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 L2AND 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