Question

Edward plans to put $75 into a savings account. He can place his money into an account represented by b(x) = 7x + 75, or into another account represented by c(x) = 75(1.06)x. Which account has the highest value in 2 years? Which account has the highest in 12 years?
A.) c(x) has the highest value in 2 years; c(x) has the highest value in 12 years

B.) b(x) has the highest value in 2 years; b(x) has the highest value in 12 years

C.) c(x) has the highest value in 2 years; b(x) has the highest value in 12 years

D.) b(x) has the highest value in 2 years; c(x) has the highest value in 12 years

Answer 1

In 2 years :
b(x) = 7x + 75....b(2) = 7(2) + 75.....b(2) = 14 + 75....b(2) = 89
c(x) = 75(1.06)(2)...c(x) = 79.5 * 2 ....c(x) = 159

in 12 years :
b(x) = 7x + 75...b(12) = 7(12) + 75....b(12) = 84 + 75....b(12) = 159
c(x) = 75(1.06)(12)....c(12) = 79.5 * 12....c(12) = 954

c(x) has the highest value in both 2 and 12 years

Answer 2

Let

x------> the time in years

b(x)----> the value in dollars of a saving account

c(x)----> the value in dollars of another account

we know that

$b(x)=7x+75$

$c(x)=75(1.06)^{x}$

Step 1

Find the value of each account for $x=2\ years$

$b(2)=7*2+75=\ 89$

$c(2)=75(1.06)^{2}=\ 84.27$

so

$b(2) > c(2)$----> b(x) has the highest value in $2$ years

Step 2

Find the value of each account for $x=12\ years$

$b(12)=7*12+75=\ 159$

$c(12)=75(1.06)^{12}=\ 150.91$

so

$b(12) > c(12)$----> b(x) has the highest value in $12$ years

therefore

the answer is the option B

b(x) has the highest value in $2$ years; b(x) has the highest value in $12$ years