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
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
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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
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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.
