SOLUTION: An exponential decay graph shows the expected depreciation for a new boat, selling for $3,500, over 10 years. a. Write an exponential function for the graph. b. Use the fun

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: An exponential decay graph shows the expected depreciation for a new boat, selling for $3,500, over 10 years. a. Write an exponential function for the graph. b. Use the fun      Log On


   

Question 628699: An exponential decay graph shows the expected depreciation for a new boat, selling for $3,500, over 10 years.
a. Write an exponential function for the graph.
b. Use the function in part a to find the value of the boat after 9.5 years.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
First, this problem has nothing to do with Trigonometry. It belongs in one of the categories related to exponents. Please post using a possibly relevant category.

Second, without the graph you were given it will be impossible for a tutor to solve this problem. I will, nevertheless, try to help as much as I can.

The general equation for exponential growth and decay is:
A+=+A%5B0%5D%2A%281%2Br%29%5Et
where
t is a number of units of time. (Often the units are years so t would be a number of years.)
A = the amount after "t" units of time.
A%5B0%5D is the starting amount (IOW the amount at t = 0. This is why the A has a subscript of zero.)
r is a decimal (or fraction) between 0 and 1 which represents the rate of change. A positive r means growth. A negative r means decay. (Note: Some books use A+=+A%5B0%5D%2Ar%5Et, merging the 1 and r. In this case if r > 1 then the equation is for growth and if r is between 0 and 1 then the equation is for decay. I prefer the first form I gave you because that percent growths or decays are easier to use with that form.)

You have been given the original amount, A%5B0%5D. So your equation so far is:
A+=+3500%2A%281%2Br%29%5Et

From this point on I can only tell you what you should do to finish:
  1. Find the value for r:
    1. Look at the graph and find the coordinates of any point (other than where t = 0) on the graph. (If the axes are not labeled t and A, then the "t" in the equation corresponds to "x" and the "A" corresponds to the "y".)
    2. Replace the t and A in your equation with the coordinates you found in step 1.
    3. Solve this equation for r.
  2. Rewrite your working equation:
    A+=+3500%2A%281%2Br%29%5Et
    substituting in the value for r you just found. This (or the equation you get if you add the 1 to r) is the answer to part a.
  3. Using the equation from part a, replace the t with 9.5 and solve for A. This will be the answer for part b.

Here's an example:
1. Solve for r
1.1 Find a point. Suppose that your graph shows that at t (or x) = 10, the A (or y) is 700.
1.2 Insert the coordinates into your working equation:
700+=+3500%2A%281%2Br%29%5E10
1.3. Solve for r.
Dividing both sides by 3500:
0.2+=+%281%2Br%29%5E10
Find the 10th root of each side. (On your calculator use 0.2^(1/10) or 0.2^0.1)
0.85133992+=+1+%2B+r
Subtract 1 from each side:
-0.14866008+=+r
(BTW: As a percent, r is an approximately 15% rate of decay.)

2. Replace the r in your working equation:
A+=+3500%2A%281%2B-0.14866008%29%5Et
This, or A+=+3500%2A%280.85133992%29%5Et, would be the answer to part a.

3. To find the value of the car after 9.5, replace the t in you equation with 9.5 and solve for A:
A+=+3500%2A%281%2B-0.14866008%29%5E%289.5%29
Simplifying...
A+=+3500%2A%280.85133992%29%5E%289.5%29
Using 0.85133992^(9.5) on our calculators:
A+=+3500%2A0.21675967
A+=+758.65884937
So after 9.5 years the car would be worth approximately $758.69.