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
x
y=1/2
·
8
3
•32=.0625b
•512=b3
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)
Finding a model.
• Cell phone subscribers 1988-1997
• t= # years since 1987
t
1
2
3
4
y
1.6
2.7
4.4
6.4
lny
5
6
7
8
9
10
8.9 13.1 19.3 28.2 38.2 48.7
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
•y =
b
ax
• Only 2 points are
needed
• Ex. 1) (2,5) & (6,9)
• 5 = a 2b
• 9 = a 6b
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
Ex. 2 Write a power function y = axb
graph passes through (4,6) & (8,15)
•
• 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.
• 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
Ex. 2 Continued
• #1
ln x
0
.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
Step #2 Graph
Step #3 (1, 1.2), (2, 5.4)
• y  ax
b
• 1.) 1.2  a1
• 2.) 5.4  a2 b
b
Ex. 2 ) Power function
y  1.2 x
2.17
Assignment