8.7 M ODELING WITH E XPONENTIAL & P OWER F UNCTIONS p. 509

J UST LIKE 2 POINTS DETERMINE A LINE, 2 POINTS DETERMINE AN EXPONENTIAL CURVE.

W RITE AN E XPONENTIAL 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 =ab 3 Then solve the system:

W RITE AN E XPONENTIAL FUNCTION, Y = AB X WHOSE GRAPH GOES THRU (1,6) & (3,24) ( CONTINUED ) 1. 6=ab 1 → a=6/b 2. 24=(6/b) b 3 24=6b 2 4=b 2 2=b a= 6/b = 6/2 = 3 So the function is Y=3·2 x

W RITE AN E XPONENTIAL FUNCTION, Y = AB X WHOSE GRAPH GOES THRU (-1,.0625) & (2,32).0625=ab -1 32=ab 2 (.0625)=a/b b(.0625)=a 32=[b(.0625)]b 2 32=.0625b 3 512=b 3 b=8 a=1/2 y=1/2 · 8 x

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)

F INDING A MODEL. Cell phone subscribers t= # years since 1987 t y lny

N OW 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= = 2.65 = – 2 7 (y -.99) =.379 (x – 2) y -.99 =.379x y =.379x LINEAR MODEL FOR (t,lny) The y values were ln’s & x’s were t so: lny =.379t now solve for y e lny = e.379t 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 · 1.46 t

Y OU CAN USE A GRAPHING CALCULATOR THAT PERFORMS EXPONENTIAL REGRESSION TO DO THIS ALSO. I T USES ALL THE ORIGINAL DATA. I NPUT INTO L1 AND L2 AND PUSH EXPONENTIAL REGRESSION

L1 & L2 here Then edit & enter the data. 2 nd quit to get out. Exp regression is 10 So the calculators exponential equation is y = 1.3 · 1.46 t which is close to what we found!

M ODELING WITH POWER FUNCTIONS y = ax b Only 2 points are needed (2,5) & (6,9) 5 = a 2 b 9 = a 6 b a = 5/2 b 9 = (5/2 b )6 b 9 = 5·3 b 1.8 = 3 b log = log 3 3 b.535 ≈ b a = 3.45 y = 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.

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