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© 2010 Pearson Prentice Hall. All rights reserved. CHAPTER 7 Algebra: Graphs, Functions, and Linear Systems.

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Presentation on theme: "© 2010 Pearson Prentice Hall. All rights reserved. CHAPTER 7 Algebra: Graphs, Functions, and Linear Systems."— Presentation transcript:

1 © 2010 Pearson Prentice Hall. All rights reserved. CHAPTER 7 Algebra: Graphs, Functions, and Linear Systems

2 © 2010 Pearson Prentice Hall. All rights reserved. 2 7.6 Modeling Data: Exponential, Logarithmic, and Quadratic Functions

3 © 2010 Pearson Prentice Hall. All rights reserved. Objectives 1.Graph exponential functions. 2.Use exponential models. 3.Graph logarithmic functions. 4.Use logarithmic models. 5.Graph quadratic functions. 6.Use quadratic models. 7.Determine an appropriate function for modeling data. 3

4 © 2010 Pearson Prentice Hall. All rights reserved. Scatter Plots & Regression Lines Data presented in a visual form as a set of points is called a scatter plot. A line that best fits the data points in a scatter plot is called a regression line. For example, the graph displays the relationship between literacy and child mortality. Each point represents a country. 4

5 © 2010 Pearson Prentice Hall. All rights reserved. Modeling with Exponential Functions 5

6 © 2010 Pearson Prentice Hall. All rights reserved. Graph: f(x) = 2 x. Solution: We start by selecting numbers for x and finding ordered pairs. Example 1: Graphing an Exponential Function We make a table: xf(x) = 2 x (x,y)(x,y) −3−3f(-3) = 2 -3 =(−3, ) −2f(-2) = 2 -2 = ¼(−2,¼) −1f(-1) = 2 -1 = ½(−1,½) 0f(0) = 2 0 = 1(0,1) 1f(1) = 2 1 = 2(1,2) 2f(2) = 2 2 = 4(2,4) 3f(3) = 2 3 = 8(3,8) 6

7 © 2010 Pearson Prentice Hall. All rights reserved. Example 1 continued Next, plot the ordered pairs and connect them with a smooth curve. 7

8 © 2010 Pearson Prentice Hall. All rights reserved. The graphs below show the world population for seven selected years from 1950 through 2008. One is a bar graph and the other is a scatter plot. Example 2: Comparing Linear and Exponential Models 8

9 © 2010 Pearson Prentice Hall. All rights reserved. After entering the data in a calculator, the graphing calculator displays the linear model, y = ax + b, and the exponential model, y = ab x, that best fit the data. a.Express each model in function notation, with numbers rounded to three decimal places. b.How well do the functions model the world population in 2008? Example 2 continued 9

10 © 2010 Pearson Prentice Hall. All rights reserved. c.By one projection, world population is expected to reach 8 billion in the year 2026. Which function serves as a better model for this prediction? Solution: a.Using the figure from the graphing calculator, the functions f(x) = 0.074x + 2.287 and g(x) = 2.566(1.017) x model world population, in billions, x years after 1949. The linear function is f and the exponential function g. Example 2 continued 10

11 © 2010 Pearson Prentice Hall. All rights reserved. b.The graph shows that the world population in 2008 was 6.8 billion. The year 2008 is 59 years after 1949. Hence, we substitute 59 for x in each function and then compare with the actual population in 2008. f(x) = 0.043x + 2.287 f(59) = 0.074(59) + 2.287 f(59) ≈ 6.7 g(x) = 2.566(1.017) x g(59) = 2.566(1.017) 59 g(59) ≈ 6.9 Example 2 continued 11

12 © 2010 Pearson Prentice Hall. All rights reserved. Since the world population was 6.8 billion, it seems both functions model world population well for 2008. c.Now, we compare the models to a world population of 8 billion in the year 2026. We use 77 for x since 2026 is 77 years after 1949. f(x) = 0.074x + 2.287 f(77) = 0.074(77) + 2.287 f(77) ≈ 8.0 Example 2 continued 12

13 © 2010 Pearson Prentice Hall. All rights reserved. g(x) = 2.566(1.017) x g(77) = 2.566(1.017) 77 g(77) ≈ 9.4 It seems the linear function f(x) serves as a better model for a projected world population of 8 billion by 2026. Example 2 continued 13

