Respuesta :
Answer:
A) increasing by 9% each year
B) Account B
Step-by-step explanation:
[tex]f(x)=1264(1.09)^x[/tex]
Part A
The amount of money in account A is increasing by 9% each year.
General form of an exponential function: [tex]f(x)=ab^x[/tex]
[tex]a[/tex] is the initial value, [tex]b[/tex] is the growth factor and [tex]x[/tex] 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
[tex]b[/tex] 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 [tex]y=ab^x[/tex]
To determine the growth factor [tex]b[/tex], divide one value of g(r) by its previous value:
[tex]\implies b=\dfrac{1512.50}{1375}=\dfrac{11}{10}=1.1[/tex]
[tex]\implies g(r)=a(1.1)^r[/tex]
To determine [tex]a[/tex], input a pair of values (r, g(r)) into the equation and solve:
[tex]\implies 1375=a(1.1)^1[/tex]
[tex]\implies a=\dfrac{1375}{1.1}=1250[/tex]
Therefore, the equation for account B: [tex]g(r)=1250(1.1)^r[/tex]
Comparing the growth factor [tex]b[/tex] of both equations, account B recorded a greater percentage change in amount of money over the previous year since 1.1 > 1.09