Question 1068162: Jennifer starts a new investment account that grows exponentially. Her financial advisor tells her the initial investment of $50,000 grows at a rate of about 15% annually.
1. Determine a function, I(t), that determines Jennifer’s investment account balance after t years. For the exponential growth function, what are the “a” and “b” values? What do those values represent? (5 points for the explanation of “a” and “b” and 5 points for the function)
Answer: 15000*(1+0.15)t
2. Calculate how much money Jennifer will have after 10 years. (10 points)
Answer: $60683.366036
3. What if Jennifer was able to deposit $100,000 as her initial investment, instead of $50,000.
Write a new function, N(t), to show this change. (3 points)
Answer: 100000*(1+0.15)t
Calculate how much money Jennifer would have after 8 years. (2 points)
Answer: $305902.286254
Using complete sentences, compare the differences in the functions and the amount of money after 8 years for the two different functions. (5 points)
Answer: y=ab^x, where a here is the amount deposited, and x is the number of compoundings. I'd round to two decimal places, since it is money. The figures are correct. The second is more than the first because more was deposited. The tripling time of money in years is ln3/interest rate as decimal number, because p/po=e^rt and ln (3), which is the ratio=rt, so ln3/r=t here in 10 years ln3/.15=7.32 years so that tripling just occurs for the second, but the first has quadrupled. Quadrupling is doubling of doubling, and the rule of 70 is used there, where 70 over the rate (in per cent) is the doubling time--70/15=4.67 years, so this doubled and again doubled. That is consistent with $60,000.
4. Graph the function I(t) and identify the y-intercept and the equation of the asymptote. (5 pts)
Answer: The y-intercept is (0,15,000)
The asymptote is y=0 (negative x, negative exponent, term goes to 0)
5. Graph the function N(t) and identify the y-intercept and the equation of the asymptote. (5 pts)\
Answer: y-intercept of (0,100000) and asymptote of y=0 (negative x, negative exponent, term goes to 0)
6. What does the asymptote mean in terms of Jennifer’s investment account? Be sure to explain your reasoning for credit. (5 pt)
Answer:??
7. Due to a worldwide recession, Jennifer’s financial advisor informs her the account is no longer growing after 4 years. He recommends she move her money to an account that is more conservative and earns only 4% annually. Determine the amount in Jennifer’s investment account after 4 years(using the $50,000 investment), rounding to the nearest dollar. It may be helpful to create a table. (10 points)
Answer:??
8. Write a new function, N(t), that represents the amount of money in Jennifer’s new account after t years. Be sure to use correct notation. (5 pts)
Answer:??
9. Did the asymptote of the new function change due to Jennifer moving her investment? Explain your reasoning. (5 pts)
Answer:??
can you please answer questions 6-9?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The asymptote is not relevant to her account, since she didn't start from 0 money, compoundings are not negative. The asymptote exists for the function itself, but negative values of x are not relevant.
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50000(1.15)^4=$87,450
Now using 4% for t years
N(t)=87450(1.04)^t
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This is a little ambiguous, because it is implied in the new function that x cannot be negative, so there is no asymptote. It is a piecewise function, made up of the first and the second, which are different curves.
Mathematically, however, the new function has an asymptote of 0 as well, since with infinitely large negative x, the function becomes 0.
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