Download 8.2 – Properties of Exponential Functions

8.2 – Properties of
Exponential Functions
Review: what is an asymptote?
“Walking halfway to the wall”
An Asymptote is a line that
a graph approaches as x
or y increases in absolute
value.
In this example, the
asymptote is the x
axis.
y  10(2)
x
Graphing y=abx when a<0
1 x
1 x
• Ex: Graph y  (2) and y   (2)
2
2
Sketch your prediction of
what the graph will look like
Where is the
asymptote?
Translating y=abx
• How does the equation change if we want to
move both graphs up 4 units? Predictions?
y
y
1 x
1
(2) and y   (2) x
2
2
1 x
1
(2)  4 and y   (2) x  4
2
2
Question: where is
the asymptote now?
To move the graph up or down, add or subtract units at the end of the
equations. No need to use inverses – if you want to go up, add; if you
want to go down, subtract.
Translating y=abx
• How does the equation change if we want to
move both graphs left 4 units? Predictions?
1 x
1
(2) and y   (2) x
2
2
1
1
y  (2) x  4 and y   (2) x  4
2
2
y
To move the graph left or right, add or subtract units to the exponent of the
equation. Reminder: use the inverse of how you want the graph to move (e.g.
x-4 will move to the right; x+4 will move to the left)
Let’s try some
• Graph each function
as a translation of
y=9(3)x
a) y  9(3)
x 1
b) y  9(3)  4
x
Make a table of values for each
Graph, from -3 to 3
c) y  9(3) x  4  1
y =9(3)x+1
y =9(3)x-4-1
y =9(3)x-4
“e = 2.718”
What is base “e” ?
e is an irrational number, approximately equal to 2.718.
Exponential
functions with a base
of e are useful for
describing
continuous growth or
decay. In the graph
below, y = e is the
asymptote to the
graph.
y=e
Graphing ex
• Using your graphing calculators,
graph y=ex. Evaluate e4 to four
decimal places.
We now need
to evaluate
where x=4
2. Press 2nd, Calc
and select 1
(value). Press
enter
3. We are
evaluating when
x=4. Enter 4 for x
and press enter.
The value of e4 is
about 54.59815
Your turn: evaluate e-3
0.0498
So, what is “e” good for???
Continuously Compounding
Interest
• A = Pert
A = amount of money in
the account
P = principal (how much
is deposited)
r = annual rate of interest
t = time (in years)
Example: Continuously
Compounded Interest Problem
• You invest $1,050 at an annual interest rate of 5.5%,
compounded continuously. How much will you have in
the account after 5 years?
•Start with:
P=$1050, r=5.5% = 0.055, t=5
A = Pert
Substitute in for p, r, and t
1050(e)0.055(5)
1050(e)0.275
Simplify they power
1050(1.316531) Evaluate e0.275 with your
calculator
A = $1382.36
Simplify
Let’s try one:
• Suppose you invest $1,300 at an annual
interest rate of 4.3%, compounded
continuously. How much will you have in
the account after three years?
Suppose you invest $1,300 at an annual interest rate of
4.3%, compounded continuously. How much will you have
in the account after three years?