The amount of money in an account may increase due to rising stock prices and decrease due to falling stock prices. Mason is studying the change in the amount of money in two accounts, A and B, over time.
The amount f(x), in dollars, in account A after x years is represented by the function below:
f(x) = 10,125(1.83)x
Part A: Is the amount of money in account A increasing or decreasing and by what percentage per year? Justify your answer. (5 points)
Part B: The table below shows the amount g(r), in dollars, of money in account B after r years.
r (number of years) 1 2 3 4
g(r) (amount in dollars) 9,638 18,794.10 36,648.50 71,464.58
Which account recorded a greater percentage change in amount of money over the previous year? Justify your answer.
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Answers
Account A: Decreasing at 8 % per year
Account B: Decreasing at 10.00 % per year
The amount f(x), in dollars, in account A after x years is represented by the function below:
f(x) = 10,125(1.83)x
Account B shows the greater percentage
change
Step-by-step explanation:
Part A: Percent change from exponential
formula
f(x) = 9628(0.92)*
The general formula for an exponential
function is
y = ab^x, where
b = the base of the exponential function.
if b < 1, we have an exponential decay
function.
f(x) decreases as x increases.
Account A is decreasing each year.
We can rewrite the formula for an
exponential decay function as:
y= a(1 – b)”, where
1- b = the decay factor
b = the percent change in decimal
form
If we compare the two formulas, we find
0.92 = 1- b
b = 1 - 0.92 = 0.08 = 8 %
The account is decreasing at an annual rate of 8%. The account is decreasing at an annual rate of 10.00%.
Account B recorded a greater percentage change in the amount of money over the previous year.
hope it helps u mate!!!!
It looks like your first account increases linearly,
unless you have a typo and meant:
f(x) = 10,125(1.83)^x
the second is
exponential and can be written as
g(r) = 9,638(1.95)^r
clearly g(r) has a higher percentage change since 1.95 > 1.83
a more interesting question might have been: When will the amount be the same?
10,125(1.83)^x = 9,638(1.95)^x
10,125(1.83)^x = 9,638(1.95)^x
(1.83)^x = .951901...(1.95)^x
take log of both sides and use log rules
xlog1.83 = log .951901... + xlog 1.95
x(log 1.83 - log 1.95) = log .951901...
x = .776..
check:
f(.776..) = 10,125(1.83)^.776.. = 1618410
g(.776..) = 9,638(1.95)^.776... = 16184.10