Below is the graph of the boa constrictor population. the x-axis is t, the time in years, and the y-axis is p(t), the snake population for any given year. in this case, t = 0 indicates the year they started keeping track of the boa population. scientists want to know how the snake population is changing every two years so they can make predictions about the next two years.

students' conjectures:

two students, sarah and ernest, agree that the graph is exponential but disagree on how the rate changes between successive time periods.

1. complete the table below to summarize each classmate's conjecture. (2 points: 1 point for each row of the chart)
classmate conjecture
ernest

sarah

2. using your knowledge of exponential functions, who do you think is correct? (1 point)

analyzing the data:

3. is this graph increasing or decreasing? does this imply that the boa constrictor population is growing or diminishing? (1 point)

4. looking at this graph, would you agree that the boa constrictor population could be a problem? (1 point)

5. what is the horizontal asymptote of this graph? (1 point)

6. what is the range? explain. (1 point)

7. what is the y-intercept? what does this mean in terms of the snake population? (1 point)

here is the function for this graph, where t is the time in years: p(t) = 5e0.89t.

8. use this formula to estimate the boa constrictor population in 2 years, 4 years, and 6 years. (6 points: 2 points for each year, including 1 point for showing your work and 1 point for the answer)
t: time in years p(t): estimated snake population at time t
2

4

6

9. find the rate of change for the snake population from year 0 to year 2. (2 points)

10. apply the rate of change you found in question 9 in order to estimate the snake population for year 4. how does your answer compare with the calculated value? (1 point)

11. if there are approximately 36,660 snakes (rounded to the nearest 10) in year 10, how many will there be in year 12? (1 point)

making a decision:

12. given your calculations above, is ernest's or sarah's conjecture correct? (2 points)

I WILL ANSWER THE QUESTION FROM NUMBER 3 BECAUSE NUMBER 1 AND 2 ARE MISSING.

3. The graph is increasing, this is because the values of P(t) increase with the value of time. Therefore we conclude that the population of boa constrictor is growing.

4. From the graph we can actually conclude that the population of boa constrictor can actually be a very big problem. this is because the population is increasing at a very fast rate, having started at P(t)=7 when t=0, and P(t)=176 at time t=4 years.

5. The horizontal asymptote  of the graph given is at point:
P(t)=7

6. Since the initial the initial population at t=0 is P(t)=7 and the population is still growing with time, the range of the graph will be:
[7,∞)

7. The y-intercept is at the point when t=0, hence at t=0, P(t)=7, therefore the y-intercept is at P(t)=7 otherwise at point (0,7)

8. From the the formula given, the boa population at the intervals given will be as follows:
i. t=2 years:
Using the formula:
P(t)=5e0.89t
plugging t=2 we obtain:
P(2)=5e(0.89×2)
simplifying this we obtain:
P(2)=29.65
Thus the population after 2 years is 29.65~30

ii. t=4 years
Using the formula:
P(t)=5e0.89t
plugging t=4 years and simplifying we obtain:
P(4)=5e(0.89×4)
P(4)=175.82

thus the population after 4 years will be P(4)=175.82~176

iii. t=6 years.
As above we use the formula:
P(t)=5e0.89t
thus plugging in our values we obtain:
P(6)=5e(0.89×6)
simplifying this gives us:
P(6)=1042.56
Thus the population after 6 years will be P(t)=1042.56~1043

The rate of change of the population form t=0 to t=2 will be given by:
[P(2)-P(0)]/(2-0)
From the formula given:
P(0)=5e(0.89*0)
P(0)=5*1=5
P(2)=30
thus:
plugging the values we obtain:
(30-5)/(2-0)
=25/2
=12.5 
Thus the rate of change is:
12.5 

10. Using our rate of change, the formula will be:
P=(rate of change)*(time)
thus, when t=4 years:
P=12.5*4=50
The answer is lower than the calculated value because the we have assumed the population grows linearly but in reality it is growing exponentially. 

11. If there are approximately 36,660 snakes(rounded to the nearest 10) in year 10, the number of snakes in year 12 will be as follows:
Using the formula:
P(t)=5e0.89t
plugging the value t=12 years, we shall have:
P(12)=5e(0.89×12)
this will give us:
P(12)=5e10.68
P(12)=217,387.75
rounding to the nearest 10, we shall have
P(12)=217, 388