Question

(04.05 MC)

The amount of money in an account may increase due to rising stock prices and decrease due to falling stock prices. Emerson 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) = 1,264(1.09)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) 1,375, 1,512.50,1,663.75,1,830.13

Which account recorded a greater percentage change in amount of money over the previous year? Justify your answer.

Answer 1

A) increasing by 9% each year

B) Account B

Explanation:

`f(x)=1264(1.09)^x`

Part A

The amount of money in account A is increasing by 9% each year.

General form of an exponential function:
`f(x)=ab^x`

`a` is the initial value,
`b` is the growth factor and
`x` is time.

• If b < 1 then the function is decreasing
• If b > 1 then the function is increasing

As b = 1.09 > 1 then the function is increasing

`b` is the decimal form of percentage change.

Therefore, for b > 1, percentage increase = b - 1

for b < 1, percentage decrease = 1 - b

So as b = 1.09 > 1, percentage increase = 1.09 - 1 = 0.09 = 9%

Part B

From inspection, the amount in account B is increasing exponentially.

Therefore, we can use
`y=ab^x`

To determine the growth factor
`b`, divide one value of g(r) by its previous value:

`\implies b=(1512.50)/(1375)=(11)/(10)=1.1`

`\implies g(r)=a(1.1)^r`

To determine
`a`, input a pair of values (r, g(r)) into the equation and solve:

`\implies 1375=a(1.1)^1`

`\implies a=(1375)/(1.1)=1250`

Therefore, the equation for account B:
`g(r)=1250(1.1)^r`

Comparing the growth factor
`b` of both equations, account B recorded a greater percentage change in amount of money over the previous year since 1.1 > 1.09