1. Slide 1-1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Graphs, Functions, and Models Chapter 1

3. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1.1 Introduction to Graphing  Plot points.  Determine whether an ordered pair is a solution of an equation.  Graph equations.  Find the distance between two points in the plane and find the midpoint of a segment.  Find an equation of a circle with a given center and radius, and given an equation of a circle, find the center and the radius.  Graph equations of circles.

4. Slide 1-4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Cartesian Coordinate System (x, y) I (+, +) II (, +) III (, ) IV (+, ) (0,0)

5. Slide 1-5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solutions of Equations  Equations in two variables have solutions (x, y) that are ordered pairs. Example: 4x + 5y = 20  When an ordered pair is substituted into the equation, the result is a true equation.

6. Slide 1-6 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Examples  Determine whether the ordered pair (  4, 2) is a solution of 3x + 4y = 2. 3(  4) + 4(2) ? 2  12 + 8 ? 2  4  2 false (  4, 2) is not a solution.  Determine whether the ordered pair (2,  1) is a solution of 3x + 4y = 2. 3(2) + 4(  1) ? 2 6 +  4 ? 2 2 = 2 true (2,  1) is a solution.

7. Slide 1-7 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphs of Equations  To graph an equation is to make a drawing that represents the solutions of that equation.

8. Slide 1-8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley x-Intercept  The point at which the graph crosses the x-axis.  An x-intercept is a point (a, 0). To find a, let y = 0 and solve for x.  Example: Find the x-intercept of 3x + 5y = 15. 3x + 5(0) = 15 3x = 15 x = 5 The x-intercept is (5, 0).

9. Slide 1-9 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley y-Intercept  The point at which the graph crosses the y-axis.  A y-intercept is a point (0, b). To find b, let x = 0 and solve for y.  Example: Find the y-intercept of 3x + 5y = 15. 3(0) + 5y = 15 5y = 15 y = 3 The y-intercept is (0, 3).

10. Slide 1-10 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example  Graph x + 2y = 6.  x-intercept: x + 2(0) = 6 x = 6 (6, 0)  y-intercept: (0) + 2y = 6 2y = 6 y = 3 (0, 3) (6, 0) (0, 3)

11. Slide 1-11 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example  Graph y = x 2 + 2x  3 = 0  Make a table of values. xy(x, y) 33 0 (  3, 0) 22 33(  2,  3) 11 44(  1,  4) 0 33(0,  3) 10(1, 0) 25(2, 5) 1. Select values for x. 2. Compute values for y.

12. Slide 1-12 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Distance Formula The distance d between any two points (x 1, y 1 ) and (x 2, y 2 ) is given by. Example: Find the distance between the points (2, 2) and (  3,  5).

13. Slide 1-13 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Midpoint Formula  If the endpoints of a segment are (x 1, y 1 ) and (x 2, y 2 ), then the coordinates of the midpoint are.  Example: Find the midpoint of a segment whose endpoints are (  5,  6) and (4, 4).

14. Slide 1-14 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Circles  A circle is the set of all points in a plane that are a fixed distance r from a center (h, k).  The equation of a circle with center (h, k) and radius r, in standard form, is (x  h) 2 + (y  k) 2 = r 2.

15. Slide 1-15 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example  Find an equation of a circle having radius 7 and center (4,  2). Using the standard form: (x  h) 2 + (y  k) 2 = r 2 [x  4] 2 + [y  (  2)] 2 = 7 2 (x  4) 2 + (y + 2) 2 = 49.

16. Slide 1-16 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example  Graph the circle (x + 4) 2 + (y  1) 2 = 9 Write the equation in standard form. [x  (  4)] 2 + [y  1] 2 = 3 2 The center is (  4, 1) and the radius is 3. (4, 1)

17. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1.2 Functions and Graphs  Determine whether a correspondence or a relation is a function.  Find function values, or outputs, using a formula.  Find the domain and the range of a function.  Determine whether a graph is that of a function.  Solve applied problems using functions.

