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Algebra 1 Warm Up 9 April 2012

Algebra 1 Warm Up 9 April 2012. State the recursive sequence (start?, how is it changing?), then find the next 3 terms. Also find the EQUATION for each. y = a∙b x 1) 12000, 10800,9720, ___, ___, ___ 2 ) 100, 105.25,110.77, ___, ___, ___ Rewrite as a fraction and decimal:

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Algebra 1 Warm Up 9 April 2012

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  1. Algebra 1 Warm Up 9 April 2012 State the recursive sequence (start?, how is it changing?), then find the next 3 terms. Also find the EQUATION for each. y = a∙bx 1) 12000, 10800,9720, ___, ___, ___ 2) 100, 105.25,110.77, ___, ___, ___ Rewrite as a fraction and decimal: 3) a) 5% b) 50% c) 5.25% Homework due Tuesday: pg. 345: 1 – 5 ADV: 12

  2. OBJECTIVE Today we will explore exponential growth and decay patterns and write exponential equations. Today we will take notes, work problems with our groups and present to the class.

  3. Once upon a time, two merchants were trying to work out a deal. For the next month, the 1st merchant was going to give $10,000 to the 2nd merchant, and in return, he would receive 1 cent the first day, 2 cents the second, 4 cents in the third, and so on, each time doubling the amount. After 1 month, who came out ahead? THINK- PAIR- SHARE

  4. Group: Money Doubling? • You have a $100.00 • Your money doubles each year. • How much do you have in 5 years? • Show work. Use a table and/or equation!

  5. 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

  6. Earning Interest • You have $100.00. • Each year you earn 10% interest. • How much $ do you have in 5 years? • Show Work. • HINT…how much is 10% of $100? HINT…..can you find a constant multiplier?

  7. 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 Can you find an equation? start at 100, CM = 110/100 = 1.1 Equation? y = 100(1.1)x y = 100(1.1)5=161.05

  8. Growth Models: Investing The equation for constant percent growth is y = A (1+ )x A = starting value (principal) r = rate of growth (÷100 to put in decimal form) x = number of time periods elapsed y = final value

  9. Using the Equation • $100.00 • 10% interest • 5 years • 100(1+ )5 = 100( 1 + 0.10)5 = 100 (1.1)5 = $161.05 10% as a fraction 10% as a decimal Constant multiplier

  10. Comparing Investmentswhich is better? • Choice 1 • $10,000 • 5.5% interest • 9 years • Choice 2 • $8,000 • 6.5% interest • 10 years

  11. Choice 1 $10,000, 5.5% interest for 9 years. Equation: y =$10,000 (1 + )9 =10,000 (1 + 0.055)9 = 10,000(1.055)9 Balance after 9 years: $16,190.94

  12. Choice 2 $8,000 in an account that pays 6.5% interest for 10 years. Equation: y=$8000 (1 + )10 =8,000 (1 + .065)10 =8,000(1 + 0.065)10  Balance after 10 years: $15,071.10

  13. Which Investment? • The first one yields more money. • Choice 1: $16,190.94 • Choice 2: $15,071.10

  14. Exponential Decay Instead of increasing, it is decreasing. Formula: y = A (1 – )x A = starting value r = rate of decrease (÷100 to put in decimal form) x= number of time periods elapsed y = final value

  15. 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

  16. Exponential Decay: Car Depreciation Assume the car was purchased for $20,000 Formula: y = a (1 – )t a = initial amount r = percent decrease t = Number of years

  17. debrief How does the exponential growth differ from linear growth? How does the difference show up in the table? How does the difference show up on the graph?

  18. Worksheet find then towards the end of page http://www.uen.org/Lessonplan/preview.cgi?LPid=24626 http://www.regentsprep.org/regents/math/algebra/AE7/ExpDecayL.htm

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