1. 8.7 Modeling with Exponential & Power Functions p. 509

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

3. 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 2. 24=ab 3 Then solve the system:

4. Write an Exponential 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

5. Write an Exponential 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

6. 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.

7. (-2, ¼) (-1, ½) (0, 1) (1, 2) (x, lny) (-2, -1.38) (-1, -.69) (0,0) (1,.69)

8. Finding a model. Cell phone subscribers 1988-1997 t= # years since 1987 t12345678910 y1.62.74.46.48.913.119.328.238.248.7 lny0.470.991.481.862.192.592.963.343.643.89

9. Now plot (x,lny) Since the points lie close to a line, an exponential model should be a good fit.

10. 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 e lny = e.379t +.233 exponentiate both sides y = (e.379t )(e.233 ) properties of exponents y = (e.233 )(e.379t) Exponential model

11. y = (e.233 )(e.379t ) y = 1.26 · 1.46 t

12. You can use a graphing calculator that performs exponential regression to do this also. It uses all the original data. Input into L1 and L2 and push exponential regression

13. 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!

14. Modeling 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 3 1.8 = log 3 3 b.535 ≈ b a = 3.45 y = 3.45x.535

15. 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.

16. Assignment