变量 x 和 y 由方程 $y = \frac{x^2}{3} + \frac{3}{x} - 4$ 关联。 下表中给出了一些对应的 x 和 y 的值（精确到两位小数）。

1. Substitute $x=1.5$ into the equation to find the missing table value.

For $x=1.5$,
$p=\frac{1.5^2}{3}+\frac{3}{1.5}-4$
$=\frac{2.25}{3}+2-4$
$=0.75+2-4$
$=-1.25$

1. Draw and label the axes using the required scale and ranges.

Use a scale of $2\text{ cm}$ to represent $1$ unit on both axes.

Draw:

$x$-axis from $0$ to $6$,

$y$-axis from $-2$ to $9$.

Label the axes and mark the scale clearly.

2. Plot all the points from the table accurately.

Plot the points:

$(0.5,2.08),\ (1,-0.67),\ (1.5,-1.25),\ (2,-1.17),\ (3,0),\ (4,2.08),\ (5,4.93),\ (6,8.50).$

The missing value from part (a), $p=-1.25$, is used for the point $(1.5,-1.25)$.

3. Join the plotted points with a smooth curve.

Join the plotted points with a smooth curve, not straight line segments, following the shape suggested by the points.

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

The curve drawn is
$y=\dfrac{x^2}{3}+\dfrac{3}{x}-4$.

Given
$\dfrac{x^2}{3}+\dfrac{3}{x}-5=0$,
we have
$\dfrac{x^2}{3}+\dfrac{3}{x}-4=1$.

So the solutions are the $x$-coordinates where the curve has $y=1$.

2. Use the graph to read off the intersection points with the horizontal line $y=1$.

Draw the horizontal line $y=1$ on the graph.

Read the $x$-coordinates where this line cuts the curve:

$x \approx 0.6$ and $x \approx 3.5$.

1. Draw the tangent at the given point on the curve.

On the graph, draw a straight line that just touches the curve at $(4, 2.08)$ and has the same direction as the curve at that point.

2. Calculate the gradient of the tangent using rise over run.

Choose two well-spaced points on the tangent and calculate

$\text{gradient}=\frac{\text{change in }y}{\text{change in }x}.$

From an accurately drawn tangent, this gives approximately

$\text{gradient}\approx 2.5.$

1. Mark the given point on the existing axes.

Plot the point $(3,2)$ on the same axes used in part (b).

2. Use the gradient to locate a second point and draw the line.

Gradient $=\dfrac{4}{3}$ means
$\frac{\text{vertical change}}{\text{horizontal change}}=\frac{4}{3}.$
From $(3,2)$, move $3$ units to the right and $4$ units up:
$(3,2)\to(6,6).$
Draw a straight line through $(3,2)$ and $(6,6)$.

1. Use the gradient and the point on the line to find the equation.

The line has gradient $\dfrac{4}{3}$ and passes through $(3,2)$.

Using $y=mx+c$,
$2=\frac{4}{3}(3)+c$
$2=4+c$
$c=-2$

So the equation is
$y=\frac{4}{3}x-2.$

1. Identify the intersection points on the graph.

From the graph, look for the points where the line $y=\dfrac{4}{3}x-2$ crosses the curve $y=\dfrac{x^2}{3}+\dfrac{3}{x}-4$ in the range $0\leq x\leq 6$.

2. Write down the coordinates of the two intersection points.

The first intersection is at the plotted point $(1,-0.67)$.
The second intersection is read from the graph as approximately $(4.9,4.5)$.