8.7 Modeling with Exponential

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8.7 Modeling with Exponential Power Functions

• p. 509

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Just like 2 points determine a line, 2 points
determine an exponential curve.
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Write an Exponential function, yabx whose graph
goes thru (1,6) (3,24)

• Substitute the coordinates into yabx to get 2
  equations.
• 1. 6ab1
• 2. 24ab3
• Then solve the system

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Write an Exponential function, yabx whose graph
goes thru (1,6) (3,24) (continued)

• 1. 6ab1 ? a6/b
• 2. 24(6/b) b3
• 246b2
• 4b2
• 2b

a 6/b 6/2 3
So the function is Y32x
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Write an Exponential function, yabx whose graph
goes thru (-1,.0625) (2,32)

• .0625ab-1
• 32ab2

• (.0625)a/b
• b(.0625)a

• 32b(.0625)b2
• 32.0625b3
• 512b3
• b8

y1/2 8x
a1/2
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• When you are given more than 2 points, you can
  decide whether an exponential model fits the
  points by plotting the natural logarithms of the
  y values against the x values. If the new points
  (x, lny) fit a linear pattern, then the original
  points (x,y) fit an exponential pattern.

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(-2, ¼) (-1, ½) (0, 1) (1, 2)
(x, lny) (-2, -1.38) (-1, -.69) (0,0) (1, .69)
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Finding a model.

• Cell phone subscribers 1988-1997
• t years since 1987

t 1 2 3 4 5 6 7 8 9 10
y 1.6 2.7 4.4 6.4 8.9 13.1 19.3 28.2 38.2 48.7
                     
lny 0.47 0.99 1.48 1.86 2.19 2.59 2.96 3.34 3.64 3.89
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Now plot (x,lny)
Since the points lie close to a line, an
exponential model should be a good fit.
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• Use 2 points to write the linear equation.
• (2, .99) (9, 3.64)
• m 3.64 - .99 2.65 .379
  9 2 7
• (y - .99) .379 (x 2)
• y - .99 .379x - .758
• y .379x .233 LINEAR MODEL FOR (t,lny)
• The y values were lns xs were t so
• lny .379t .233 now solve for y
• elny e.379t .233 exponentiate both
  sides
• y (e.379t)(e.233) properties of exponents
• y (e.233)(e.379t) Exponential model

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• y (e.233)(e.379t)
• y 1.26 1.46t

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You can use a graphing calculator that performs
exponential regression to do this also. It uses
all the original data.Input into L1 and L2and
push exponential regression
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L1 L2 here Then edit enter the data. 2nd
quit to get out.
Exp regression is 10
So the calculators exponential equation is y
1.3 1.46t which is close to what we found!
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Modeling with POWER functions
a 5/2b 9 (5/2b)6b 9 53b 1.8 3b log31.8
log33b .535 b a 3.45 y 3.45x.535

• y axb
• Only 2 points are needed
• (2,5) (6,9)
• 5 a 2b
• 9 a 6b

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• You can decide if a power model fits data points
  if
• (lnx,lny) fit a linear pattern
• Then (x,y) will fit a power pattern
• See Example 5, p. 512
• You can also use power regression on the
  calculator to write a model for data.

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Assignment