7.7AWriteExponentialFunctions.ppt

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: 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,.0625) & (2,32) .0625=ab-1 32=ab2 (.0625)=a/b b(.0625)=a 32=[b(.0625)]b2 32=.0625b3 512=b3 b=8 a=1/2 y=1/2 · 8x 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, ?) (-1, ?) (0, 1) (1, 2) (x, lny) (-2, -1.38) (-1, -.69) (0,0) (1, .69) 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