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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 - PowerPoint PPT Presentation
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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=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=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 isY=3·2x
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 a=1/2
y=1/2 · 8x
• 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.
(-2, ¼) (-1, ½) (0, 1) (1, 2)
(x, lny)(-2, -1.38) (-1, -.69) (0,0) (1, .69)
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
Now plot (x,lny)
Since the points lie close to a line, an exponential model should be a good fit.
• 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
• y = (e.233)(e.379t)
• y = 1.26 · 1.46t
You can use a graphing calculator that performs
exponential regression to do this also. It uses all the
original data.Input into L1 and L2
and push exponential regression
L1 & L2 hereThen edit & enter the data. 2nd quit toget 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
•y = axb
• Only 2 points are needed
• (2,5) & (6,9)
• 5 = a 2b
• 9 = a 6b
a = 5/2b
9 = (5/2b)6b
9 = 5·3b
1.8 = 3b
log31.8 = log33b
.535 ≈ ba = 3.45y = 3.45x.535
• 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.
Assignment