§7.7 Exponential Growth and Decay

1. §7.7 Exponential Growth and Decay

2. Warm-Up Use the formula I = prt to find the interest for principal p, interest rate r, and time t in years. 1. principal: $1000; interest rate: 5%; time: 2 years 2. principal: $360; interest rate: 6%; time: 3 years 3. principal: $2500; interest rate: 4.5%; time: 2 years 4. principal: $1680; interest rate: 5.25%; time: 4 years 5. principal: $1350; interest rate: 4.8%; time: 5 years

3. Solutions 1.l = prt = ($1000)(0.05)(2) = $50(2) = $100 2.l = prt = ($360)(0.06)(3) = $21.60(3) = $64.80 3.l = prt = ($2500)(0.045)(2) = $112.50(2) = $225 4.l = prt = ($1680)(0.0525)(4) = $88.20(4) = $352.80 5.l = prt = ($1350)(0.048)(5) = $64.80(5) = $324

4. Rule: Exponential Growth:is modeled with the function y = a•bx for a > 0 and b > 1. y = a• bx Exponential Growth: Exponential Growth:is modeled with the function y = a•bx for a > 0 and b > 1. Exponential Growth:is modeled with the function y = a•bx for a > 0 and b > 1. starting amount (when x = 0) y = a• bx Exponential Growth:is modeled with the function y = a•bx for a > 0 and b > 1. starting amount (when x = 0) y = a• bx The base, which is greater than 1, is the growth factor. Exponential Growth:is modeled with the function y = a•bx for a > 0 and b > 1. starting amount (when x = 0) y = a• bx exponent The base, which is greater than 1, is the growth factor.

5. Relate:  y = a • bx Use an exponential function. Let x = the number of years since 1998. Let y = the population of the town at various times. Let a = the initial population in 1998, 13,000 people. Let b = the growth factor, which is 100% + 1.4% = 101.4% = 1.014. Define: Example 1: Modeling Exponential Growth In 1998, a certain town had a population of about 13,000 people. Since 1998, the population has increased about 1.4% a year. a. Write an equation to model the population increase. Write: y = 13,000 • 1.014x

6. 2006 is 8 years after 1998, so substitute 8 for x. y = 13,000 • 1.0148 Use a calculator. Round to the nearest whole number. 14,529 Example 1: Modeling Exponential Growth (continued) b. Use your equation to find the approximate population in 2006. y = 13,000 • 1.014x The approximate population of the town in 2006 is 14,529 people.

7. Key Terms: Compound Interest:interest paid on both the principal and the interest that has already been paid. Interest Period: Compound Interest: Compound Interest:interest paid on both the principal and the interest that has already been paid. Compound Interest:interest paid on both the principal and the interest that has already been paid. Interest Period:the length of time over which interest is calculated.

8. Example 2a: Compound Interest Relate:  y = a • bx Use an exponential function. Define: Let x = the number of interest periods. Let y = the balance. Let a = the initial deposit, $1000 Let b = 100% + 7.2% = 107.2% = 1.072. Once a year for 5 years is 5 interest periods. Substitute 5 for x. = 1000 • 1.0725 Use a calculator. Round to the nearest cent. 1415.71 Suppose you deposit $1000 in a college fund that pays 7.2% interest compounded annually. Find the account balance after 5 years. Write: y = 1000 • 1.072x The balance after 5 years will be $1415.71.

9. Example 2b: Compound Interest Relate:  y = a • bx Use an exponential function. Define: Let x = the number of interest periods. Let y = the balance. Let a = the initial deposit, $1000 Let b = 100% + There are 4 interest periods in 1 year, so divide the interest into 4 parts. = 1 + 0.018 = 1.018 7.2% 4 Suppose the account in the Example 2 paid interest compounded quarterly instead of annually. Find the account balance after 5 years.

10. Example 2b: Compound Interest Four interest periods a year for 5 years is 20 interest periods. Substitute 20 for x. = 1000 • 1.01820 Use a calculator. Round to the nearest cent. 1428.75 (continued) Write: y = 1000 • 1.018x The balance after 5 years will be $1428.75.

11. Example ¥: Real-World Problem Solving Technetium-99 has a half-life of 6 hours. Suppose a lab has 80 mg of technetium-99. How much technetium-99 is left after 24 hours? In 24 hours there are four 6-hour half lives. After one half-life, there are 40 mg. After two half-lives, there are 20 mg. After three half-lives, there are 10 mg. After four half-lives, there are 5 mg.

12. Rule: Exponential Decay: Exponential Decay:the function y = a • bx models exponential decay for a > 0 and 0 < b < 1. Exponential Decay:the function y = a • bx models exponential decay for a > 0 and 0 < b < 1. y = a• bx Exponential Decay:the function y = a • bx models exponential decay for a > 0 and 0 < b < 1. starting amount (when x = 0) y = a• bx Exponential Decay:the function y = a • bx models exponential decay for a > 0 and 0 < b < 1. starting amount (when x = 0) y = a• bx The base, which is between 0 and 1, is the decay factor. Exponential Decay:the function y = a • bx models exponential decay for a > 0 and 0 < b < 1. starting amount (when x = 0) y = a• bx exponent The base, which is between 0 and 1, is the decay factor.

13. Relate:  y = a • bx Use an exponential function. Define: Let x = the number of years since 1999 Let y = the number of animals that remain Let a = 60, the initial population in 1999 Let b = the decay factor, which is 100% - 2.4 % = 97.6% = 0.976 Example 3: Modeling Exponential Decay Suppose the population of a certain endangered species has decreased 2.4% each year. Suppose there were 60 of these animals in a given area in 1999. a. Write an equation to model the number of animals in this species that remain alive in that area. Write: y = 60 • 0.976x

14. 2005 is 6 years after 1999, so substitute 6 for x. y = 60 • 0.9766 52 Use a calculator. Round to the nearest whole number. Example 3: Modeling Exponential Decay (continued) b. Use your equation to find the approximate number of animals remaining in 2005. y = 60 • 0.976x The approximate number of animals of this endangered species remaining in the area in 2005 is 52.

15. Assignment: Pg. 464 9-27 Left