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 = .3