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8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
8.2 Exploring exponential models
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8.2 Exploring exponential models

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  • 1. 8.2 Writing Exponential Models
  • 2. What is an exponential equation? An exponential equation has the general form y=abx where a ≠ 0, b〉 0 and b ≠ 1
  • 3. Growth Factor, Decay Factor Given the general form y=abx  When b > 1, b is the growth factor  When 0 < b < 1, b is the decay factor
  • 4. Growth or Decay??? y = 10(1.2) y = 5(.9) x Growth x Decay y = 50(1.54) y = 5.2(.70) y = 4(2) x y = 100(.07) x x Growth Decay Growth x Decay
  • 5. Writing Exponential Equations  Find the exponential equation passing through the points (3,20) and (1,5). y = ab 20 = ab 20 =a 3 b x 3 Start with the general form. Choose a point. Substitute for x and y using (3, 20) Solve for a 20 1 5= 3 b b 1−3 5 = 20b Substitute x and y using (1, 5) and a using Division property of exponents 2 =a b3
  • 6. Writing Exponential Equations  Find the exponential equation passing through the points (3,20) and (1,5). 5 = 20b −2 Simplify 20 5= 2 b 20 2 b = =4 Solve for b 5 b=2 20 20 20 5 a= 3 = 3 = = b 2 8 2 Go back to the equation for a; substitute in b and solve for a
  • 7. Writing Exponential Equations  Find the exponential equation passing through the points (3,20) and (1,5). y = ab x 5 x y = (2) 2 Going back to the general form, substitute in a and b The exponential equation passing through 5 the points (3,20) and (1,5) is y = ( 2) x 2
  • 8. Let’s Try One  Find the exponential equation passing through the points (2,4) and (3,16). y = ab x 4 = ab 4 =a 2 b 2 4 3 16 = 2 b b 16 = 4b 3− 2 Start with the general form. Choose a point. Substitute for x and y using (2, 4) Solve for a Substitute x and y using (3, 16) and a using Division property of exponents 4 =a b2
  • 9. Writing Exponential Equations 16 = 4b 1 Simplify b=4 y = ab Solve for b x 4 1 a = 2 = = 0.25 4 4 y = 0.25(4) x Go back to the equation for a; substitute in b and solve for a Going back to the general form, substitute in a and b The exponential equation passing through x the points (2,4) and (3,16) is y = 0.25(4)
  • 10. Putting it all together . . .  Find the equation of the exponential function that goes through (1,6) and (0,2). Graph the function.
  • 11. Modeling Growth with an Exponential Equation  The growth factor can be found in word problems using b=1+r where r = rate or amount of increase. You can substitute your new b into your general equation to find the exponential function.
  • 12.  EX- a guy puts $1000 into a simple 3% interest account. What is the exponential equation? r = rate 3% (write as 0.03) b = 1 + r = 1.03 x = time a = amount put into the account ($1,000) y = ab x y = 1000 (1.03) x
  • 13.  EX – a colony of 1000 bacteria cells doubles every hour. What is the exponential equation? r = 1 (why not 2?) b=r+1=2 x = time (in hours) a = the original number in the colony (1,000 bacteria ) y = ab x y = 1000 (2) x b = r + 1, where r is the amount of increase. We are increasing by 100% each time something doubles, so r = 1
  • 14.  EX- a $15000 car depreciates at 10% a year. What is the exponential equation? r = - 10% (the car is worth 10% less each year) b = 1 - r = 1 – 0.1 = 0.9 x = time (in years) a = amount put into the account ($15,000) y = ab x y = 15000 (0.9) x
  • 15. Compound Interest  The formula for compound interest: nt  r A(t ) = P 1 + ÷  n Where n is the number of times per year interest is being compounded and r is the annual rate.
  • 16. Compound Interest - Example  Which plan yields the most interest? Plan A: A $1.00 investment with a 7.5% annual rate compounded monthly for 4 years  Plan B: A $1.00 investment with a 7.2% annual rate compounded daily for 4 years  12(4) A:11 + 0.075  ≈1.3486  ÷ 12   $1.35 365(4)  B:  0.072  11 + ≈1.3337 ÷ 365    $1.34

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