University:
High School
Course:
High School Mathematics
Academic year:

2022
Views:

129

Pages:

4
Author:

garrotar4i3

6.1 Exponential Growth & Decay Functions A. VOCABULARY & CONCEPTS 1. Exponential Function y = ab^x a = growth or decay factor b > 1 (growth) 0 < b < 1 (decay) (a) Growth graph showing exponential growth with an asymptote at y = 0 (b) Decay graph showing exponential decay with an asymptote at y = 0 2. Exponential Models (a) Growth y = a(1+r)^t (b) Decay y = a(1-r)^t a = initial amount r = increase/decrease (decimal) t = time B. EXAMPLES Evaluate the expression for x = -2 and x = 3 a) 5 * 2^x 5 * 2^-2 = 5 * 1/4 = 5/4 5 * 2^3 = 5 * 8 = 40 b) 4^x - 2 4^-2 - 2 = 1/16 - 2 = -31/16 4^3 - 2 = 64 - 2 = 62 Growth or decay? Then graph. a) y = 2^x b > 1 Exponential Growth x | y -2 | 2^-2 = 1/4 -1 | 2^-1 = 1/2 0 | 2^0 = 1 1 | 2^1 = 2 2 | 2^2 = 4 3. The value of a car (y, in thousands of dollars) can be approximated by the model y = 25(0.85)^t, where t is the number of years since the car was new. a) Tell whether the model represents exponential growth or exponential decay. y = 25(0.85)^t 0 < b < 1 Exponential decay b) Identify the annual percent of increase or decrease in the value of the car. y = 25(0.85)^t Exponential decay model y = a(1 - r)^t Rewrite 0.85 as 1 - r The percent of decrease is 15%. c) Estimate when the value of the car will be $8000. In 8 years: y = 25(0.85)^8 = 25(0.272) = 6.8 thousand In 7 years: y = 25(0.85)^7 = 25(0.321) = 8.01 thousand In approximately 7 years, the value of the car will be $8,000.