1. Algebra 1 Section 8.6

2. Exponential Growth The value of a function has the same percent increase during each unit of time. In the fish example, growth is 50% per year; the next year’s population is 150% of the current population.

3. Exponential Decay The value of a function has the same percent decrease during each unit of time. In the fish example, population decreases 25% per year; the next year’s population is 75% of the current population.

4. Definition Given an initial amount, A, and the growth rate, r, the exponential function f(t) = A(1 + r) t can be used to model both exponential growth (r > 0) and exponential decay (-1 < r < 0). The quantity (1 + r) is called the growth factor.

5. Example 1a Initial amount: Growth rate (r): Growth factor (b): Type: 500 -0.75 or -75% 0.25 decay f(t) = 500(1 – 0.75) t Notice that r < 0.

6. Example 1b Initial amount: Growth rate (r): Growth factor (b): Type: 100 2.5 or 250% 3.5 growth y = 100(1 + 2.5) x Notice that r > 0.

7. Example 1c Initial amount: Growth rate (r): Growth factor (b): Type: 1 1 or 100% 2 growth f(d) = 2 d Rewrite as f(d) = 1(1 + 1) d

8. Example 1d Initial amount: Growth rate (r): Growth factor (b): Type: 10,000 -½ or -50% ½ decay y = 10,000(½) x Rewrite as y = 10,000(1 – ½) x

9. Example 2 a) f(n) = 64(1 – ) n 1 2 b) f(3) = 64(1 – ) 3 1 2 = 64( ) 3 1 2 = 64( ) 1 8 = 8 teams

10. Example 3 f(d) = 5(1 + 2) d or f(d) = 5(3) d f(7) = 5(1 + 2) 7 f(7) = 5(3) 7 f(7) = 10,935 g

11. Compound Interest A = P(1 + r) x, where A = account balance P = principal amount x = number of periods r = periodic interest rate

12. Compound Interest A = P(1 + r) x If there are n compounding periods in each of t years, then r = x = nt APR n

13. Example 4 Initial amount: A = 6000 Quarterly interest rate: 0.04 4 r = = 0.01 f(q) = 6000(1 + 0.01) q q is the number of quarterly periods

14. Example 4 f(q) = 6000(1 + 0.01) q 10 years = 40 quarterly periods f(40) = 6000(1 + 0.01) 40 ≈ $8933.18

15. Depreciation The value of many items decreases by the same percent each year. This depreciation can be modeled by an exponential decay function.

16. Example 5 Initial amount: A = 30,000 Growth rate: r = -0.12 f(t) = 30,000(1 – 0.12) t = 30,000(0.88) t

17. Example 5 f(t) = 30,000(0.88) t f(5) = 30,000(0.88) 5 ≈ $15,800 f(10) = 30,000(0.88) 10 ≈ $8400 f(15) = 30,000(0.88) 15 ≈ $4400

18. Example 5 30,000 – 15,800 = $14,200 15,800 – 8400 = $7400 8400 – 4400 = $4000

19. Homework: pp. 361-364