8 7 modeling with exponential power functions
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8.7 Modeling with Exponential & Power Functions

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8.7 Modeling with Exponential & Power Functions. p. 509. Just like 2 points determine a line, 2 points determine an exponential curve. Write an Exponential function, y=ab x whose graph goes thru (1,6) & (3,24). Substitute the coordinates into y=ab x to get 2 equations. 1. 6=ab 1

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write an exponential function y ab x whose graph goes thru 1 6 3 24
Write an Exponential function, y=abx whose graph goes thru (1,6) & (3,24)
  • Substitute the coordinates into y=abx to get 2 equations.
  • 1. 6=ab1
  • 2. 24=ab3
  • Then solve the system:
write an exponential function y ab x whose graph goes thru 1 6 3 24 continued
Write an Exponential function, y=abx whose graph goes thru (1,6) & (3,24) (continued)
  • 1. 6=ab1→ a=6/b
  • 2. 24=(6/b) b3
  • 24=6b2
  • 4=b2
  • 2=b

a= 6/b = 6/2 = 3

So the function is

Y=3·2x

write an exponential function y ab x whose graph goes thru 1 0625 2 32
Write an Exponential function, y=abx whose graph goes thru (-1,.0625) & (2,32)
  • .0625=ab-1
  • 32=ab2
  • (.0625)=a/b
  • b(.0625)=a
  • 32=[b(.0625)]b2
  • 32=.0625b3
  • 512=b3
  • b=8

y=1/2 · 8x

a=1/2

slide6
When you are given more than 2 points, you can decide whether an exponential model fits the points by plotting the natural logarithmsof 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.
slide7
(-2, ¼) (-1, ½) (0, 1) (1, 2)

(x, lny)

(-2, -1.38) (-1, -.69) (0,0) (1, .69)

finding a model
Finding a model.
  • Cell phone subscribers 1988-1997
  • t= # years since 1987
now plot x lny
Now plot (x,lny)

Since the points lie close to a line, an exponential model should

be a good fit.

slide10
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 ln’s & x’s 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
slide11
y = (e.233)(e.379t)
  • y = 1.26 · 1.46t
slide12
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
slide13
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!

modeling with power functions
Modeling with POWER functions

a = 5/2b

9 = (5/2b)6b

9 = 5·3b

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
slide15
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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