18. Slide 1-18 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Function A function is a correspondence between a first set, called the domain, and a second set, called the range, such that each member of the domain corresponds to exactly one member of the range. FunctionNot a Function  749  24 7 57 00 63  24 8 2

19. Slide 1-19 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Relation A relation is a correspondence between the first set, called the domain, and a second set, called the range, such that each member of the domain corresponds to at least one member of the range. Which of the following relations is a function? {(8,  2), (8, 4), (7, 3)}Not a function {(  6, 4), (1, 4), (7, 4)}Function

20. Slide 1-20 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Notation for Functions  The inputs (members of the domain) are values of x substituted into the equation. The outputs (members of the range) are the resulting values of y.  f(x) is read “f of x,” or “f at x,” or “the value of f at x.”  Example: Given f(x) = 3x 2  4, find f(6). f(6) = 3(6) 2  4 = 3(36)  4 = 104

21. Slide 1-21 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphs of Functions  We select values for x and find the corresponding values of f(x). Then we plot the points and connect them with a smooth curve.  The Vertical-Line Test If it is possible for a vertical line to cross a graph more than once, then the graph is not the graph of a function.

22. Slide 1-22 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example: Does the graph represent a function? The graph is a function because we cannot find a vertical line that crosses the graph more than once.

23. Slide 1-23 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example: Does the graph represent a function? The graph is not a function. We can find a vertical line that crosses the graph in more than one point.

24. Slide 1-24 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Finding Domains of Functions Find the indicated function values and determine whether the given values are in the domain of the function. f(1) and f(5), for f(1) = Since f(1) is defined, 1 is in the domain of f. f(5) = Since division by 0 is not defined, the number 5 is not in the domain of f.

25. Slide 1-25 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Another Example Find the domain of the function Solution: We can substitute any real number in the numerator, but we must avoid inputs that make the denominator 0. Solve x 2  3x  28 = 0. (x  7)(x + 4) = 0 x  7 = 0 or x + 4 = 0 x = 7 or x =  4 The domain consists of the set of all real numbers except  4 and 7 or {x|x   4 and x  7}.

26. Slide 1-26 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Visualizing Domain and Range Keep the following in mind regarding the graph of a function:  Domain = the set of a function’s inputs, found on the horizontal axis.  Range = the set of a function’s outputs, found on the vertical axis.

27. Slide 1-27 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph the function. Then estimate the domain and range. Domain = [1,  ) Range = [0,  )

28. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1.3 Linear Functions, Slope, and Applications  Determine the slope of a line given two points on the line.  Solve applied problems involving slope and linear functions.

29. Slide 1-29 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Linear Functions A function f is a linear function if it can be written as f(x) = mx + b, where m and b are constants. If m = 0, the function is a constant function f(x) = b. If m = 1 and b = 0, the function is the identity function f(x) = x.

30. Slide 1-30 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Examples  Linear Functions  Nonlinear Functions y = x y = 3 y = x 2 + 1

31. Slide 1-31 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Horizontal and Vertical Lines  Horizontal lines are given by equations of the type y = b or f(x) = b. They are functions.  Vertical lines are given by equations of the type x = a. They are not functions. y =  2 x =  2

32. Slide 1-32 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slope  The slope m of a line containing the points (x 1, y 1 ) and (x 2, y 2 ) is given by

33. Slide 1-33 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example  Graph the function and determine its slope. Calculate two ordered pairs, plot the points, graph the function, and determine its slope. Rise  2 Run 4

34. Slide 1-34 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Horizontal and Vertical Lines  A horizontal line has a slope of 0. Graph y = 4 and determine its slope. Choose any number for x; y must be 4. x y  4 4  2 4 1 4  A vertical line has an undefined slope because we cannot divide by zero. Graph x = 2 and determine its slope. Choose any number for y; x must be 2. x y 2 4 2 1 2  2

