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8.2 Exponential Decay P. 474
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Exponential Decay Has the same form as growth functions f(x) = ab x Where a > 0 BUT: 0 < b < 1 (a fraction between 0 & 1)
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Recognizing growth and decay functions State whether f(x) is an exponential growth or decay function f(x) = 5(2/3) x b=2/3, 0<b<1 it is a decay function. f(x) = 8(3/2) x b= 3/2, b>1 it is a growth function. f(x) = 10(3) -x rewrite as f(x)=10(1/3) x so it is decay
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Recall from 8.1: The graph of y= ab x Passes thru the point (0,a) (the y intercept is a) The x-axis is the asymptote of the graph a tells you up or down D is all reals (the Domain) R is y>0 if a>0 and y<0 if a<0 (the Range)
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Graph: y = 3(1/4) x Plot (0,3) and (1,3/4) Draw & label asymptote Connect the dots using the asymptote Domain = all reals Range = reals>0 y=0
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Graph y = -5(2/3) x Plot (0,-5) and (1,-10/3) Draw & label asymptote Connect the dots using the asymptote y=0 Domain : all reals Range : y < 0
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Now remember: To graph a general Exponential Function: y = a b x-h + k Sketch y = a b x h= ??? k= ??? Move your 2 points h units left or right …and k units up or down Then sketch the graph with the 2 new points.
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Example graph y=-3(1/2) x+2 +1 Lightly sketch y=-3·(1/2) x Passes thru (0,-3) & (1,-3/2) h=-2, k=1 Move your 2 points to the left 2 and up 1 AND your asymptote k units (1 unit up in this case)
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y=1 Domain : all reals Range : y<1
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Using Exponential Decay Models When a real life quantity decreases by fixed percent each year (or other time period), the amount y of the quantity after t years can be modeled by: y = a(1-r) t arWhere a is the initial amount and r is the percent decrease expressed as a decimal. The quantity 1-r is called the decay factor
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Ex: Buying a car! You buy a new car for $24,000. The value y of this car decreases by 16% each year. Write an exponential decay model for the value of the car. Use the model to estimate the value after 2 years. Graph the model. Use the graph to estimate when the car will have a value of $12,000.
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Let t be the number of years since you bought the car. The model is: y = a(1-r) t = 24,000(1-.16) t = 24,000(.84) t Note:.84 is the decay factor When t = 2 the value is y=24,000(.84) 2 ≈ $16,934
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Now Graph The car will have a value of $12,000 in 4 years!!!
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Assignment
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