14 © 2010 Pearson Prentice Hall. All rights reserved. The Role of e in Applied Exponential Functions An irrational number, symbolized by the letter e, appears as a base in many exponential functions. This irrational number e ≈ 2.72 or more accurately e ≈ 2.71828… The number e is called the natural base. The function f(x) = e x is called the natural exponential function. Graph of f(x) = e xGraph of f(x) = e x 14

15 © 2010 Pearson Prentice Hall. All rights reserved. Medical research indicates that the risk of having a car accident increases exponentially as the concentration of alcohol in the blood increases. The risk is modeled by R = 6 e 12.77x, where x is the blood alcohol concentration and R, given as a percent, is the risk of having a car accident. In many states, it is illegal to drive with a blood alcohol concentration at 0.08 or greater. What is the risk of a car accident with a blood alcohol concentration at 0.08? Example 3: Alcohol and Risk of a Car Accident 15

16 © 2010 Pearson Prentice Hall. All rights reserved. Solution: We substitute 0.08 for x in the function. R = 6 e 12.77x R = 6 e 12.77(0.08) Putting this in the calculator, we get an approximation of 16.665813. Rounding to one decimal place, the risk of getting in a car accident is approximately 16.7% with a blood alcohol concentration at 0.08. Example 3 continued 16

17 © 2010 Pearson Prentice Hall. All rights reserved. Definition of e Euler’s number e ≈ 2.7128… Discovered by Jacob Bernoulli (in late 17 th century) An account that starts with $1.00 and pays 100 percent interest per year. If the interest is credited once a year, the value is $2.00; but if the interest is computed and added twice in the year, the $1 is multiplied by 1.5 twice, yielding $1.00×1.5² = $2.25. Compounding quarterly yields $1.00×1.25 4 = $2.4414..., and compounding monthly yields $1.00×(1.0833...) 12 = $2.613035.... 1

18 © 2010 Pearson Prentice Hall. All rights reserved. What is e? x(1 + 1/x) x f(x) 1(1+1/1)2 2(1+1/3) 2 2.25 3(1+1/3) 3 2.37 4(1+1/3) 4 2.44 5(1+1/3) 5 2.49 ……… 10(1+1/3) 10 2.59 ……… 20(1+1/3) 20 2.65 ……… 100(1+1/3) 100 2.70 ……… ∞ (1+1/3) ∞ 2.7128… = e 1 e = Graph of (1+1/x) x

19 © 2010 Pearson Prentice Hall. All rights reserved. Modeling with Logarithmic Functions 19

20 © 2010 Pearson Prentice Hall. All rights reserved. Graph: y = log 2 x. Solution: Because y = log 2 x means 2 y = x, we will use the exponential form of the equation to obtain the function’s graph. Example 4: Graphing a Logarithmic Function x = 2 y y(x,y)(x,y) 2 -2 = ¼−2−2(¼,−2) 2 -1 = ½−1−1(½,−1) 2 0 = 10(1,0) 2 1 = 21(2,1) 2 2 = 42(4,2) 2 3 = 83(8,3) 20

21 © 2010 Pearson Prentice Hall. All rights reserved. When the outside air temperature is anywhere from 72° to 96°F, the temperature in an enclosed vehicle climbs by 43°in the first hour. The bar graph and scatter plot are given below. Example 5: Dangerous Heat: Temperature in an Enclosed Vehicle 21

22 © 2010 Pearson Prentice Hall. All rights reserved. After entering data in a graphing calculator, the calculator displays the logarithmic model y = a + b ln x, where ln x is called the natural logarithm. a.Express the model in function notation, with numbers rounded to one decimal place. b.Use the function to find the temperature increase, to the nearest degree, after 50 minutes. How well does the model resemble the actual increase? Example 5 continued 22

23 © 2010 Pearson Prentice Hall. All rights reserved. Solution: a.Using the calculator figure and rounding to one decimal place, the function f(x) = −11.6 + 13.4 ln x models the temperature increase after x minutes.. Example 5 continued 23

24 © 2010 Pearson Prentice Hall. All rights reserved. Solution: b.We substitute 50 for x and evaluate the function. f(x) = −11.6 + 13.4 ln x f(50) = −11.6 + 13.4 ln 50 f(50) ≈ 41 Since the actual temperature increases 41° after 50 minutes, the function models the actual increase well. Example 5 continued 24

