advertisement

Applications that Apply to Me! Exponential Function What do we know about exponents? What do we know about functions? Exponential Functions Always involves the equation: bx Example: 23 = 2 · 2 · 2 = 8 Group investigation: x Y = 2 Create an x,y table. Use x values of -1, 0, 1, 2, 3, Graph the table What do you observe. The Table: Results X -1 0 1 2 3 F(x) = 2x 2-1 = ½ 20 = 1 21 = 2 22 = 4 23 = 8 The Graph of y = x 2 Observations What did you notice? What is the pattern? What would happen if x= -2 What would happen if x = 5 What real-life applications are there? Group: Money Doubling? You have a $100.00 Your money doubles each year. How much do you have in 5 years? Show work. Money Doubling Year 1: $100 · 2 = $200 Year 2: $200 · 2 = $400 Year 3: $400 · 2 = $800 Year 4: $800 · 2 = $1600 Year 5: $1600 · 2 = $3200 Earning Interest on You have $100.00. Each year you earn 10% interest. How much $ do you have in 5 years? Show Work. Earning 10% results Year 1: $100 + 100·(.10) = $110 Year 2: $110 + 110·(.10) = $121 Year 3: $121 + 121·(.10) = $133.10 Year 4: $133.10 + 133.10·(.10) = $146.41 Year 5: $146.41 + 1461.41·(.10) = $161.05 Growth Models: Investing The Equation is: A = P (1+ t r) P = Principal r = Annual Rate t = Number of years Using the Equation $100.00 10% interest 5 years 100(1+ 100·(.10))5 = $161.05 What could we figure out now? Comparing Investments Choice 1 $10,000 5.5% interest 9 years Choice 2 $8,000 6.5% interest 10 years Choice 1 $10,000, 5.5% interest for 9 years. Equation: $10,000 (1 + .055)9 Balance after 9 years: $16,190.94 Choice 2 $8,000 in an account that pays 6.5% interest for 10 years. Equation: $8,000 (1 + .065)10 Balance after 10 years: $15,071.10 Which Investment? The first one yields more money. Choice 1: $16,190.94 Choice 2: $15,071.10 Exponential Decay Instead of increasing, it is decreasing. Formula: y = a (1 – r)t a = initial amount r = percent decrease t = Number of years Real-life Examples What is car depreciation? Car Value = $20,000 Depreciates 10% a year Figure out the following values: After 2 years After 5 years After 8 years After 10 years Exponential Decay: Car Depreciation Assume the car was purchased for $20,000 Depreciation Rate 10% Value after 2 years $16,200 Value after 5 years $11,809.80 Value after 8 years $8609.34 Formula: y = a (1 – r)t a = initial amount r = percent decrease t = Number of years Value after 10 years $6973.57 What Else? What happens when the depreciation rate changes. What happens to the values after 20 or 30 years out – does it make sense? What are the pros and cons of buying new or used cars. Assignment 2 Worksheets: Exponential Growth: Investing Worksheet (available at ttp://www.uen.org/Lessonplan/preview.cgi?LPid=24626) Exponential Decay: Car Depreciation