9.6 Notes Part I Exponential Growth and Decay...exponential decay function to model this situation....
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9.6 Notes Part I Exponential Growth
and Decay

I. Exponential Growth
(1 )ty C r
Final Amount
Initial Amount
Rate of Change
Time

Ex 1: The original value of a painting is $9000 and the value increases by 7% each year. Write an exponential growth function to model this situation. Then find the painting’s value in 15 years.
Step 1 Write the exponential growth function for this situation.
y = a(1 + r)t
= 9000(1 + 0.07)t
= 9000(1.07)t
Substitute 9000 for a and 0.07 for r.
Simplify.
Write the formula.
Step 2 Find the value in 15 years.
y = 9000(1.07)t
= 9000(1.07)15
≈ 24,831.28
The value of the painting in 15 years is $24,831.28.
Substitute 15 for t.
Use a calculator and round to the
nearest hundredth.

Ex 2: An investment is increasing in value at a rate of 8% per year, and its value in 2000 was $1200. Write an exponential growth function to model this situation. Then find the investment’s value in 2006.
Step 1 Write the exponential growth function for this situation.
y = a(1 + r)t
= 1200(1.08)t
Write the formula.
Substitute 1200 for a and 0.08 for r.
Simplify.
= 1200(1 + 0.08)t
Step 2 Find the value in 6 years.
y = 1200(1.08)t
= 1200(1.08)6
≈ 1,904.25
The value of the painting in 6 years is $1904.25.
Substitute 6 for t.
Use a calculator and round to the
nearest hundredth.

II. Exponential Decay
(1 )ty C r
Final Amount
Initial Amount
Rate of Change
Time

Ex 3: The population of a town is decreasing at a rate of 3% per year. In 2000 there were 1700 people. Write an exponential decay function to model this situation. Then find the population in 2012.
Step 1 Write the exponential decay function for this situation.
y = a(1 – r)t
= 1700(1 – 0.03)t
= 1700(0.97)t
Write the formula.
Substitute 1700 for a and 0.03 for r.
Simplify.
Step 2 Find the population in 2012.
≈ 1180
Substitute 12 for t. y = 1700(0.97)12
Use a calculator and round to the
nearest whole number.
The population in 2012 will be approximately 1180 people.

Ex 4: The fish population in a local stream is decreasing at a rate of 3% per year. The original population was 48,000. Write an exponential decay function to model this situation. Then find the population after 7 years.
Step 1 Write the exponential decay function for this situation.
y = a(1 – r)t
= 48,000(1 – 0.03)t
= 48,000(0.97)t
Write the formula.
Substitute 48,000 for a and 0.03 for
r.
Simplify.
Step 2 Find the population in 7 years.
≈ 38,783
Substitute 7 for t. y = 48,000(0.97)7
Use a calculator and round to the
nearest whole number.
The population after 7 years will be approximately 38,783 people.

A common application of exponential decay is halflife. The halflife of a substance is the time it takes for onehalf of the substance to decay into another substance.
III. HalfLife

Ex 1: Astatine218 has a halflife of 2 seconds.
Find the amount left from a 500 gram sample of astatine218 after 10 seconds.
Step 1 Find t, the number of halflives in the given time period.
Divide the time period by the halflife. The
value of t is 5.
Write the formula.
= 500(0.5)5 Substitute 500 for P and 5 for t.
= 15.625 Use a calculator.
There are 15.625 grams of Astatine218 remaining after 10 seconds.
Step 2 A = P(0.5)t

Ex 2: Astatine218 has a halflife of 2 seconds.
Find the amount left from a 500gram sample of Astatine218 after 1 minute.
Step 1 Find t, the number of halflives in the given time period.
Divide the time period by the halflife. The
value of t is 30.
1(60) = 60 Find the number of seconds in 1 minute.
Step 2 A = P(0.5)t Write the formula.
= 500(0.5)30 Substitute 500 for P and 30 for t.
= 0.00000047 g Use a calculator.
There are 0.00000047 grams of Astatine218 remaining after 60 seconds.

Ex 3: Cesium137 has a halflife of 30 years. Find the amount of Cesium137 left from a 100 milligram sample after 180 years.
Step 1 Find t, the number of halflives in the given time period.
Divide the time period by the halflife. The
value of t is 6.
Step 2 A = P(0.5)t Write the formula.
= 100(0.5)6
Use a calculator.
There are 1.5625 milligrams of Cesium137 remaining after 180 years.
Substitute 100 for P and 6 for
t.
= 1.5625 mg

Ex 4: Bismuth210 has a halflife of 5 days.
Find the amount of Bismuth210 left from a 100gram sample after 5 weeks. (Hint: Change 5 weeks to days.)
Step 1 Find t, the number of halflives in the given time period.
Divide the time period by the halflife. The
value of t is 5.
5 weeks = 35 days Find the number of days in 5 weeks.
Step 2 A = P(0.5)t Write the formula.
= 100(0.5)7
= 0.78125 g Use a calculator.
There are 0.78125 grams of Bismuth210 remaining after 5 weeks.
Substitute 100 for P and 7 for t.

Lesson Quiz: Part I
1. The number of employees at a certain company is 1440 and is increasing at a rate of 1.5% per year. Write an exponential growth function to model this situation. Then find the number of employees in the company after 9 years.
y = 1440(1.015)t; 1646
Write a compound interest function to model each situation. Then find the balance after the given number of years.
2. $12,000 invested at a rate of 6% compounded quarterly; 15 years
A = 12,000(1.015)4t, $29,318.64

Lesson Quiz: Part II
4. The deer population of a game preserve is decreasing by 2% per year. The original population was 1850. Write an exponential decay function to model the situation. Then find the population after 4 years.
y = 1850(0.98)t; 1706
5. Iodine131 has a halflife of about 8 days. Find the amount left from a 30gram sample of Iodine131 after 40 days.
0.9375 g