25 © 2010 Pearson Prentice Hall. All rights reserved. Modeling with Quadratic Functions A quadratic function is any function of the form y = ax 2 + bx + c or f(x) = ax 2 + bx + c, where a, b, and c are real numbers, with a ≠ 0. The graph of any quadratic function is called a parabola. The vertex of the parabola is the lowest point or the highest point on the graph. 25

26 © 2010 Pearson Prentice Hall. All rights reserved. Modeling with Quadratic Functions 26

27 © 2010 Pearson Prentice Hall. All rights reserved. Modeling with Quadratic Functions Vertex of a Parabola 27

28 © 2010 Pearson Prentice Hall. All rights reserved. Graphing Quadratic Equations 28

29 © 2010 Pearson Prentice Hall. All rights reserved. Graph the quadratic function: y = x 2 – 2x – 3. Solution: We follow the steps: 1.Determine how the parabola opens. Since a is the coefficient of x 2 and a = 1 in this case, then the parabola opens upward. 2.Find the vertex. Example 6: Graphing a Parabola 29

30 © 2010 Pearson Prentice Hall. All rights reserved. We use the formula to find the x-coordinate: We plug x = 1 into the original function to find the y-coordinate: y = x 2 – 2x – 3 y = (1) 2 – 2(1) – 3 y = −4 The vertex is (1,−4). Example 6 continued 30

31 © 2010 Pearson Prentice Hall. All rights reserved. 3.Find the x-intercepts. Replace y with 0 in y = x 2 – 2x – 3. y = x 2 – 2x – 3 0 = x 2 – 2x – 3 0 = (x – 3)(x + 1) x – 3 = 0 or x + 1 = 0 x = 3 or x = 1 The x-intercepts are 3 and −1. The parabola passes through (3,0) and (−1,0). Example 6 continued 31

32 © 2010 Pearson Prentice Hall. All rights reserved. 4. Find the y-intercept. Replace x with 0 in y = x 2 – 2x – 3. y = x 2 – 2x – 3 y = 0 2 – 2(0) – 3 = −3 The parabola passes through the point (0,−3). Example 6 continued 32

33 © 2010 Pearson Prentice Hall. All rights reserved. Steps 5. and 6. Plot the intercepts and vertex. Connect these points with a smooth curve. Example 6 continued 33

34 © 2010 Pearson Prentice Hall. All rights reserved. The graphs below show the U.S. adult wine consumption, in gallons per person, for selected years from 1980 through 2005. Example 7: Modeling Wine Consumption 34

35 © 2010 Pearson Prentice Hall. All rights reserved. After entering the data into a graphing calculator, we get the display: a.Express the model in function notation, with numbers rounded to three decimal places. b.According to the function in part (a), what was the U.S. adult wine consumption in 2005? Round to one decimal place. How does this model the value shown by the bar graph? Example 7 continued 35

36 © 2010 Pearson Prentice Hall. All rights reserved. Solution: a.Using the calculator display and rounding to three decimal places, we get f(x) = 0.004x 2 – 0.094x + 2.607 to model the wine consumption after x years after 1980. Example 7 continued 36

37 © 2010 Pearson Prentice Hall. All rights reserved. Solution: b.The graph shows the wine consumption in 2005 was 2.8 gallons per adult. The year 2005 is 25 years after 1980. Because the rounded value is the same as the 2.8 gallons shown in the bar graph, the function describes the data extremely well. Example 7 continued 37

38 © 2010 Pearson Prentice Hall. All rights reserved. Determine an Appropriate Function for Modeling Data Description of Data Points in a Scatter Plot Model Lie on or near a lineLinear function y = mx + b or f(x) = mx + b Increasing more and more rapidlyExponential function y = b x, or f(x) = b x, b > 1 Increasing, although rate of increase is slowing down Logarithmic function, y = log b x, b > 1 y = log b x means b y = x. Decreasing and then increasing Quadratic Function y = ax 2 + bx + c or f(x) = ax 2 + bx + c, a > 0. The vertex is a minimum. Increasing and then decreasing Quadratic Function y = ax 2 + bx + c or f(x) = ax 2 + bx + c, a < 0. The vertex is a maximum. 38


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