8.7 Modeling with Exponential & Power Functionsp. 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=ab12. 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/b2. 24=(6/b) b3

24=6b2 4=b2 2=ba= 6/b = 6/2 = 3So the function isY=32x

Write an Exponential function, y=abx whose graph goes thru (-1,.0625) & (2,32).0625=ab-132=ab2

(.0625)=a/bb(.0625)=a32=[b(.0625)]b232=.0625b3512=b3b=8a=1/2y=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, ) (-1, ) (0, 1) (1, 2)(x, lny)(-2, -1.38) (-1, -.69) (0,0) (1, .69)

Finding a model.Cell phone subscribers 1988-1997t= # 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 - .758y = .379x + .233 LINEAR MODEL FOR (t,lny)The y values were lns & xs were t so:lny = .379t + .233 now solve for yelny = e.379t + .233 exponentiate both sidesy = (e.379t)(e.233) properties of exponentsy = (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 10So the calculators exponential equation is

y = 1.3 1.46t

which is close to what we found!

Modeling with POWER functionsy = axbOnly 2 points are needed(2,5) & (6,9)5 = a 2b9 = a 6ba = 5/2b

9 = (5/2b)6b9 = 53b1.8 = 3blog31.8 = log33b.535 ba = 3.45y = 3.45x.535

You can decide if a power model fits data points if:(lnx,lny) fit a linear patternThen (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