8.8 Exponential Growth and Decay Exponential Growth –Modeled with the function: y = a b x for a > 0 and b > 1. y = a b x a = the starting amount (when.

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8.8 Exponential Growth and Decay Exponential Growth –Modeled with the function: y = a b x for a > 0 and b > 1. y = a b x a = the starting amount (when x = 0) b = the base, which Is greater than 1, is the growth factor x = exponent

Modeling Exponential Growth Since 1985, the daily cost of patient care in community hospitals in the United States has increased about 8.1% per year. In 1985, such hospital costs were an average of $460 per day. a. Write an equation to model the cost of hospital care. Relate: y = a b x Define: Let x = the number of years since 1985 Let y = the cost of community hospital care at various times Let a = the initial cost in 1985, $460 Let b = the growth factor, which is 100% + 8.1% = 108.1% = Write: y = x

Modeling Exponential Growth b. Use your equation to find the approximate cost per day in y = x y = y ≈ 1480 The average cost per day in 2000 was about $1480.

Compound Interest Suppose your parents deposited $1500 in an account paying 6.5% interest compounded annually (once a year) when you were born. Find the account balance after 18 years. Relate: y = a b x Define: Let x = the number of interest periods Let y = the balance Let a = the initial deposit, $1500 Let b = 100% + 6.5% = 106.5% = Write: y = x = ≈ The balance after 18 years will be $

Annual Interest Rate of 8% CompoundedPeriods per Year Interest Rate per Period Annually18% every year Semi-annually24% every 6 months Quarterly42% every 3 months Monthly120.6% every month

Compound Interest Suppose the account in the other example paid interest compounded quarterly instead of annually. Find the account balance after 18 years. Relate: y = a b x Define: Let x = the number of interest periods Let y = the balance Let a = the initial deposit, $1500 Let b = 100% + 6.5% 4 = = Write: y = x = ≈ The balance after 18 years will be $