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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

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Modeling with Exponential & Power Functions

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  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.

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