Question

The exponential decay graph shows the expected depreciation for a new boat, selling for $3500, over 10 years. (Will attach graph) 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 1

need help i dont understand this problem

Answer 2

I think I can help you with this.

Answer 3

thank you that would be nice if you could

Answer 4

I'm still here - just working on the formula.

Answer 5

Exponential decay and the exponential growth curves both have the formula:
y = a(b)^x where a and b are positive.
If 0 < b < 1 it is an exponential decay.
If b > 1 it is an exponential growth.

Answer 6

Here we have two good points:
when x = 0, y = 3500
when x = 2, y = 2000
y = a(b)^x
when x = 0, y = 3500
3500 = a(b)^0 = a
a = 3500
y = 3500(b)^x
when x = 2, y = 2000
2000 = 3500(b)^2
b^2 = 2000 / 3500 = 0.57143
b = 0.7559

y = 3500(0.7559)^x

Answer 7

y = 3500(0.7559)^x
when x = 9.5, y = 3500(0.7559)^9.5
y = $245

Answer 8

Okay I determined that the "half-life" of the price is 2.5 years.
Basically, the price of the boat decreases by 2 every 2.5 years.
For example, after 5 years the price is reduced 4 times.

The equation for the price of the boat is
price = 3,500 / 2^ (years/2.5)
and here is the yearly price for the boat:

Years Price
0 3,500
1 2,653
2 2,010
3 1,523
4 1,155
5 875
6 663
7 503
8 381
9 289
10 219
11 166
12 126
13 95

The price of the boat after 9.5 years?
price = 3,500 / 2^ (years/2.5)
price = 3,500 / 2^ (9.5/2.5)
price = 3,500 / 2^ (3.8)
price = 3,500 / 13.93
price = $251

Oh and a graph is attached

Answer 9

ranga - my equation gives the price after 9.5 years as 251 whereas you said it's 245

I'd say that is pretty darned good agreement !!!

Answer 10

Yes it is. And I think they will accept either answer as long as it is in the ball park because reading the y values from this graph is an estimation.