Exponential Functions Percent Increase and Decrease

The value of a car is $15,000 and depreciates at a rate of 8% per year. What is the decay factor?

• ^.80

• ^1.08

• .92

• 8

Tags

CCSS.HSF-IF.C.8B

Which of the following functions shows an initial amount of $15 and an increase of 35% each year?

• y = 15(35)^x

• y = 15(1.35)^x

• y = 15(0.35)^x

• y = 35(1.15)^x

Tags

CCSS.HSF.LE.A.2

Which equation is representative of the graph?

• y=2(3)^x

• y=3(2)^x

• y=2(6)^x

Tags

CCSS.HSF.LE.A.2

The growth factor for this table is 

• 2

• 4

• 12

• 144

Tags

CCSS.HSF-LE.A.1A

Rhonda deposited $3000 in an account in the Merrick National Bank, earning 4.2% interest, compounded annually. She made no deposits or withdrawals. Write an equation that can be used to find B, her account balance after t years.

• B = 3000(1 – 4.2)^t

• B = 3000(1 + 4.2)^t

• B = 3000(1 – 0.042)^t

• B = 3000(1 + 0.042)^t

Marilyn collects old dolls. She purchases a doll for $450. Research shows this doll's value will increase by 2.5% each year. Write an equation that determines the value, V, of the doll t years after purchase.

• V = 450(1 + 0.025)^t

• V = 450(1 – 0.025)^t

• V = 450(1 + 2.5)^t

• V = 450(1 – 2.5)^t

Tags

CCSS.HSF.LE.A.2

A car was purchased for $25,000. Research shows that the car has an average yearly depreciation rate of 18.5%. Create a function that will determine the value, V(t), of the car t years after purchase.

• V(t) = 25000(1 – 0.185)^t

• V(t) = 25000(1 + 0.185)^t

• V(t) = 25000(1 – 18.5)^t

• V(t) = 25000(1 + 18.5)^t

Tags

CCSS.HSF.LE.A.2

The bear population in a given area is currently 1580. They anticipate the bear population to decrease by 2% each year. Which function represents the population of bears, B, after t years.

• B = 1580(0.02)

• B = 1580(1 – 0.02)

• B = 1580(1 + 0.02)

• B = 1580(1 – 0.2)

Tags

CCSS.HSF.LE.A.2

The population of a town is currently 6342 and is increasing by a rate of 1.3% each year. Which function represents the population of people, P, after t years.

• P = 6342(1 + 1.3)^t

• P = 6342(1 – 1.3)^t

• P = 6342(1 + .013)^t

• P = 6342(1 – .013)^t

Tags

CCSS.HSF.LE.A.2