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Do Now Three years ago you bought a Lebron James card for $45. It has appreciated (gone up in value) by 20% each year since then. How much is worth today?

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Applications that Apply to Me!

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Exponential Functions Always involves the equation: b x Example: 2 3 = 2 2 2 = 8

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Group investigation: Y = 2 x Create an x,y table. Use x values of -1, 0, 1, 2, 3, Graph the table What do you observe.

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The Table: Results XF(x) = 2 x 2 -1 = 02 0 = 1 12 1 = 2 22 2 = 4 32 3 = 8

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The Graph of y = 2 x

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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?

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Group: Money Doubling? You have a $100.00 Your money doubles each year. How much do you have in 3 years? Show work.

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Money Doubling Year 1: $100 2 = $200 Year 2: $200 2 = $400 Year 3: $400 2 = $800

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Earning Interest on You have $100.00. Each year you earn 10% interest. How much $ do you have in 3 years? Show Work.

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Earning 10% results Year 1: $100 + 100(.10) = $110 Year 2: $110 + 110(.10) = $121 Year 3: $121 + 121(.10) = $133.10

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Exponential Growth Model The Formula is: y = a(1+r) t a = Initial Amount r = Growth Rate (1+r) = Growth Factor t = Time Period

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Using the Equation $100.00 10% interest 5 years 100(1+ (.10)) 5 = $161.05 What could we figure out now?

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Exponential Decay Model Instead of increasing, it is decreasing. Formula: y = a(1r) t a = Initial Amount r = Decay Rate (1-r) = Decay Factor t = Time Period

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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

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Exponential Decay: Car Depreciation Depreciation Rate Value after 2 years Value after 5 years Value after 8 years Value after 10 years 10% $16,200$11,809.80$8609.34$6973.57 Assume the car was purchased for $20,000 Formula: y = a (1 r) t a = initial amount r = percent decrease t = number of years

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Homework Textbook: Page 481 #24-28 Page 489 #26-30