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1 Sec 3.1 Exponential Functions and Their Graphs Exponential Function - the independent variable is in the exponent. Model situations with constant percentage change exponential growth exponential decay Domain: Range: asymptote: x-intercept: y-intercept: For each function, state the domain, range, asymptote and intercept. Then draw the graph. Answers are on next slide. Graphs follow.

2 For each function, state the domain, range, asymptote and intercept. Then draw the graph. D: all real numbers R: y < 0 asymptote: y = 0 intercept: (0, -1) D: all real numbers R: y > 1 asymptote: y = 1 intercept: (0, 2) D: all real numbers R: y > 0 asymptote: y = 0 intercept: (0, 1/8) D: all real numbers D: all real numbers R: y > 0 R: y > 0 asymptote: y = 0 asymptote: y = 0 intercept: (0, 1) intercept: (0, 1)

3 The population of Westfield is given by the function y = 65(1.03) x where y is the population, in thousands, x years after a.) Find the population in b.) Predict the population in c.) Find the population in 2010 d.) What meaning does 65 have in the situation? e.) What meaning does 1.03 have in the situation? the pop was growing by 3% growth rate is 3% growth factor is 1.03

4 y = ab x Exponential functions model situations involving constant percentage change. a is the initial quantity b is the growth(decay) factor b = 1 + r where r is the growth rate (r is negative for decay) If r = -.25 b = =.75 y-intercept is (0, a) Worksheet on application word problems. Suppose you invest $400 at 3% annual interest compounded annually. How much money will you have after 5 years, assuming you make no deposits or withdrawals? Show work. 400(1 +.03) 5 = What if the interest is compounded monthly? Then how much money would you have at the end of one year? after five years? A(5) = 400( ) 60 = $464.65

5 Can you write a formula for this type of problem using the variables below? P = principal (amount of money invested) r = annual interest rate as a decimal n = the number of times interest is compounded in a year t = the number of years the money was invested. The compound interest formula: A(t) = P(1 + ) nt total # of times interest is compounded % interest earned each time interest is compounded (only a fraction of the annual rate) P = principal (amount of money invested) r = annual interest rate as a decimal n = the number of times interest is compounded in a year t = the number of years the money was invested.

6 1.) Suppose you invest $5000 at 4.5% interest. How much money will you have after 6 years, assuming no deposits or withdrawals during this time if interest is compounded a.) semi-annually? b.) daily 2.) Suppose you invest $15,000 at 6% interest compounded quarterly. How much money would you have after 10 years, assuming no deposits or withdrawals? Suppose you want to invest $1500 at an annual interest rate of 5%. How much money will you have after 10 years if interest is compounded: annually monthly daily The more often you compound, the faster the money grows.

7 Suppose you invest $3000 AT 5% APR compounded monthly. Calculate the value of the investment after 10 years 20 years 30 years 40 years $ $ $13, $22, The annual percentage yield (APY) is the percentage that your money effectively earns in one year due to compounding. Find the annual percentage yield on an account that pays 3% APR with monthly compounding. Invest $1 for 1 year and see what it grows to. increase Find the APY if interest is compounded continuously.

8 (APY) Find the annual percentage yield on an account that pays 8% interest compounded daily. Find the annual percentage yield (APY) on an account which pays 4.3% annual interest compounded continuously. APY = 4.39% Find the value of $1 invested at 100% interest for 1 year compounded: annually semi-annually quarterly daily hourly (8760) every second (31,536,000) The value does not increase without bound. It approaches a limit. That limit is the number we call e.

9 e is an irrational number y = e x is called the natural exponential function It has many applications in science, population growth radioactive decay, and continuous compounding of interest. Derivation of the continuous change formula A(t) = P(1 + ) nt

10 Continuous Change Formula A(t) = Pe rt r = interest rate as a decimal (do not add 1) P = principal t = number of years of the investment Invest $3000 at 5% interest compounded continuously. How much money will you have after 10 years? A(10) = 3000e.05(10) = $ You can use either y = ab x or y = Pe rt (contiuous change formula) to represent situations involving constant percentage change. 1.) The population of a small city was 50,400 in 1995 and growing at a rate of 3% each year. a.) Use y = ab x to write a function for the population x years after Then find the population in f(x) = 50,400(1.03) x f(7) = 50,400(1.03) 7 = $61,986 b.) Use the continuous change formula to write a function for the population x years after Then find the population in f(x) = 50,400e.03x f(7) = 50,400e.03(7) = $62,177 Note that the continuous change formula gives a slightly larger result. We can convert our continuous change formula to the form y = ab x y = 50,400e.03x e.03 = so, y = 50,400(1.0305) x y = ab x form Notice that the multiplier is larger than 1.03 in part (a). Therefore, the result is a little bigger.

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