1. 7.7A Write Exponential Functions
Algebra II

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

3. 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. 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

5. 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.

6. 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

7. Assignment

8. 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.

9. (-2, ¼) (-1, ½) (0, 1) (1, 2)
(x, lny)
(-2, -1.38) (-1, -.69) (0,0) (1, .69)

10. Cell phone subscribers 1988-1997
Finding a model.
Cell phone subscribers
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

11. Now plot (x,lny)
Since the points lie close to a line, an exponential model should
be a good fit.

12. Use 2 points to write the linear equation.
(2, .99) & (9, 3.64)
m= = 2.65 = –
(y - .99) = .379 (x – 2)
y = .379x
y = .379x LINEAR MODEL FOR (t,lny)
The y values were ln’s & x’s were t so:
lny = .379t now solve for y
elny = e.379t exponentiate both sides
y = (e.379t)(e.233) properties of exponents
y = (e.233)(e.379t) Exponential model

13. y = (e.233)(e.379t)
y = 1.26 · 1.46t

14. 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

15. 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!

16. 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

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

18. 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.

19. 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

20. 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

21. Step #2 Graph

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

23. Ex. 2 ) Power function

24. Assignment