Vehicle 1 Price: $25, 000 Depreciation: $3000 per year Vehicle 2 Price: $25, 000 Depreciation: 15%...

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