Question 1005042: Your parents decide to start saving money to buy you a new car. They begin with $10 and say they will double the balance every month. Write an equation in function notation to represent the amount of money that your parents saved as a function of the number of months m.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! month 1 is 10 dollars.
month 2 is 2 * 10 = 20 dollars.
month 2 is 2 * 20 = 40 dollars.
month 4 is 2 * 40 = 80 dollars.
if you look at what's happening to your original deposit, you get:
month 1 = 10 * 2^0 = 10 * 1 = 10
month 2 = 10 * 2^1 = 10 * 2 = 20
month 3 = 10 * 2^2 = 10 * 4 = 40
month 4 = 10 * 2^3 = 10 * 8 = 80
month 5 = 10 * 2^4 = 10 * 16 = 160
month n would therefore be equal to 10 * 2^(n-1)
this looks very much like a geometric series where A1 is the initial amount, and An is the amount in time point n.
the general formula for a geometric series is An = A1 * r^(n-1)
An is the amount in time point n.
A1 is the original amount in time point 1.
r is the common ratio
n is the time point.
using this formula, you get:
A1 = 10 * 2^(1-1) = 10 * 2^0 = 10
A2 = 10 * 2^(2-1) = 10 * 2^1 = 10 * 2 = 20
A3 = 10 * 2^(3-1) = 10 * 2^2 = 10 * 4 = 40
A4 = 10 * 2^(4-1) = 10 * 2^3 = 10 * 8 = 80
A5 = 10 * 2^(5-1) = 10 * 2^4 = 10 * 16 = 160
money is doubling every month.
looks like the formula is what you're looking for.
in functional notation, you would say:
f(n) = 10 * 2^(n-1)
your function name is f.
the argument to your function is n.
your independent variable is n.
your dependent variable is f(n).
give it a different function name, such as g, and your function becomes:
g(n) = 10 * 2^(n-1)
your function name is g.
the argument to your function is n.
your independent variable is n.
your dependent variable is g(n).
give it a different argument name, such as x, and your function becomes:
g(x) = 10 * 2^(x-1)
your function name is g.
the argument to your function is x.
your independent variable is x.
your dependent variable is g(x).
f(n) = 10 * 2^(n-1) will give you the same result as g(x) = 10 * 2^(x-1) as long as x = n.
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