vehicle 1 price: $25, 000 depreciation: $3000 per year vehicle 2 price: $25, 000 depreciation: 15%...
TRANSCRIPT
6.5 CONSTRUCT & APPLY EXPONENTIAL MODELS
THINGS ARE NOT ALWAYS WHAT THEY SEEM…
THINGS ARE NOT ALWAYS WHAT THEY SEEM…
CONSTRUCT & APPLY MODELS
Vehicle 1 Price: $25, 000
Depreciation: $3000 per year
Vehicle 2 Price: $25, 000
Depreciation: 15% per year
So you want to buy a car?
APPLYING EXPONENTIAL MODELS Every year, Vehicle 1 depreciates $3000 Every year, Vehicle 2 depreciates 15%
Years (n) Vehicle 1 Vehicle 2
0 $25 000 $25 000
1 $22 000 $21 250
2 $19 000 $18 062.50
3 $16 000 $15 353.13
VEHICLE 1: v = 25 000 – 3000n
VEHICLE 2: v = 25 000
Which vehicle is a better deal when it comes to depreciation?
APPLYING EXPONENTIAL MODELS
0 1 2 3 4 5 6 7 8 90
5000
10000
15000
20000
25000
30000
Vehicle Depreciations
Vehicle 1Vehicle 2
Time (years)
Valu
e o
f V
ehic
le
Vehicle 2 depreciates less in the long run! Stunning result…
SIMPLE & COMPOUND INTEREST Edgar has $500 to invest and is
considering two investment options.Option A: A treasury bond that pays 8% simple interest. The amount, A, after n years is given by the equation A = 500 + 40n.Option B: A savings account that pays 6.5% per year, compounded annually. The amount, A, after n years is given by the equation A = 500(1.065)n.
Which is a better investment? Why?
SIMPLE & COMPOUND INTEREST Option A: A = 500 + 40n Option B: A = 500(1.065)n
0 2 4 6 8 10 12 140
200
400
600
800
1000
1200
Investment Options
Option AOption B
Time (years)
Investm
ent
Am
ount
($)
SIMPLE & COMPOUND INTEREST
Option A ahead at the start Option B catches up and surpasses A Also, notice that Option A is linear, but
Option B is exponential.
0 2 4 6 8 10 12 140
200
400
600
800
1000
1200
Investment Options
Option AOption B
Time (years)
Investm
ent
Am
ount
($)
LINEAR, QUADRATIC, EXPONENTIAL, OR NONE?
LINEAR
LINEAR, QUADRATIC, EXPONENTIAL, OR NONE?
EXPONENTIAL