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Park rangers tagged fish in a state park to track their population. The function f(x)=200(1.05)x can be used to find the population of fish after x years.

1. What is the initial population of fish? What feature of the graph does this represent?

2.At what rate is the fish population growing or declining? How do you know?

3. Create a graph and table that represent the situation. The axes and asymptotes of your graph should be labeled and your table should include the y-intercept.

4.What are the domain and range for this situation? What do they represent in the context of the problem?

5.What is the population of fish after 11 years? Explain how you found your answer.

as soon as possible

Sagot :

The given exponential function can be used to determine the

population of the fish at different times.

Response:

1. 200

2. 0.05

3. Please find attached the graph of the exponential function and the table

[tex]\begin{array}{|c|c|c}x&f(x)\\0&200 & y-intercept\\1&210\\3&231.525\\4&243.1\\5&255.26\end{array}[/tex]

4. Range; [0, ∞)

Domain: [200, ∞)

5. 346 fishes (approximate value)

How is the population of the fish obtained from the exponential function?

The given function is f(x) = 200·(1.05)ˣ

1. The initial population is given by the value of the function when x = 0,

which is given as follows;

At x = 0, f(0) = 200·(1.05)⁰ = 200

  • The initial population of the fish is 200

2. The rate of growth of the fish from the exponential function is given as follows;

The growth factor = 1.05

  • The growth rate = 0.05

3. Please find attached the graph of the exponential function created

with MS Excel.

The asymptote of the exponential function is the line, y = 0

The table of values is presented as follows;

[tex]\begin{array}{|c|c|c}x&f(x)\\0&200 & y-intercept\\1&210\\3&231.525\\4&243.1\\5&255.26\end{array}[/tex]

4. The domain is the possible x-values which is; 0 ≤ x < ∞

The range is the possible y-values which is; 200 ≤ f(x) < ∞

Which gives;

  • Domain; [0, ∞)
  • Range: [200, ∞)

5. The population of the fish after 11 years is f(11) = 200·(1.05)¹¹

f(11) = 200 × (1.05)¹¹ ≈ 342

  • The population of the fish after 11 years is 346 (fishes)

Learn more about exponential function here:

https://brainly.com/question/15662160

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