7.7A Write Exponential Functions Algebra II

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

Ex. 1)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.) So the function is y=3·2x Write an Exponential function, y=abx whose graph goes thru (1,6) & (3,24) (continued) 1. 6=ab1 → a=6/b Solve for a, then substitute into other eq. 2. 24=(6/b) b3 24=6b2 4=b2 2=b 3.) a= 6/b = 6/2 = 3 4.) So the function is y=3·2x

Ex. 2 Note: Please ignore the textbook directions to draw a scatter plot. Find an exponential model for the data. (1,18), (2, 36), (3, 72), (4,144),(5, 288) (When you are given more than 2 points, you can decide what the exponential model is by choosing two points from the given information & following the same steps as we did in Example 1.) So, you try it.

Ex. 3) Write an Exponential function, y=abx whose graph goes thru (-1, b(.0625)=a 32=[b(.0625)]b2 32=.0625b3 512=b3 b=8 y=1/2 · 8x a=1/2

Assignment

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)

Cell phone subscribers 1988-1997 Finding a model. Cell phone subscribers 1988-1997 t= # years since 1987 t 1 2 3 4 5 6 7 8 9 10 y 1.6 2.7 4.4 6.4 8.9 13.1 19.3 28.2 38.2 48.7   lny 0.47 0.99 1.48 1.86 2.19 2.59 2.96 3.34 3.64 3.89

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 - .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

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 L2 and push exponential regression

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!

7.7B Solving 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 Ex. 1) (2,5) & (6,9) 5 = a 2b 9 = a 6b

Ex. 2 Write a power function y = axb graph passes through (4,6) & (8,15)

1.) Find ln x and ln y for the table. 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. Steps - 1.) Find ln x and ln y for the table. 2.) Draw a scatterplot for ln x and ln y. 3.) If the scatterplot can be made into straight line, then use 2 points from original table and follow the same process as in previous example. You can also use power regression on the calculator to write a model for data.

Ex. 2 x 1 2 3 4 5 6 7 y 1.2 5.4 9.8 14.3 25.6 41.2 65.8 #39 from assignment

Ex. 2 Continued ln x .693 1.099 1.386 1.609 1.791 1.946 x 1 2 3 4 5 6 7 y 1.2 5.4 9.8 14.3 25.6 41.2 65.8 ln y .182 1.686 2.282 2.66 3.243 3.718 4.187 #1 1 2 3 4 5 6 7 y 1.2 5.4 9.8 14.3 25.6 41.2 65.8

Step #2 Graph

Step #3 (1, 1.2), (2, 5.4) 1.) 2.)

Ex. 2 ) Power function

Assignment