一个菌群中的细菌数量每小时翻倍。
该菌群初始有：50个细菌。
下表显示了时间 t 之后菌群中的细菌数量。

时间（小时，t）	0	1	2	3	4	5	6	7
细菌数量 (n)	50	100	200	400	800	1600	3200	6400

1. Identify the coordinates to be plotted from the table

From the table, the graph of $n$ against $t$ uses the ordered pairs
$(t,n)=(0,50),(1,100),(2,200),(3,400),(4,800),(5,1600),(6,3200),(7,6400).$

2. Plot the points accurately on the grid

On the horizontal axis, use $t=0,1,2,3,4,5,6,7$.

On the vertical axis, use the corresponding values of $n$: $50,100,200,400,800,1600,3200,6400$.

Plot each point carefully according to the grid scale.

3. Join the plotted points with the correct graph shape

Join the plotted points with a smooth increasing curve, not with separate straight line segments, for $0\le t\le 7$.

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

Locate $t=5.5$ on the horizontal axis and mark the corresponding point on the curve. Draw a straight line which just touches the curve at this point and follows the direction of the curve there.

2. Find the gradient of the tangent

Using two well-spaced points on the tangent,
$\text{gradient}=\frac{\text{vertical change}}{\text{horizontal change}}=\frac{\Delta n}{\Delta t}.$
From an accurately drawn tangent, this gives approximately
$\text{gradient}\approx 1600.$

1. State the meaning of the gradient in context

The graph is of number of bacteria, $n$, against time, $t$.
So the gradient at $t=5.5$ represents the instantaneous rate of change of $n$ with respect to $t$ at $t=5.5$.

Therefore, it is the rate at which the number of bacteria is increasing at $t=5.5$ hours.

1. Find two points on the straight-line graph within the given interval

For $n=4000-500t$:

When $t=0$,
$n=4000-500(0)=4000.$

When $t=7$,
$n=4000-500(7)=500.$

So two points are $(0,4000)$ and $(7,500)$.

2. Plot the points and draw the line segment for the required domain

Plot $(0,4000)$ and $(7,500)$ on the grid used in part (a).

Join the two points with a straight line segment, only for $0\le t\le 7$.

1. Read the time-coordinate of the intersection of the two graphs

From the graph, locate the point where the curve for the first colony intersects the straight line $n=4000-500t$. The corresponding value on the $t$-axis is approximately

$t \approx 4.9.$

1. Use the initial value to find the constant $k$

At $t=0$, $n=50$.

Substitute into $n=k2^t$:

$50=k2^0$

Since $2^0=1$,

$50=k$

Therefore, $k=50.$