Algebra 1 Section 8 6 Exponential Growth The

Algebra 1 Section 8. 6

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.

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.

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.

Example 1 a f(t) = 500(1 – 0. 75)t Initial amount: 500 Growth rate (r): -0. 75 or -75% Growth factor (b): 0. 25 Type: decay Notice that r < 0.

Example 1 b y = 100(1 + 2. 5)x Initial amount: 100 Growth rate (r): 2. 5 or 250% Growth factor (b): 3. 5 Type: growth Notice that r > 0.

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

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

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

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

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

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

Example 4 Initial amount: A = 6000 Quarterly interest rate: 0. 04 r= = 0. 01 4 f(q) = 6000(1 + 0. 01)q q is the number of quarterly periods

Example 4 f(q) = 6000(1 + 0. 01)q 10 years = 40 quarterly periods f(40) = 6000(1 + 0. 01)40 ≈ $8933. 18

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

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

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

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

Homework: pp. 361 -364