Question

Victoria has some money from her birthday and the amount is modeled by the function h(x) = 200. She read about a bank that has savings accounts that accrue interest according to the function s(x) = (1.05)^x - 1. Victoria is thinking about putting her money into the savings account to gain interest. Using complete sentences, explain to Victoria how she can combine her functions to create a new function, and explain what this new function means.

Answer 1

s(x) is the function for how much percent of interest she will gain?

Answer 2

I just don't know how to combine the functions like do we multiply divide add subtract???

Answer 3

wait, so the \(\large\color{black}{ s(x) }\) is basically saying that for the first year it will be \(\large\color{black}{ s(1) }\) , for the 2nd year it will be \(\large\color{black}{ s(2) }\) , and on.... like this?

Answer 4

Yes I do believe that x is the number of years...

Answer 5

but it wouldn't be that s(2) is the combined interest after the 1st 2 years (altogether) ?

Answer 6

I would assume that it is not so. and will see how it comes out.

Answer 7

I think that if we do 2 years we would make the function Sn=200+1.05^2-1

Answer 8

it depends in whether the entire interest for 1st 2 years would be \(\large\color{black}{s(1)+s(2) }\) , or just \(\large\color{black}{s(2) }\) . I am sure about this from the wording of the question.

Answer 9

That went to 210 it kinda seems right??

Answer 10

I think you would add the 200 since she got that money and then we have to add the interest that she receives

Answer 11

you multiply.
lets say your increase is + 0.04 of the sum, then you would do sum * 1.04

Answer 12

I still got 210 as my answer???

Answer 13

you are asked to make one function that would be giving you the interest based on the s(x), and with the initial sum of 200 (as in h(x) ).

Answer 14

no that was illogical just now,

\(\large\color{black}{s(x \times h) =200[~(1.05)^x - 1~] }\) would be giving you the function for the just additional money.

Answer 15

am I interpreting the s(x) correctly, maybe x-1 is the exponent?

Answer 16

\[s(x)=1.05^(x-1)\]

Answer 17

\(\large\color{black}{s(x) =(1.05)^{x-1} }\) like this?

Answer 18

yeah

Answer 19

oh that is better

Answer 20

\(\large\color{black}{s(x) =(1.05)^{x-1} }\) is just a geometric sequence where:

~\(\large\color{black}{r=1.05 }\)
~\(\large\color{black}{a_1=1 }\)

Answer 21

Now how do we combine them that is my problem

Answer 22

\(\large\color{black}{s(x \times h) }\)

Answer 23

do you know how to write \(\large\color{black}{s(x \times h) }\) if:

~ \(\large\color{black}{s( h)=200 }\)
~ \(\large\color{black}{s(x) =(1.05)^{x-1} }\)

Answer 24

How would you solve that?

Answer 25

If I had:
\(\large\color{black}{s(h ) =50 }\)
\(\large\color{black}{s(x) =(2.07)^{x-2} }\)

then \(\large\color{black}{s(x \times h) =50(2.07)^{x-2} }\)

Answer 26

now, if I have:
\(\large\color{black}{s(h ) =200 }\)
\(\large\color{black}{s(x) =(1.05)^{x-1} }\)

then \(\large\color{black}{s(x \times h) =? }\)

Answer 27

I don't know how to solve the exponent to get it down

Answer 28

now, if I have:
\(\large\color{black}{s(h ) =200 }\)
\(\large\color{black}{s(x) =(1.05)^{x-1} }\)

then \(\large\color{black}{s(x \times h) =200(1.05)^{x-1} }\)

Answer 29

see what i am doing>?

Answer 30

Yes

Answer 31

I think I got it thanks

Answer 32

yw