8.2 – Properties of Exponential Functions

Review: what is an asymptote?

x)2(10y “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.

Graphing y=abx when a<0

• Ex: Graph xx yy )2(

2

1 and )2(

2

1

Where is the asymptote?

Sketch your prediction of what the graph will look like

Translating y=abx

• How does the equation change if we want to move both graphs up 4 units? Predictions?

xx yy )2(2

1 and )2(

2

1

4)2(2

1 and 4)2(

2

1 xx yy

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?

xx yy )2(2

1 and )2(

2

1

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)

44 )2(2

1 and )2(

2

1 xx yy

Let’s try some

• Graph each function as a translation of y=9(3)x

1)3(9 c)

4)3(9 b)

)3(9 a)

4

1

x

x

x

y

y

y

Make a table of values for each

Graph, from -3 to 3

y =9(3)x+1

y =9(3)x-4

y =9(3)x-4-1

“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 = PertA = 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:

A = Pert

1050(e)0.055(5)

1050(e)0.275

1050(1.316531)

A = $1382.36

Substitute in for p, r, and t

Simplify they power

Evaluate e0.275 with your calculator

Simplify

P=$1050, r=5.5% = 0.055, t=5

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?