35. Slide 1-35 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Types of Slopes  Positive—line slants up from left to right  Negative—line slants down from left to right  Zero—horizontal line  Undefined—vertical line zero positive negative undefined

36. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1.4 Equations of Lines and Modeling  Find the slope and the y–intercept of a line given the equation y = mx + b, or f(x) = mx + b.  Graph a linear equation using the slope and the y-intercept.  Determine equations of lines.  Given the equations of two lines, determine whether their graphs are parallel or whether they are perpendicular.  Model a set of data with a linear function.  Fit a regression line to a set of data; then use the linear model to make predictions.

37. Slide 1-37 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slope-Intercept Equation The linear function f given by f(x) = mx + b has a graph that is a straight line parallel to y = mx. The constant m is called the slope, and the y-intercept is (0, b). (0, 0) (0, b) y = mx f(x) = mx + b

38. Slide 1-38 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find the slope and y-intercept of the line with equation y =  0.36x + 4.2. Solution: y =  0.36x + 4.2 Slope =  0.36; y-intercept = (0, 4.2)

39. Slide 1-39 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example  Find the slope and y-intercept of the line with equation 4x + 3y  12 = 0. Solution: We solve for y: Thus, the slope is and the y-intercept is (0, 4).

40. Slide 1-40 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example  A line has slope 3 and contains the point (  2, 5). Find an equation of the line. Solution: We use the slope-intercept equation, y = mx + b, and substitute for m. y = 3x + b. Using the point (  2, 5), we substitute for x and y and solve for b. 5 = 3(  2) + b 5 =  6 + b 11 = b The equation of the line is y = 3x + 11.

41. Slide 1-41 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example  Graph Solution: The equation is in slope-intercept form, y = mx + b. The y-intercept is (0,  2).

42. Slide 1-42 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Point-Slope Equation The point-slope equation of the line with slope m passing through (x 1, y 1 ) is y  y 1 = m(x  x 1 ). Example: Find the equation of the line containing the points (2, 7) and (  1,  8). Solution: First determine the slope: Using the point-slope equation, substitute 5 for m and either of the points for (x 1, y 1 ):

43. Slide 1-43 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Parallel Lines  Vertical lines are parallel. Nonvertical lines are parallel if and only if they have the same slope and different y-intercepts. y = 3x + 2 y = 3x  4

44. Slide 1-44 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Perpendicular Lines  Two lines with slopes m 1 and m 2 are perpendicular if and only if the product of their slopes is  1: m 1 m 2 =  1.  Lines are also perpendicular if one is vertical (x = a) and the other is horizontal (y = b). y = 3x  4

45. Slide 1-45 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Examples  Determine whether each of the following pairs of lines is parallel, perpendicular, or neither. a) y  4x =  3, 4y  8 =  x(perpendicular) b) 2x + 3y = 4, 3x + 2y = 8(neither) c) 2y = 4x + 12, y  8 = 2x(parallel)

46. Slide 1-46 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example  Write equations of the lines (a) parallel and (b) perpendicular to the graph of the line 3y + 4 = 18x and containing the point (1,  2). Solve the equation for y: (a) The line parallel to the given line will have the same slope. We use either the slope-intercept or point- slope equation for the line.

47. Slide 1-47 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Substitute and solve the equation.

48. Slide 1-48 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued (b) For a line perpendicular: m =

49. Slide 1-49 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Curve Fitting  In general, we try to find a function that fits, as well as possible, observations (data), theoretical reasoning, and common sense.  Example: Model the data given in the table on foreign travel on the next slide with two different linear functions. Then with each function, predict the number of U.S. travelers to foreign countries in 2005. Of the two models, which appears to be the better fit?

