1.A scientist is studying the growth of a particular species of plant. He writes the following equation to show the height of the plant f(n), in cm, after n days:
f(n) = 12(1.03)n
Part A: When the scientist concluded his study, the height of the plant was approximately 16.13 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(n) represent? (2 points)
Part C: What is the average rate of change of the function f(n) from n = 3 to n = 10, and what does it represent? (4 points)
2.The price of products may increase due to inflation and decrease due to depreciation. Derek is studying the change in the price of two products, A and B, over time.
The price f(x), in dollars, of product A after x years is represented by the function below:
f(x) = 72(1.25)x
Part A: Is the price of product A increasing or decreasing and by what percentage per year? Justify your answer. (5 points)
Part B: The table below shows the price f(t), in dollars, of product B after t years:
t (number of years)
1 2 3 4
f(t) (price in dollars)
65 84.5 109.85 142.81
Which product recorded a greater percentage change in price over the previous year? Justify your answer.
3.Belinda wants to invest $1000. The table below shows the value of her investment under two different options for three different years:
Number of years
1 2 3
Option 1 (amount in dollars)
1100 1210 1331
Option 2 (amount in dollars) 1100 1200 1300
Part A: What type of function, linear or exponential, can be used to describe the value of the investment after a fixed number of years using option 1 and option 2? Explain your answer. (2 points)
Part B: Write one function for each option to describe the value of the investment f(n), in dollars, after n years. (4 points)
Part C: Belinda wants to invest in an option that would help to increase her investment value by the greatest amount in 20 years. Will there be any significant difference in the value of Belinda's investment after 20 years if she uses option 2 over option 1? Explain your answer, and show the investment value after 20 years for each option.
Part A: The function f(n) = 12(1.03)^n represents the height of the plant after n days. Since the height of the plant was approximately 16.13 cm at the end of the study, we can use this information to determine a reasonable domain to plot the growth function.
To find the domain, we need to solve the equation f(n) = 16.13. Plugging in the given values, we have:
16.13 = 12(1.03)^n
Divide both sides of the equation by 12:
1.3442 = (1.03)^n
To solve for n, we can take the logarithm of both sides of the equation with base 1.03:
log₁.₀₃(1.3442) = log₁.₀₃((1.03)^n)
Using logarithm properties, we get:
n ≈ log₁.₀₃(1.3442)
Using a calculator, we find n ≈ 20. Therefore, a reasonable domain to plot the growth function would be from n = 0 to n = 20, representing the initial growth and the height of the plant at the end of the study.
Part B: The y-intercept of the graph of the function f(n) represents the initial height of the plant. To find the y-intercept, we set n = 0 in the equation:
f(0) = 12(1.03)^0 = 12(1) = 12
Therefore, the y-intercept of the graph represents the initial height of the plant, which is 12 cm.
Part C: The average rate of change of the function f(n) from n = 3 to n = 10 can be calculated by finding the difference in the function values divided by the difference in the input values:
Average rate of change = (f(10) - f(3)) / (10 - 3).
Using the given equation, we can evaluate the function values:
f(10) = 12(1.03)^10
f(3) = 12(1.03)^3
Substituting these values into the formula, we can calculate the average rate of change.
2.The price of product A can be determined using the function f(x) = 72(1.25)^x.
Part A: To determine if the price of product A is increasing or decreasing, we need to examine the value of the growth factor, 1.25. Since the growth factor is greater than 1, the price of product A is increasing over time.
To determine the percentage increase per year, we subtract 1 from the growth factor and then multiply by 100:
Percentage increase per year = (1.25 - 1) * 100 = 25%.
Therefore, the price of product A is increasing by 25% per year.
Part B: To determine which product recorded a greater percentage change in price over the previous year, we need to compare the percentage changes of product A and product B.
In the table given, we can calculate the percentage change in price for each year by dividing the difference in prices by the previous year's price and multiplying by 100.
Percentage change for product A:
Year 1: (84.5 - 65) / 65 * 100 ≈ 30.0%
Year 2: (109.85 - 84.5) / 84.5 * 100 ≈ 30.0%
Year 3: (142.81 - 109.85) / 109.85 * 100 ≈ 30.0%
Percentage change for product B:
Year 1: (84.5 - 65) / 65 * 100 ≈ 30.0%
Year 2: (109.85 - 84.5) / 84.5 * 100 ≈ 30.0%
Year 3: (142.81 - 109.85) / 109.85 * 100 ≈ 30.0%
Both product A and product B recorded the same percentage change in price over the previous year, which is approximately 30.0%. Therefore, there is no significant difference in the percentage change for both products.
3.For option 1, the values are increasing by a fixed amount each year, which indicates a linear function. For option 2, the values are increasing exponentially with a constant ratio, indicating an exponential function.
Part B: The linear function for option 1 can be represented as:
f(n) = 1100 + 100n
The exponential function for option 2 can be represented as:
f(n) = 1100(1.10)^n
Here, n represents the number of years.
Part C: To determine if there will be any significant difference in the value of Belinda's investment after 20 years using option 2 over option 1, we can calculate the value using each option.
Option 1 after 20 years:
f(20) = 1100 + 100(20) = 1100 + 2000 = 3100 dollars
Option 2 after 20 years:
f(20) = 1100(1.10)^20 ≈ 1100(6.7275) ≈ 7490.25 dollars
There is a significant difference in the value of Belinda's investment after 20 years if she uses option 2 over option 1. The value of her investment using option 2 is approximately 7490.25 dollars, while the value using option 1 is only 3100 dollars. Option 2 provides a much higher return on investment after 20 years.