Find the derivative of the curve y = x² − 2x + 6.

The gradient function, $\frac{dy}{dx}$, is found by differentiating the equation with respect to $x$.
$$
\frac{dy}{dx} = \frac{d}{dx}(x^2 - 2x + 6)
$$
$$
\frac{dy}{dx} = 2x - 2
$$

$$
2x - 2 = 4
$$
$$
2x = 4 + 2
$$
$$
2x = 6
$$
$$
x = \frac{6}{2}
$$
$$
x = 3
$$

$$
y = x^2 - 2x + 6
$$
$$
y = (3)^2 - 2(3) + 6
$$
$$
y = 9 - 6 + 6
$$
$$
y = 9
$$

The coordinates of the point are $(x, y) = (3, 9)$.  
The coordinates are $\boxed{(3, 9)}$.

Let $n$ be the number of Sh 50 notes.  
Let $c$ be the number of Sh 20 coins.

Condition 1: She has 3 times as many Sh 20 coins as Sh 50 notes.
$$
c = 3n \quad (1)
$$
Condition 2: Altogether she has Sh 3400.  
The total value is the sum of the value of notes and coins.
$$
50n + 20c = 3400 \quad (2)
$$

Substitute equation (1) into equation (2):
$$
50n + 20(3n) = 3400
$$
$$
50n + 60n = 3400
$$
$$
110n = 3400
$$
$$
n = \frac{3400}{110}
$$
$$
n = \frac{340}{11}
$$
$$
n = 30.909...
$$
There seems to be an issue with the numbers provided in the question, as the number of notes/coins must be an integer. Let's re-check the calculation.  
If $n$ must be an integer, then 3400 must be divisible by 110. $3400/110 = 340/11$, which is not an integer.  
Assuming there might be a typo in the total amount or the ratio, and proceeding with the given numbers to show the method. If the total amount was, for example, Sh 3300, then $n = 3300/110 = 30$.  
Let's assume the total amount was intended to be Sh 3300 for integer results.  
If $110n = 3300$, then $n = 30$.  
Then $c = 3n = 3(30) = 90$.  
Number of Sh 50 notes = 30, Number of Sh 20 coins = 90.  
Total value = $50(30) + 20(90) = 1500 + 1800 = 3300$.

**Given the exact numbers in the problem, an integer solution for the number of notes/coins is not possible. This indicates a potential error in the problem statement itself.**  
However, if we must provide an answer based on the given numbers:
$$
n = \frac{3400}{110} \approx 30.91
$$
$$
c = 3n = 3 \times \frac{3400}{110} = \frac{10200}{110} \approx 92.73
$$
Since the number of notes and coins must be whole numbers, this problem as stated does not yield an integer solution. If we assume the total amount was Sh 3300:  
Number of Sh 50 notes: $\boxed{30}$  
Number of Sh 20 coins: $\boxed{90}$  
**(Note: This assumes a slight adjustment to the total amount to allow for integer counts of notes/coins.)**

Initial workers = 45  
Initial days to complete = 10
$$
\text{Total Work} = 45 \text{ workers} \times 10 \text{ days} = 450 \text{ worker-days}
$$

Workers for the first 4 days = 45  
Days worked = 4
$$
\text{Work Done} = 45 \text{ workers} \times 4 \text{ days} = 180 \text{ worker-days}
$$

$$
\text{Remaining Work} = \text{Total Work} - \text{Work Done}
$$
$$
\text{Remaining Work} = 450 \text{ worker-days} - 180 \text{ worker-days} = 270 \text{ worker-days}
$$

Initial workers = 45  
Workers laid off = 15
$$
\text{Remaining Workers} = 45 - 15 = 30 \text{ workers}
$$

$$
\text{Days Needed} = \frac{\text{Remaining Work}}{\text{Remaining Workers}}
$$
$$
\text{Days Needed} = \frac{270 \text{ worker-days}}{30 \text{ workers}} = 9 \text{ days}
$$
The remaining workers would take $\boxed{9 \text{ days}}$ to complete the work.