50. Slide 1-50 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Curve Fitting continued Model I: Choose any two points to determine the equation. Predict the number of travelers:

51. Slide 1-51 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Curve Fitting continued Model II: Choose any two points to determine the equation. Predict the number of travelers:

52. Slide 1-52 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Curve Fitting continued Using model I, we predict that there will be about 6.92 million U.S. foreign travelers in 2006, and using model II, we predict about 7.51 million. Since it appears from the graphs that model II fits the data more closely, we would choose model II over model I.

53. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1.5 More on Functions  Graph functions, looking for intervals on which the function is increasing, decreasing, or constant, and estimate relative maxima and minima.  Given an application, find a function that models the application; find the domain of the function and function values, and then graph the function.  Graph functions defined piecewise.

54. Slide 1-54 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Increasing, Decreasing, and Constant Functions On a given interval, if the graph of a function rises from left to right, it is said to be increasing on that interval. If the graph drops from left to right, it is said to be decreasing. If the function values stay the same from left to right, the function is said to be constant.

55. Slide 1-55 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Definitions A function f is said to be increasing on an open interval I, if for all a and b in that interval, a < b implies f(a) < f(b).

56. Slide 1-56 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Definitions continued A function f is said to be decreasing on an open interval I, if for all a and b in that interval, a f(b).

57. Slide 1-57 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A function f is said to be constant on an open interval I, if for all a and b in that interval, f(a) = f(b). Definitions continued

58. Slide 1-58 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Relative Maximum and Minimum Values Suppose that f is a function for which f(c) exists for some c in the domain of f. Then: f(c) is a relative maximum if there exists an open interval I containing c such that f(c) > f(x), for all x in I where x  c; and f(c) is a relative minimum if there exists an open interval I containing c such that f(c) < f(x), for all x in I where x  c.

59. Slide 1-59 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Functions Defined Piecewise  Graph the function defined as: a) We graph f(x) =  3 only for inputs x less than or equal to 0. b) We graph f(x) =  3 + x 2 only for inputs x greater than 0 and less than or equal to 2. c) We graph f(x) = only for inputs x greater than 2. f(x) =  3, for x  0 f(x) =  3 + x 2, for 0 < x  2

60. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1.6 The Algebra of Functions  Find the sum, the difference, the product, and the quotient of two functions, and determine the domains of the resulting functions.  Find the difference quotient for a function.  Find the composition of two functions and the domain of the composition; decompose a function as a composition of two functions.

61. Slide 1-61 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Sums, Differences, Products, and Quotients of Functions  If f and g are functions and x is in the domain of each function, then

62. Slide 1-62 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example  Given that f(x) = x + 2 and g(x) = 2x + 5, find each of the following. a) (f + g)(x)b) (f + g)(5) Solution: a) b) We can find (f + g)(5) provided 5 is in the domain of each function. This is true. f(5) = 5 + 2 = 7g(5) = 2(5) + 5 = 15 (f + g)(5) = f(5) + g(5) = 7 + 15 = 22 or (f + g)(5) = 3(5) + 7 = 22

63. Slide 1-63 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Another Example  Given that f(x) = x 2 + 2 and g(x) = x  3, find each of the following. a) The domain of f + g, f  g, fg, and f/g b) (f  g)(x) c) (f/g)(x) Solution: a) The domain of f is the set of all real numbers. The domain of g is also the set of all real numbers. The domains of f + g, f  g, and fg are the set of numbers in the intersection of the domains—that is, the set of numbers in both domains, or all real numbers. For f/g, we must exclude 3, since g(3) = 0.

64. Slide 1-64 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Another Example continued b) (f  g)(x) = f(x)  g(x) = (x 2 + 2)  (x  3) = x 2  x + 5 c) (f/g)(x) = Remember to add the stipulation that x  3, since 3 is not in the domain of (f/g)(x).

