Answer:
A) increasing by 9% each year
B) Account B
Step-by-step explanation:
![f(x)=1264(1.09)^x](https://tex.z-dn.net/?f=f%28x%29%3D1264%281.09%29%5Ex)
<u>Part A</u>
The amount of money in account A is increasing by 9% each year.
General form of an exponential function: ![f(x)=ab^x](https://tex.z-dn.net/?f=f%28x%29%3Dab%5Ex)
is the initial value,
is the growth factor and
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
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%
<u>Part B</u>
From inspection, the amount in account B is increasing exponentially.
Therefore, we can use ![y=ab^x](https://tex.z-dn.net/?f=y%3Dab%5Ex)
To determine the growth factor
, divide one value of g(r) by its previous value:
![\implies b=\dfrac{1512.50}{1375}=\dfrac{11}{10}=1.1](https://tex.z-dn.net/?f=%5Cimplies%20b%3D%5Cdfrac%7B1512.50%7D%7B1375%7D%3D%5Cdfrac%7B11%7D%7B10%7D%3D1.1)
![\implies g(r)=a(1.1)^r](https://tex.z-dn.net/?f=%5Cimplies%20g%28r%29%3Da%281.1%29%5Er)
To determine
, input a pair of values (r, g(r)) into the equation and solve:
![\implies 1375=a(1.1)^1](https://tex.z-dn.net/?f=%5Cimplies%201375%3Da%281.1%29%5E1)
![\implies a=\dfrac{1375}{1.1}=1250](https://tex.z-dn.net/?f=%5Cimplies%20a%3D%5Cdfrac%7B1375%7D%7B1.1%7D%3D1250)
Therefore, the equation for account B: ![g(r)=1250(1.1)^r](https://tex.z-dn.net/?f=g%28r%29%3D1250%281.1%29%5Er)
Comparing the growth factor
of both equations, account B recorded a greater percentage change in amount of money over the previous year since 1.1 > 1.09