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8.7 Modeling with Exponential & Power FunctionsPowerPoint Presentation

8.7 Modeling with Exponential & Power Functions

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## PowerPoint Slideshow about ' 8.7 Modeling with Exponential & Power Functions' - eve-pearson

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Just like 2 points determine a line, 2 points determine an exponential curve.

Write an Exponential function, y=ab exponential curve.x 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 exponential curve.x 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 exponential curve.x 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

- 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.

(-2, ¼) (-1, ½) (0, 1) (1, 2) whether an exponential model fits the points by plotting the

(x, lny)

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

Finding a model. whether an exponential model fits the points by plotting the

- Cell phone subscribers 1988-1997
- t= # years since 1987

Now plot (x,lny) whether an exponential model fits the points by plotting the

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

be a good fit.

- Use 2 points to write the linear equation. whether an exponential model fits the points by plotting the
- (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 whether an exponential model fits the points by plotting the .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 L2and push exponential regression

L1 & L2 here regression to do this also. It uses all the original data.

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 regression to do this also. It uses all the original data.

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

- You can decide if a power model fits data points if: regression to do this also. It uses all the original data.
- (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 regression to do this also. It uses all the original data.

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