ALGEBRA I Chapter 6 Section 6 - Winston-Salem/Forsyth ...



ALGEBRA I Chapter 10 Section 6

EXPONENTIAL FUNCTIONS: GROWTH AND DECAY

1. TWO EXAMPLES OF EXPONENTIAL FUNCTIONS

|EQUATION |Find I and a. |TABLE |GRAPH (scale appropriately) |

|y = 16(3/2)x |I = 16 |x |0 |

| | | | |

| |a = 3/2 | | |

|y = 64(1/2)x |I = 64 |x |0 |

| | | | |

| |a = 1/2 | | |

2. Observations about Exponential Growth and Decay

|Exponential GROWTH also called |Exponential DECAY also called |

|Appreciation |Depreciation |

|Increase |Decrease |

|Initial value still used |Initial value still used |

|a > 1 |a< 1 |

|Graph goes from small to big |Graph goes big to small |

Looking at our First Two Examples:

• The number without the exponent is still the __INTIAL VALUE and Y-INTERCEPT____.

• The number with the exponent is __a, the number we are multiplying by each time/ PATTERN__.

o Growth: a > 1 and Decay: a < 1

3. We replace a of y = ax with the concept of RATE or PERCENT OF CHANGE, r:

|Exponential GROWTH |Exponential DECAY |

|a = 1 + r |a = 1 - r |

Increase by 8%: 1 + .08 = 1.08. Exponential growth Decrease by 8%: 1 - .08 = .92. Exponential Decay

Increase by 12%: _1+.12 = 1.12__ Decrease by 12%: __1- .12 = 0.88___

4. Equations:

r = Rate or Percent of Change (%) t = TIME

I = Initial Value or y-intercept y = Final Amount (predicted value)

|Exponential GROWTH |Exponential DECAY |

|y = I* (1 + r)t |y = I* (1 - r)t |

$50,000 increases by 7% for 15 years=

50,000 * (1 +.07)15 = $137,951.58

$45,000 appreciates by 10% for 12 years =

45,000 * (1 +.1)12 = $141,229.28

$50,000 decreases by 7% for 15 years =

50,000 * (1 - .07)15 = $16,835.04

$35,000 depreciates by 15% for 8 years =

35,000 * (1 - .15)8 = $9,537.17

Exponential Growth and Decay Problems

1. The equation for the price of a car that was bought for $10,000 and has depreciated 10% yearly is given as y = (10000) * (1-.1)t, where t = number of years since it was originally bought. Find the price of the car 8 years later.

a. Answer: _$4304.67__

b. What is the rate of decrease? __10%____

2. The equation for the price of a baseball card that was bought for 5 dollars and has appreciated 5% yearly is given as y = (5) * (1 + .05)t, where t = number of years since it’s original purchase. Find the value of the card 25 years later.

a. Answer: ___$16.93__

b. What is the rate of increase? ___5%__________

3. A town of 1000, grows at a rate of 25% every year. Find the equation for the population growth of this city and find the size of the city in 10 years.

Equation: ___y = 1000(1+0.25)x__________ Answer: ___ y = 1000(1+0.25)10 = 9313.22

4. A city of 100,000 is having pollution problems and is decreasing in size 1% annually (every year). Find the equation for the population of this city and find the size of the city in 100 years.

Equation: ___ y = 10,000(1-0.01)x________ Answer: __ y = 10,000(1-0.01)100 = 3660.32

5. The price of a gallon of milk is given by the equation P(t) = .5(1.03)t, where t = the years since 1939.

a. What is the price of a gallon of milk in 1939? _$0.50______

b. What was the rate of inflation (growth)? ___3%_____

c. What is the predicted value of a gallon of milk in 2009? __.5(1.03)70 = $3.96____

6. In 1982, the number of Starbucks was 5 shops. It has exponentially grown by 21% yearly. Let t = the number of years since 1982. Find an equation for this growth and find the number of Starbucks predicted in 2015.

Equation: ___y = 5(1 + 0.21)x_________ Answer: __ y = 5(1.21)33 = 2697.04________

7. A computer’s value declines about 7% yearly. Sally bought a computer for $800 in 2005. How much is it worth in 2009.

Answer: ___ y = 800 (1 – 0.07)4 = 5988.16_____

8. The use of free weights for fitness has increased in popularity. In 1997, there were 43.2 million people who used free weights. Assume the use of free weights increases 6% annually.

a. Write an equation that describes the number of people using free weights in t years since 1997.

43.2(1 + 0.06)t

b. Predict the number of people using free weights in 2007.

43.2(1 + 0.06)10 = 77.36 million

9. The depreciation of the value for a car is modeled by the equation y = 100,000(.85)x for x years since 2000.

a. If the value of the car was $61,412.5 in 2003, then what is the value of the car in 2004?

61,412.5(.85) = $52200.63 or 100,000(.85)4 = $52200.63

b. In what year, will the value of the car reach ¼ of its original value.

¼ (100,000) = 25,000 = 100,000(.85)x; x = 8.5 years or between 2008 and 2009

SPECIAL CASE OF EXPONENTIAL GROWTH:

COMPOUND INTEREST:[pic]

A represents the amount of the investment

P is the principal or initial amount of the investment

r represents the annual rate of interest as a decimal

n represents the number of times the interest is compounded each year

t represents the number of years since the initial investment

EXAMPLE PROBLEMS:

1) If the money the Native Americans received for Manhattan ($24 in 1626) had been invested at 6% per year compounded semiannually (twice a year), how much money would there be in the year 2026?

[pic]nearly 447 billion dollars

2) Michelle plans to invest $1000 into a savings account which has a 12% annual interest rate compounded monthly (12 times a year), how much money would there be in 10 years?

[pic]

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