copyright © 2015, 2008, 2011 pearson education, inc. section 4.4, slide 1 chapter 4 exponential...

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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide Chapter 4 Exponential Functions

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Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 3 Example: Finding an Equation of an Exponential Curve An exponential curve contains the points listed in the table below. Find an equation of the curve.

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Page 1: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1

Chapter 4Exponential Functions

Page 2: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 2

4.4 Finding Equations of Exponential Functions

Page 3: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 3

Example: Finding an Equation of an Exponential Curve

An exponential curve contains the points listed in the table below. Find an equation of the curve.

Page 4: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 4

Solution

For f(x) = abx, recall that the y-intercept is (0, a). We see from the table that the y-intercept is (0, 3), so a = 3. As the value of x increases by 1, the value of y is multiplied by 2. By the base multiplier property, b = 2. Therefore, and equation of the curve is

f(x) = 3(2)x

Page 5: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 5

Solution

Check the result with a graphing calculator.

Page 6: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 6

Solving Equations of the Form bn = k for b

To solve an equation of the form bn = k for b,

1. If n is odd, the real-number solution is

2. If n is even and k ≥ 0, the real number solutions are

3. If n is even and k < 0, there is no real-number solution.

1 .nk

1 .nk

Page 7: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 7

Example: One-Variable Equations Involving Exponents

Find all real-number solutions. Round any results to the second decimal place.

1. 5.42b6 – 3.19 = 43.74 2.9

4703

bb

Page 8: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 8

Solution

1. 65.42 3.19 43.74b 65.42 43.74 3.19b 65.42 46.93b 6 46.93

5.42b

1 646.935.42

b 1.43b

Page 9: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 9

Solution

2.9

4703

bb

5 703

b

1 5703

b

1.88b

Page 10: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 10

Example: Finding an Equation of an Exponential Curve

Find an approximate equation y = abx of the exponential curve that contains the points (0, 3) and (4, 70). Round the value of b to two decimal places.

Page 11: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 11

SolutionSince the y-intercept is (0, 3), the equation has the form y = 3bx. Next, substitute (4, 70) in the equation and solve for b:

470 3b43 70b 4 70

3b

1 4703

b

2.20b

Page 12: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 12

Solution

So, our equation is y = 3(2.20)x; its graph contains the given point (0, 3). Since we rounded the value of b, the graph of the equation comes close to, but does not pass through, the given point (4, 70). We use a graphing calculator to verify our work.

Page 13: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 13

Dividing Left Sides and Right Sides of Two Equations

If a = b, c = d, c ≠ 0, and d ≠ 0, then

In words, the quotient of the left sides of two equations is equal to the quotient of the right sides.

a bc d

Page 14: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 14

Example: Finding the Equation of an Exponential Curve

Find an approximate equation y = abx of the exponential curve that contains (2, 5) and (5, 63). Round the values of a and b to two decimal places.

Page 15: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 15

Solution

Since both of the ordered pairs (2, 5) and (5, 63) must satisfy the equation y = abx, we have the following system of equations:

25 ab563 ab

Page 16: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 16

Solution

It will be slightly easier to solve this system if we switch the equations to list the equation with the greater exponent of b first:

25 ab

563 ab

Page 17: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 17

Solution

We divide the left sides and divide the right sides of the two equations to get the following result for nonzero a and b:

5

2635

baab

3635

b

Page 18: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 18

Solution

We can now solve for b by finding the cube root:

3 635

b

1 3635

b

2.33b

Page 19: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 19

Solution

Substitute 2.33 for the constant b in the equation y = abx:

To find a, substitute the coordinates of the given point (2, 5) into the equation:

2.33 xy a

25 2.33a

25

2.33a

0.92a

Page 20: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 20

Solution

So, an equation that approximates the exponential curve that passes through (2, 5) and (5, 63) is y = 0.92(2.33)x.

We use a graphing calculator to verify our work.

Page 21: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 21

Exponential Function

We can find an equation of an exponential function by using the base multiplier property or by using two points. Both methods give the same result.