1 / 15

Modeling with Exponential & Power Functions

Modeling with Exponential & Power Functions. Math 3 Keeper 32. 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

Download Presentation

Modeling with Exponential & Power Functions

An Image/Link below is provided (as is) to download presentation 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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Modeling with Exponential & Power Functions Math 3 Keeper 32

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

  3. 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:

  4. 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

  5. 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

  6. 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.

  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

  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 • elny = 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 · 1.46t

  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 L2and push exponential regression

  13. 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!

  14. 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

  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.

More Related