Question

Need help! Will give medal and fan.
. The exponential decay graph shows the expected depreciation for a new boat, selling for $3500, over 10 years.

a.Write an exponential function for the graph. b.Use the function to find the value of the boat after 9.5 years. Show your work
Here's the graph:
http://assets.openstudy.com/updates/attachments/50ec4d8be4b07cd2b648e682-swiftskier96-1357663636612-graph.jpg

Answer 1

@phi

Answer 2

Do you know the equation of a "generic" exponential function ?

Answer 3

I dont'. I do know this one A(t)=a(1+r)^t. Does it have anything to do with the problem?

Answer 4

yes, that would work.

Your first job is pick some points off the graph, and collect some (t, A(t) ) pairs
(where A(t) is the y value )

Answer 5

Ok. So there's (3,1500) and (2,2000) and (1,2500). It decreases 500 each time right?

Answer 6

It doesn't *quite* decrease by 500. the value at t=1 looks a bit over 2500.
I would use (0,3500) (t= 0 is always a nice choice)

so let's use (0,3500) and (3,1500)

Answer 7

use the first point, and sub t=0 and A(0) = 3500 into your
A(t)=a(1+r)^t

Answer 8

So it will be A(0)=3500(1+2)^0?

Answer 9

I meant 1+r

Answer 10

the "a" is unknown. 3500 is the A(0) (A is the cost of the boat)

Answer 11

So 3500=a(1+r)?

Answer 12

yes, except you should have (1+r)^0

Answer 13

anything to the zero power (except 0) becomes 1.
\[ 3500=a(1+r)^0 \\
3500=a\cdot 1 \\
a= 3500
\]

Answer 14

that is one of the "unknowns" is now known.
Instead of solving for r, let's solve for 1+r, so that we get an equation that looks like
\[ A(t) = 3500 x^t \]
where x is (1+r) (some number)

Use your 2nd point, (3,1500) in this equation

Answer 15

So 1500=3500x^3
then 0.428571429=x^3
x=0.753947?
Did I do that right?

Answer 16

yes. Now your equation looks like this
\[ A(t) = 3500 (0.753947)^t \]
If we test this with t=7 we get about $485, which looks reasonable.

They want A(9.5) so replace t with 9.5

Answer 17

Then it will be A(t)=3500(0.753947)^9.5
A(t)=3500(0.06835)
A(t)=$239??

Answer 18

yes, that looks good.
But I would replace t with 9.5 in A(t) to show you are find A when t is 9.5 years
in other words, write it as
A(9.5) = $239

Answer 19

OMG. Thank you soooo much. You're the best. I never thought I wasn't going to figure this out

Answer 20

another way is use the form
\[ A(t) = a\ e^{kt} \]
where e is the base of the "natural logarithm" (about 2.71828)
we would find a is still 3500. and you must find k.
using the point (3,1500)
\[1500 = 3500 e^{3k} \\
\frac{3}{7} = e^{3k}
\]
take the ln (natural log) of both sides
\[ -0.84729786 = 3 k
\\ k = -0.84729786/3 \\
k = -0.2824 \]
and the equation is
\[ A(t) = 3500 \ e^{-0.2824 t} \]
if we use t= 9.5 we will get A(9.5) = $239

Answer 21

That form seems a little simpler. I'll remember that. Thanks :D

Answer 22

of course
\[ e^{-0.2824} = 0.753947 \]
and
\[ e^{-0.2824t} =\left( e^{-0.2824}\right)^t = (0.753947)^t \]
so the 2 forms are really the same equation...

Answer 23

Oh ok. Thanks again for your help :)