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 1200 1300 option 2 (amount in dollars) 1100 1210 1331 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 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

Step-by-step explanation:

Part A:

Option 1 is exponential and increases 130% per year

Option 2 increases $300 per year and it's linear 

Part B:

f(n)=1000(1.30)^n option 1

f(n)=1000+300n option 2

Part C:

f(20)=1000+300(20)

f(20)=1000+6000

f(20)=7,000 after 20 years for option 1

f(20)=1000(1.30)^20

f(20)=1000(190.04)

f(20)=190,050 after 20 years for option 2

We can see that there would be a significant difference in Belinda's investment if she uses option 2 over option 1.

There is a huge difference of 190,050 - 7,000 = 183,050 dollars.

Part A.

Option 1: linear function

Option 2: exponential function

This is because for Option 1 we can readily see that the money is constantly increasing by 100 every year hence linear. While for Option 2 the increase changes every year hence exponential.

Part B.

Option 1: Linear function has general form of:

y = mx + b

where y is the amount of money while x is the number of years

Calculating for slope m by using the data points in year 1 and 2 ( you can use any 2 data pairs)

m = (y2 – y1) / (x2 – x1)

m = (1200 – 1100) / (2 – 1)

m = 100

While b is the value of y when x = 0, hence b = 1000

Therefore:

y = 100 x + 1000

Option 2: Exponential function has general form of:

y = y0 (1+r)^x

where y0 is the initial amount and r is the growth rate

Solving for r using data at time x = 1

1100 = 1000 (1 + r)^1

r = 0.1

Therefore:

y = 1000 (1.1)^x

Part C. Solving for y for each options when x = 20

Option 1:

y = 100 (20) + 1000 = 3000

Option 2:

y = 1000 (1.1)^20 = 6727.50

Therefore Belinda should invest her money in Option 2 since it has greater return after 20 years.

7200

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