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
Option 1 is exponential and increases 130% per year
Option 2 increases $300 per year and it's linear
f(n)=1000(1.30)^n option 1
f(n)=1000+300n option 2
f(20)=7,000 after 20 years for option 1
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.
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.
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
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
y = 1000 (1.1)^x
Part C. Solving for y for each options when x = 20
y = 100 (20) + 1000 = 3000
y = 1000 (1.1)^20 = 6727.50
Therefore Belinda should invest her money in Option 2 since it has greater return after 20 years.
Part A: Option 1 is exponential because it does not increase by a constant rate. Option 2 is linear because it increases by a constant rate.
For option 1: a(b)^x
a=1000 and b=1100/1000=1.1
the function will be:f(n)=1000(1.1)^x
For option 2:
Since the price increases by $100 each year, its linear.
Here would be the function: f(n)=1000+100(n)
Part C: Invest for 20 years in option 1: Yes.
y=6727.50, which is the price of the investment in 20 years.
For option 2:
=3000, which is the price the of the investment in 20 years.
If she uses option 2 rather than option 1, she will loose $3727.5
In option two, the investment increases by a common ratio of 1.1, or increases by 10%, each year, and can thus be modeled by an exponential equation.
The exponential function will grow much larger versus the linear function because the rate of change increases every year...after 20 years
f2(20)=1000(1.1^20)=$6727.50 (to nearest cent)
So the exponential function is over 242% more than the linear function over this time period.
The function that will be used to model the situation is exponential function, This is because the increment in the values of the the investment are not linear increment. The appropriate function is exponential function.
Exponential function is given by:
a= initial value
for Option A:
thus the function will be:
for option B
Suppose Belinda invested in :
Option A, the value after 20 years will be:
Option B, the value after 20 years will be:
thus option A has a greater value
Exponential Growth is much greater than linear growth.