# 8.7 Modeling with Exponential & Power Functions - PowerPoint PPT Presentation

1 / 16

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

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

8.7 Modeling with Exponential & Power Functions

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

## 8.7 Modeling with Exponential & Power Functions

p. 509

### 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 is

Y=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

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)

(x, lny)

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

### Finding a model.

• Cell phone subscribers 1988-1997

• t= # years since 1987

### 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 L2and push exponential regression

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

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:

• (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.