(3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a...

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Transcript of (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a...

Page 1: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

I CAN GRAPH AND WRITE EXPONENTIAL FUNCTIONS

Page 2: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

REVIEW

3

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x

wx

34

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fe

ef

(3 s-5 t3 s2)2

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q

q

Page 3: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

PROPERTIES OF EXPONENTS

1. x0=12. xa•xb = xa+b

3. xa = xa-b

xb

4. (xa)b = xab

5. x-a = 1/xa

6. xa/b= (b√x)a

Page 4: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

These same properties hold for irrational exponents.

Simplify 1. (2√5)√5

2. (2√5) (2√5)3. 2√(3) + 2

2√(3) - 2

Page 5: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

Some exponential equations can be solved by re- writing each side with a common base.

Solve.1. 9x = 27

2. 2x = 4√2

Page 6: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

The exponential function with base b is denoted by f(x) = bx. Note that b represents a number. The variable is in the exponent in an exponential function. Often in applications we will use f(x) = a•bx. (b >0, b≠1)

Page 7: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

Graph.1. y = 2x

2. y = 3x

3. y = (1/2)x

List the domain, range, intercepts and asymptotes. Decribe the end behavior and whether the function is increasing or decreasing.

How do the functions differ? How are they alike?

Page 8: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

Domain all realsRange y>0y-intercept (0,1)x-intercepts noneHA y = 0

They differ in how quickly they “grow”. The first two increase from left to right, the third decreases.

Page 9: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

Now graph these. Describe the transformation.

1. y = -2x

2. y = 2-x (which of the previous graphs is this equivalent to?)

3. y = 2x+3

4. y = 2x - 2

Page 10: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

THE NATURAL BASE (EULER’S NUMBER)

For many applications it is convenient to use the irrational number e ≈ 2.718…

Graph y = ex.

This is derived from lim(1 + 1/x)x as x → ∞.

More about that later….

Page 11: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

COMPOUND INTEREST

How much money would you have after x years if you started with $100 and added $5 per year?

Is this linear or exponential growth?

What if you earned 5% interest per year?Is this linear or exponential growth?

What if you earned 5% interested compounded quarterly?

Page 12: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

INTEREST FORMULAS

CompoundA = P(1 + r/n)nt

ContinuousA = Pert

A balanceP principalR interest rate (as a decimal)t time in yearsn number of times compounded per year

Page 13: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

EXPONENTIAL MODELS N = N0(1 + r)t

what happens if r<0? what if the population is doubling?

In 2000 there were 125 deer in a wooded area. In 2010 there were 264 deer. Write an exponential function to model the yearly growth of the population. (Do you have enough information to decide whether this is a good model?)Predict when the population will reach 500 deer.

Answer: N = 125(1.078)t

about 18.5 years

Page 14: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

CONTINUOUS MODELS (GROWTH OR DECAY)

Use model N = N0ekt (A = Pert is an example)

*N0 will always be positive, k can be pos or neg

*N0 will represent the starting value, k the relative growth rate, t time

N is like y, t is like xYou need to know or find N0 and k

Page 15: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

At the beginning of an experiment you have 1200 bacteria, 4 hours later you have 2200. Write a continuous exponential growth function to model the situation.

k≈.1515, so N = 1200e.1515t

Page 16: (3 s -5 t 3 s 2 ) 2 1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a.

The half-life of a substance is the time it takes for 50% of the sample to disintegrate. (It is a property of the substance, it does not depend on the initial amount.)

For example, iodine-131 has a half-life of 8 days. If you started with 2 grams of this substance, graph the amount remaining after 0, 8 ,16, and 24 days. Then write a continuous exponential decay function to model the situation.

Answer: N = 2●e-.0866t