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Exponential Growth and Decay MGSE9-12.A.REI.10 & 11 MGSE9-12.F.LE.1-5 Where a = initial amount b = growth/decay factor x = time y = ending amount Growth When a > 0 and b > 1, the function models growth. (b is called the growth factor) (a represents the initial amount) Decay a > 0 and 0 < b < 1, When the function models decay. (b is called the decay factor) (a represents the initial amount) Let's look at examples of these exponential functions at work. 1. Population: The population of the popular town of Smithville in 2003 was estimated to be 35,000 people with an annual rate of increase (growth) of about 2.4%. a.) What is the growth factor for Smithville? After one year the population would be 35,000 + 0.024(35000). By factoring we see that this is 35,000(1 + 0.024) or 35,000(1.024). The growth factor is 1.024. (Remember that the growth factor is greater than 1.) b.) Write an equation to model future growth. where y is the population; x is the number of years since 2003 c.) Use your equation to estimate the population in 2007 to the nearest hundred people. Exponential Growth and Decay MGSE9-12.A.REI.10 & 11 MGSE9-12.F.LE.1-5 2 Money: . Marisa invests $300 at a bank that offers 5% compounded annually. a.) What is the growth factor for the investment? The growth factor is 1.05. (Remember that the growth factor is greater than 1.) b.) Write an equation to model the growth of the investment. where y is the money; x is the number of years since the initial investment c.) How many years will it take for the initial investment to double? Algebraic solution: Graphical solution: and solve for x. Answer: approximately 14.2 years Exponential Growth and Decay MGSE9-12.A.REI.10 & 11 MGSE9-12.F.LE.1-5 3. Depreciation: Matt bought a new car at a cost of $25,000. The car depreciates approximately 15% of its value each year. a.) What is the decay factor for the value of this car? The decay factor is 1 - 0.15 = 0.85. (Remember that the decay factor is 0 < b < 1.) b.) Write an equation to model the decay value of this car. where y is the value of the car; x is the number of years since new purchase c.) What will the car be worth in 10 years? Answer: $4,921.86 Exponential Growth and Decay MGSE9-12.A.REI.10 & 11 MGSE9-12.F.LE.1-5 Where y0 = initial amount k = constant of proportionality t = time y = ending amount Growth Decay models growth since e > 1. Exponential functions with the base e are often used to describe continuous growth or decay. models decay since between 0 and 1 which is 4 More Money: Most banks compound interest more than once a year. When . interest is compounded n times per year for t years at an interest rate of r, the principal, P, grows to the amount A given by the formula: If you allow n to get larger and larger, If compounding takes place you will discover that as n increases, without interruption (called compounded continuously), this formula becomes: approaches e. Compound Quarterly: Jose invests $500 at a bank offering 10% compounded quarterly. Find the amount of the investment at the end of 5 years (if untouched). Compound Continuously: Tamika invests $500 at a bank offering 10% compounded continuously. Find the amount of the investment at the end of 5 years (if untouched). Exponential Growth and Decay MGSE9-12.A.REI.10 & 11 MGSE9-12.F.LE.1-5 5. Half-Life: Radium-226, a common isotope of radium, has a half-life of 1620 years. Professor Korbel has a 120 gram sample of radium-226 in his laboratory. a.) Find the constant of proportionality for radium-226. It should make sense that k is a negative value, since this is an example of decay. b.) How many grams of the 120 gram sample will remain after 100 years? The "official" equation for radium-226 is now written using the value for k. Substitute the time and solve. Exponential Growth and Decay MGSE9-12.A.REI.10 & 11 MGSE9-12.F.LE.1-5 6. Bacteria Growth: A certain strain of bacteria that is growing on your kitchen counter doubles every 5 minutes. Assuming that you start with only one bacterium, how many bacteria could be present at the end of 96 minutes? Now, form your equation using this k value, and solve the problem using the time of 96 minutes.