65. Slide 1-65 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Difference Quotients Example: For the function f given by f(x) = x 2 + 2x  3, find the difference quotient. Solution: We first find f(x + h):

66. Slide 1-66 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Difference Quotients continued

67. Slide 1-67 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Composition of Functions Definition:

68. Slide 1-68 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Given that f(x) = 3x  1 and g(x) = x 2 + x  3, find: a) b)

69. Slide 1-69 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Decomposing a Function as a Composition In calculus, one needs to recognize how a function can be expressed as the composition of two functions. Example: If h(x) = (3x  1) 4, find f(x) and g(x) such that Solution: The function h(x) raises (3x  1) to the fourth power. Two functions that can be used for the composition are: f(x) = x 4 and g(x) = 3x  1.

70. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1.7 Symmetry and Transformations  Determine whether a graph is symmetric with respect to the x-axis, the y-axis, and the origin.  Determine whether a function is even, odd, or neither even nor odd.  Given the graph of a function, graph its transformation under translations, reflections, stretchings, and shrinkings.

71. Slide 1-71 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Symmetry Algebraic Tests of Symmetry x-axis: If replacing y with  y produces an equivalent equation, then the graph is symmetric with respect to the x-axis. y-axis: If replacing x with  x produces an equivalent equation, then the graph is symmetric with respect to the y-axis. Origin: If replacing x with  x and y with  y produces an equivalent equation, then the graph is symmetric with respect to the origin.

72. Slide 1-72 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Test x = y 2 + 2 for symmetry with respect to the x-axis, the y-axis, and the origin. x-axis: We replace y with  y: The resulting equation is equivalent to the original so the graph is symmetric with respect to the x-axis. y-axis: We replace x with  x: The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis.

73. Slide 1-73 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Origin: We replace x with  x and y with  y: The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.

74. Slide 1-74 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Even and Odd Functions If the graph of a function f is symmetric with respect to the y-axis, we say that it is an even function. That is, for each x in the domain of f, f(x) = f(  x). If the graph of a function f is symmetric with respect to the origin, we say that it is an odd function. That is, for each x in the domain of f, f(  x) =  f(x).

75. Slide 1-75 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Determine whether the function is even, odd, or neither. 1. We see that h(x)  h(  x). Thus, h is not even. 2. We see that h(  x)   h(x). Thus, h is not odd.

76. Slide 1-76 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Vertical Translation For b > 0, the graph of y = f(x) + b is the graph of y = f(x) shifted up b units; the graph of y = f(x)  b is the graph of y = f(x) shifted down b units. y = 3x 2

77. Slide 1-77 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Horizontal Translation For d > 0, the graph of y = f(x  d) is the graph of y = f(x) shifted right d units; the graph of y = f(x + d) is the graph of y = f(x) shifted left d units. y = 3x 2

78. Slide 1-78 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Reflections  The graph of y =  f(x) is the reflection of the graph of y = f(x) across the x-axis.  The graph of y = f(  x) is the reflection of the graph of y = f(x) across the y-axis.  If a point (x, y) is on the graph of y = f(x), then (x,  y) is on the graph of y =  f(x), and (  x, y) is on the graph of y = f(  x).

79. Slide 1-79 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example  Reflection of the graph y = 3x 3  4x 2 across the x-axis. y = 3x 3  4x 2 y =  3x 3 + 4x 2

80. Slide 1-80 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Vertical Stretching and Shrinking The graph of y = af(x) can be obtained from the graph of y = f(x) by stretching vertically for |a| > 1, or shrinking vertically for 0 < |a| < 1. For a < 0, the graph is also reflected across the x-axis. (The y-coordinates of the graph of y = af(x) can be obtained by multiplying the y-coordinates of y = f(x) by a.)

81. Slide 1-81 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Examples

82. Slide 1-82 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Horizontal Stretching or Shrinking The graph of y = f(cx) can be obtained from the graph of y = f(x) by shrinking horizontally for |c| > 1, or stretching horizontally for 0 < |c| < 1. For c < 0, the graph is also reflected across the y-axis. (The x-coordinates of the graph of y = f(cx) can be obtained by dividing the x-coordinates of the graph of y = f(x) by c.)

83. Slide 1-83 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Examples