Question

Belinda wants to invest $1,000. 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) 1300 1690 2197
Option 2 (amount in dollars) 1300 1600 1900
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.
Part B: Write one function for each option to describe the value of the investment f(n) in dollars, after n years.
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.
Plz explain good

Answer 1

Answer:

  A) 1: exponential; 2: linear

  B) 1: f(n) = 1000(1.3^n); 2: f(n) = 300n +1000

  C) yes there is a significant difference. Option 1 is by far the better choice with its value of about $190,050 vs. $7,000 for Option 2.

Step-by-step explanation:

When the x-values are sequential (1, 2, 3, ...), the y-values will have a common difference for a linear function, and a common ratio for an exponential function.

For the two investment options, we notice Belinda earns 1300 -1000 = 300 the first year for either option. The difference is the same the next year for Option 2 (1600 -1300 = 300), but is not the same for Option 1. For that option, the ratio is the same for the second year as it was for the first year:

  1690/1300 = 1.3 = 1300/1000

Part A:

Option 1 is described by an exponential function with a common ratio of 1.3. Option 2 is described by a linear function with a rate of change of ...

  (1300 -1000)/(1 -0) = 300 . . . dollars per year

__

Part B:

The generic form of an exponential function is ...

  f(n) = a·b^n

The value 'a' is the initial value, when n=0. For Belinda's investment, it is $1000. The value 'b' is the annual multiplier when n is in number of years. We determined that to be 1.3.

  Option 1: f(n) = 1000·1.3^n

The generic form of an exponential function is ...

  f(n) = a·n + b

where 'a' is the rate of change (the amount of change when n changes by 1 unit), and 'b' is the initial value. For Option 2, we determined the rate of change to be $300 per year, and we know the initial value is $1000.

  Option 2: f(n) = 300n +1000

__

Part C:

From our previous experience, we know that an exponential function will eventually outpace any linear function over a long-enough period. After 20 years, the values of the two options will be ...

  Option 1: f(20) = 1000·1.3^20 = 190,049.64 . . . dollars

  Option 2: f(20) = 300(20) +1000 = 7,000 . . . dollars

The result from using Option 2 will be significantly less than the result from using Option 1 over a period of 20 years.

Option 1 would give Belinda a balance about 27 times a much as option 2.