Question
A marine biologist is studying the growth of a particular species of fish. She writes the following equation to show the length of the fish, f(m), in cm, after m months:
f(m) = 3(1.09)m
Part A: When the marine biologist concluded her study, the length of the fish was approximately 5.98 cm. What is a reasonable domain to plot the growth function? (4 points)
Part B: What does the y-intercept of the graph of the function f(m) represent? (2 points)
Part C: What is the average rate of change of the function f(m) from m = 2 to m = 8, and what does it represent? (4 points)
f(m) = 3(1.09)m
Part A: When the marine biologist concluded her study, the length of the fish was approximately 5.98 cm. What is a reasonable domain to plot the growth function? (4 points)
Part B: What does the y-intercept of the graph of the function f(m) represent? (2 points)
Part C: What is the average rate of change of the function f(m) from m = 2 to m = 8, and what does it represent? (4 points)
Answers
The m by the (1.09) is an exponent: f(m) = 3(1.09)^m
a) you would be solving:
5.98 = 3(1.09)^m
1.9333... = 1.09^m
take logs of both sides and use log rules ..
log 1.933... = m(log1.09)
m = appr 8.004
I would use a domain axis , time ? , of 0 to 10
b) at y-intercept, m = 0, so f(0) = 3 <----- the length of the fish at the start of the experiment
c)
f(2) = 3(1.09)^2 = appr 3.5643
f(8) = 3(1.09)^8 = appr 5.9777
avg rate of change or slope = (5.9777 - 3.5643)/(8-2) = .3017 cm/month
5.98 = 3(1.09)^m
1.9333... = 1.09^m
take logs of both sides and use log rules ..
log 1.933... = m(log1.09)
m = appr 8.004
I would use a domain axis , time ? , of 0 to 10
b) at y-intercept, m = 0, so f(0) = 3 <----- the length of the fish at the start of the experiment
c)
f(2) = 3(1.09)^2 = appr 3.5643
f(8) = 3(1.09)^8 = appr 5.9777
avg rate of change or slope = (5.9777 - 3.5643)/(8-2) = .3017 cm/month