Belinda wants to invest $1000. The table below shows the value of her investment under two different options for two different years:
Number of years 1 2 3
Option 1 (amount in dollars) 1300 1600 1900
Option 2 (amount in dollars) 1120 1254.40 1404.93
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. Which option should Belinda choose? Explain your answer, and show the investment value after 20 years for each option. (4 points)
A)
Option 1 is linear, $300 per year
Option 2 , exponential is compound interest, a geometric sequence
an = an-1 * 1.12
B)
Option 1
A = 1300 + 300(n-1) = P +.23 P (n-1)
Option 2
A = 1120 * 1.12^(n-1) = P *1.12^(n-1)
C) after 20 years
Option 1
P + .23*19 P = 5.37 P
Option 2
P* 1.12^19 = 8.61 P
so despite the lower percentage, the compounding wins big time.
Just change some words
Part A: To determine whether the function that describes the value of the investment after a fixed number of years is linear or exponential, we need to analyze the pattern in the given values.
For Option 1, when the number of years increases by 1, the value of the investment increases by a fixed amount (e.g., 300). This consistent increase indicates a linear relationship.
For Option 2, when the number of years increases by 1, the value of the investment does not increase by a fixed amount. Instead, it increases at a varying rate that is not constant. This indicates an exponential relationship.
Therefore, Option 1 can be described by a linear function, whereas Option 2 can be described by an exponential function.
Part B:
Option 1: Since Option 1 is linear, we can use the slope-intercept form of a linear function: f(n) = mx + b.
The slope (m) is the increase in value (change in y) divided by the increase in years (change in x): m = (1600 - 1300)/(2 - 1) = 300.
To find the y-intercept (b), which is the initial value, we can use any given value and solve for b. Let's use the first-year value: 1300 = 300(1) + b. Solving for b, we get b = 1000.
Therefore, the function for Option 1 is: f(n) = 300n + 1000.
Option 2: Since Option 2 is exponential, we use the general form of an exponential function: f(n) = ab^n.
Let's calculate the constant ratio (b^n) for n = 2 and n = 3: (1254.40/1120) ≈ 1.118 and (1404.93/1254.40) ≈ 1.119.
The values are close, but not exactly equal. This suggests that the constant ratio is slightly increasing. As a result, we can't directly determine the base (b) for the exponential function. However, we can approximate the function using a regression model or other statistical methods.
Part C: To determine which option would increase Belinda's investment value by the greatest amount after 20 years, we can substitute n = 20 into the functions we derived in Part B and compare the results.
For Option 1: f(20) = 300(20) + 1000 = 6000 + 1000 = 7000.
For Option 2: Since we don't have an exact exponential function, we can't calculate a precise value like we did for Option 1.
However, we can estimate the value by assuming a steady rate of increase. Based on the given data, the average constant ratio is approximately 1.1185. Using this ratio, we can approximate the value for 20 years:
f(20) = 1120(1.1185^20) ≈ 1120 * 9.668 ≈ 10,841.6.
Comparing the two results, we see that Option 2 would help to increase Belinda's investment value by the greatest amount in 20 years, as it is approximately $3,841.60 higher than Option 1's value.