Objective: to use exponential functions to model growth applications Do-Now: Grab the exponential function half sheet on the back table Get out textbook assignment

Exponential Growth Section 8-8 b >1

Rounding Rules….

Example #1 In 1996 North Carolina’s population was about 13 million people. Since then, the state’s population has grown about 1.7% each year. Since the population is growing at a constant rate in relation to each previous year, it is growing exponentially. Using this information we can predict what the population of North Carolina in 2015.

Just a reminder… Just a reminder… y = ab x is the generic formy = ab x is the generic form a is the “starting point”a is the “starting point” b is the rate at which we are changing – growth rateb is the rate at which we are changing – growth rate x is the number of times we change.x is the number of times we change. let y = final population let y = final population let x = number of years = 19 (2015 – 1996) let x = number of years = 19 (2015 – 1996) 100% + 1.7% growth rate …. So the base is y = 13 · x y = 13 · The population for 2015 is approximately 17.9 million. VOCAB: “growth factor” – just the base!

Example #2…You try Since 1985, the daily cost of patient care in community hospitals in the U.S. has increased about 8.1% per year. In 1985 the average cost of a hospital visit was about $460 per day. Set up and solve an equation to model the average cost of a hospital visit in let y = average cost of hospital visit let y = average cost of hospital visit let x = number of years let x = number of years y = 460 x x y = 460 x x y = 460 x y = 460 x y ≈ $ y ≈ $

Compound Interest When a bank pays interest on both the principal and interest an account has already earned. When a bank pays interest on both the principal and interest an account has already earned. An interest period is the length of time (in years) over which interest is calculated. Number of times is dependent on HOW interest is compounded.

Compounding Frequency Number of Compounding Periods Yearly1 Semi-Annually2 Quarterly4 Monthly12

Example 3 – Compound Interest Suppose your parents deposited $1,500 in an account paying 6.5% interest compounded annually the year you were born. Find the account after 18 years.

Example 3 – Compound Interest

Example 4 – Compound Interest Suppose your parents deposited $1,500 in an account paying 6.5% interest compounded quarterly the year you were born. Find the account after 18 years.

Example 4 – Compound Interest

Reflect Reflect  Did you earn more interest compounding annually or quarterly?  How do you think compounding monthly would compare?  Compute it! Example #5 Suppose your parents deposited $1500 in an account paying 6.5% interest compounded monthly the year you were born. Find the account balance after 18 years.

Example 6 … You try! You deposit $200 into an account earning 5% monthly. How much will be in the account after 1 year? After 2 years? After 5 years? 1 year:y = 200(1+.05 ÷ 12) 12 y ≈ $ years: y = 200(1+.05 ÷ 12) 24 y ≈ $ y ≈ $ years: y = 200(1+.05 ÷ 12) 60 y ≈ $ y ≈ $256.67

Homework pp # 1-4, 16-17, odd, 36-43, 47

Exponential Decay 8-8

Objective: to use exponential functions to model growth applications Do-Now: Grab the Warm-Up on the back table. Please work on the FRONT-SIDE ONLY! Get out textbook assignment

Example #1 Since 1980, the number of gallons of milk each person in the U.S. drinks each year has decreased 4.1% each year. In 1980, each person drank an average of 16.5 gallons of milk per year. Set up an equation to model the number of gallons of milk each person drinks in a year. Determine the amount of milk each person will drink in 2000.

Just a reminder… Just a reminder… y = ab x is the generic formy = ab x is the generic form a is the “starting point”a is the “starting point” b is the rate at which we are changing – decay factorb is the rate at which we are changing – decay factor x is the number of times we changex is the number of times we change let y = the amount of milk drank in gallons let y = the amount of milk drank in gallons let x = number of years since 1980 = 20 let x = number of years since 1980 = % - 4.1% decay rate …. So the base is y = 16.5 · x y = 16.5 · y ≈ 7.14 The average person drank 7.14 gallons of milk in VOCAB: “decay factor” – just the base!

Example # 2 … You try! In 1990, the population of Washington, D.C., was about 604,000 people. Since then the population had decreased about 1.8% each year. Set up an equation and find out about how many people will live in D.C. in let y = population let x = number of years since 1990 = 20 y = 604,000(.982) x y = 604,000(.982) 20 ≈ 420,017 people

Example # 3 - Depreciation You buy a car for $35,000. Its value depreciates at a rate of 7.5% each year. What will the value of your car be in five years? let y = car value let x = number of years y = 35,000(.925) x y = 35,000(.925) x y = 35,000(.925) 5 y ≈ $23,701.55

Half-Life The half-life of a radioactive substance is the length of time it takes for ½ of the substance to decay. Ex: The half-life of iodine-131 is 8 days. How many half life periods occur in … 8 days? 16 days? 80 days?

Example # 4 – Half-Life The half-life of iodine-131 is 8 days. If a patient receives a 12-mCi (millicuries) treatment. How much iodine is left in the patient after 16 days? Let y = amount of iodine Let x = number of half-life periods

Example # 4 – Half-Life

Example #5 … You try! The half-life of Zn-71, an isotope of Zinc, is 2.4 minutes. How many grams would remain from a 100,000 gram sample after one hour and 12 minutes have elapsed? Let y = amount of Zn-71 Let x = number of half-life periods

Example #5 … You try! The half-life of Zn-71, an isotope of Zinc, is 2.4 minutes. How many grams would remain from a 100,000 gram sample after one hour and 12 minutes have elapsed?

Example 6 … You try ! Some people are frightened of certain medical tests because the tests involve the injection of radioactive materials. A hepatobiliary scan of a patient’s gallbladder involves an injection of 0.5 cc's (or about one-tenth of a teaspoon) of Technetium-99m, which has a half-life of almost exactly 6 hours. While undergoing the test, the technician said that "in twenty-four hours, you'll be down to background radiation levels." Figure out just how much radioactive material remained in the gallbladder after twenty-four hours.

Lesson Review y = ab x is the generic form of all exponential functions (growth and decay) where a is our starting point, b is the rate at which we are changing, and x is the number of times changed. y = ab x is the generic form of all exponential functions (growth and decay) where a is our starting point, b is the rate at which we are changing, and x is the number of times changed. If our problem is one of exponential growth, then b > 1. If our problem is one of exponential decay, then 0<b<1

Homework pp # 20-30, 32, 46, 56 pp # 20-30, 32, 46, 56