on the tangent, for example approximately $(1, -1)$ and $(2, -3)$,

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

$=\frac{-3-(-1)}{2-1}$

$=-2$

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

$2y=12-3x$
$y=6-\frac{3}{2}x$
So the line has $y$-intercept $6$ and gradient $-\frac{3}{2}$.

2. Choose two points on the straight line and draw it on the given axes.

Using $y=6-\frac{3}{2}x$:

When $x=0$, $y=6$, giving $(0,6)$.

When $y=0$,
$0=6-\frac{3}{2}x$
$\frac{3}{2}x=6$
$x=4,$
giving $(4,0)$.

Plot $(0,6)$ and $(4,0)$, then join them with a straight line on the same axes.

1. Identify the points where the curve and straight line intersect.

From the graph, locate the two points where the curve $y=2x+\dfrac{9}{x}-11$ meets the straight line $2y=12-3x$.

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

Reading vertically down from the two intersection points to the $x$-axis gives
$x \approx 0.6 \quad \text{and} \quad x \approx 4.3.$

1. Form the equation satisfied at the intersection points.

At the points where the graphs meet,
$2x+\frac{9}{x}-11=6-1.5x.$

2. Rearrange the equation into the form $7x^2+Ax+B=0$ and compare coefficients.

$2x+\frac{9}{x}-11=6-1.5x$
$3.5x+\frac{9}{x}-17=0$
Multiply by $2x$:
$7x^2+18-34x=0$
$7x^2-34x+18=0$
Comparing with $7x^2+Ax+B=0$,
$A=-34,\qquad B=18.$