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Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

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exponential function base asymptote exponential growth exponential decay Vocabulary

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Page 1: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Warm UpEvaluate.

1. 100(1.08)20

2. 100(0.95)25

3. 100(1 – 0.02)10

4. 100(1 + 0.08)–10

≈ 466.1≈ 27.74≈ 81.71≈ 46.32

Page 2: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Write and evaluate exponential expressions to model growth and decay situations.

Objective

Page 3: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

exponential functionbaseasymptoteexponential growthexponential decay

Vocabulary

Page 4: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Moore’s law, a rule used in the computer industry, states that the number of transistors per integrated circuit (the processing power) doubles every year. Beginning in the early days of integrated circuits, the growth in capacity may be approximated by this table.

Page 5: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Growth that doubles every year can be modeled by using a function with a variable as an exponent. This function is known as an exponential function. The parent exponential function is f(x) = bx, where the base b is a constant and the exponent x is the independent variable.

Page 6: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

The graph of the parent function f(x) = 2x is shown. The domain is all real numbers and the range is {y|y > 0}.

Page 7: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Notice as the x-values decrease, the graph of the function gets closer and closer to the x-axis. The function never reaches the x-axis because the value of 2x cannot be zero. In this case, the x-axis is an asymptote. An asymptote is a line that a graphed function approaches as the value of x gets very large or very small.

Page 8: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

A function of the form f(x) = abx, with a > 0 and b > 1, is an exponential growth function, which increases as x increases. When 0 < b < 1, the function is called an exponential decay function, which decreases as x increases.

Page 9: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Negative exponents indicate a reciprocal. For example:

Remember!

In the function y = bx, y is a function of x because the value of y depends on the value of x.

Remember!

Page 10: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Tell whether the function shows growth or decay. Then graph.

Example 1A: Graphing Exponential Functions

Step 1 Find the value of the base.

3104

x

f x

The base , ,is less than 1.

This is an exponential decay

function.

34 3

410

x

f x

Page 11: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Example 1A ContinuedStep 2 Graph the function by using a table of values.

x 0 2 4 6 8 10 12f(x) 10 5.6 3.2 1.8 1.0 0.6 0.3

Page 12: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Tell whether the function shows growth or decay. Then graph.

Example 1B: Graphing Exponential Functions

Step 1 Find the value of the base.

g(x) = 100(1.05)x

The base, 1.05, is greater than 1. This is an exponential growth function.

g(x) = 100(1.05)x

Page 13: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Example 1B Continued

Step 2 Graph the function by using a graphing calculator.

Page 14: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Tell whether the function p(x) = 5(1.2x) shows growth or decay. Then graph.

Step 1 Find the value of the base.

Check It Out! Example 1

p(x) = 5(1.2 x) The base , 1.2, is greater than 1. This is an exponential growth function.

Page 15: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Step 2 Graph the function by using a table of values.

x –12 –8 –4 0 4 8 10f(x) 0.56 1.2 2.4 5 10.4 21.5 30.9

Check It Out! Example 1 Continued

Page 16: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

You can model growth or decay by a constant percent increase or decrease with the following formula:

In the formula, the base of the exponential expression, 1 + r, is called the growth factor. Similarly, 1 – r is the decay factor.

Page 17: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Clara invests $5000 in an account that pays 6.25% interest per year. After how many years will her investment be worth $10,000?

Example 2: Economics Application

Step 1 Write a function to model the growth in value of her investment.

f(t) = a(1 + r)t Exponential growth function.

Substitute 5000 for a and 0.0625 for r.

f(t) = 5000(1 + 0.0625)t

f(t) = 5000(1.0625)t Simplify.

Page 18: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Example 2 Continued

Step 2 When graphing exponential functions in an appropriate domain, you may need to adjust the range a few times to show the key points of the function.

Step 3 Use the graph to predict when the value of the investment will reach $10,000. Use the feature to find the t-value where f(t) ≈ 10,000.

Page 19: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Example 2 Continued

Step 3 Use the graph to predict when the value of the investment will reach $10,000. Use the feature to find the t-value where f(t) ≈ 10,000.

The function value is approximately 10,000 when t ≈ 11.43 The investment will be worth $10,000 about 11.43 years after it was purchased.

Page 20: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

In 1981, the Australian humpback whale population was 350 and increased at a rate of 14% each year since then. Write a function to model population growth. Use a graph to predict when the population will reach 20,000.

P(t) = a(1 + r)t

Substitute 350 for a and 0.14 for r.

P(t) = 350(1 + 0.14)t

P(t) = 350(1.14)t Simplify.

Exponential growth function.

Check It Out! Example 2

Page 21: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Graph the function. Use to find when the population will reach 20,000.

It will take about 31 years for the population to reach 20,000.

Check It Out! Example 2 Continued

Page 22: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32

Homework!

Holt Chapter 7 Section 1Page 493-495 #1-17, 21, 22

Page 23: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32
Page 24: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32
Page 25: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32
Page 26: Warm Up Evaluate. 1. 100(1.08) 20 2. 100(0.95) 25 3. 100(1 – 0.02) 10 4. 100(1 + 0.08) –10 ≈ 466.1 ≈ 27.74 ≈ 81.71 ≈ 46.32