Trigonometry 8e lial

lial_22737_ISM_TTL_CPY.qxd 7/1/04 7:45 AM Page INSTRUCTOR’S SOLUTIONS MANUAL HEIDI A HOWARD Florida Community College at Jacksonville to accompany TRIGONOMETRY EIGHTH EDITION Margaret L Lial American River College John Hornsby University of New Orleans David I Schneider University of Maryland www.elsolucionario.net CONTENTS TRIGONOMETRIC FUNCTIONS 1.1 Angles 1.2 Angle Relationships and Similar Triangles 1.3 Trigonometric Functions 1.4 Using the Definitions of the Trigonometric Functions Chapter 1: Review Exercises Chapter 1: Test Chapter 1: Quantitative Reasoning 17 27 36 44 45 ACUTE ANGLES AND RIGHT TRIANGLES 2.1 Trigonometric Functions of Acute Angles 2.2 Trigonometric Functions of Non-Acute Angles 2.3 Finding Trigonometric Function Values Using a Calculator 2.4 Solving Right Triangles 2.5 Further Applications of Right Triangles Chapter 2: Review Exercises Chapter 2: Test Chapter 2: Quantitative Reasoning 47 57 66 74 86 97 108 111 RADIAN MEASURE AND CIRCULAR FUNCTIONS 3.1 Radian Measure 3.2 Applications of Radian Measure 3.3 The Unit Circle and Circular Functions 3.4 Linear and Angular Speed Chapter 3: Review Exercises Chapter 3: Test Chapter 3: Quantitative Reasoning 113 117 126 138 144 152 155 GRAPHS OF THE CIRCULAR FUNCTIONS 4.1 Graphs of the Sine and Cosine Functions 4.2 Translations of the Graphs of the Sine and Cosine Functions 4.3 Graphs of the Other Circular Functions Summary Exercises on Graphing Circular Functions 4.4 Harmonic Motion Chapter 4: Review Exercises Chapter 4: Test Chapter 4: Quantitative Reasoning www.elsolucionario.net 157 170 191 217 222 227 239 244 TRIGONOMETRIC IDENTITIES 5.1 Fundamental Identities 5.2 Verifying Trigonometric Identities 5.3 Sum and Difference Identities for Cosine 5.4 Sum and Difference Identities for Sine and Tangent 5.5 Double-Angle Identities 5.6 Half-Angle Identities Summary Exercises on Verifying Trigonometric Identities Chapter 5: Review Exercises Chapter 5: Test Chapter 5: Quantitative Reasoning 245 256 269 278 292 303 315 322 337 339 INVERSE CIRCULAR FUNCTIONS AND TRIGONOMETRIC EQUATIONS 6.1 Inverse Circular Functions 6.2 Trigonometric Equations I 6.3 Trigonometric Equations II 6.4 Equations Involving Inverse Trigonometric Functions Chapter 6: Review Exercises Chapter 6: Test Chapter 6: Quantitative Reasoning 341 358 373 384 395 406 410 APPLICATIONS OF TRIGONOMETRY AND VECTORS 7.1 Oblique Triangles and the Law of Sines 7.2 The Ambiguous Case and the Law of Sines 7.3 The Law of Cosines 7.4 Vectors, Operations, and the Dot Product 7.5 Applications of Vectors Chapter 7: Review Exercises Chapter 7: Test Chapter 7: Quantitative Reasoning 411 422 430 448 458 469 479 481 COMPLEX NUMBERS, POLAR EQUATIONS, AND PARAMETRIC EQUATIONS 8.1 Complex Numbers 8.2 Trigonometric (Polar) Form of Complex Numbers 8.3 The Product and Quotient Theorems 8.4 De Moivre’s Theorem; Powers and Roots of Complex Numbers 8.5 Polar Equations and Graphs 8.6 Parametric Equations, Graphs, and Applications Chapter 8: Review Exercises Chapter 8: Test Chapter 8: Quantitative Reasoning www.elsolucionario.net 483 490 498 505 521 540 554 564 568 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 9.1 Exponential Functions 9.2 Logarithmic Functions 9.3 Evaluating Logarithms; Equations and Applications Chapter 9: Review Exercises Chapter 9: Test Chapter 9: Quantitative Reasoning 569 583 592 603 609 612 APPENDIX A EQUATIONS AND INEQUALITIES 613 APPENDIX B GRAPHS OF EQUATIONS 623 APPENDIX C FUNCTIONS 633 APPENDIX D GRAPHING TECHNIQUES 639 www.elsolucionario.net www.elsolucionario.net Al Rubaee (a never never ending resource), I wish to thank Faiz Al John Samons (a consummate professional), and Amanda Nunley (a budding mathematician) for all their contributions to this supplement I also want to acknowledge the endless efforts of Joanne Ha, Sandra Scholten, Sheila Spinney, and Joe Vetere (aka Mr Bender) Thanks to you all www.elsolucionario.net www.elsolucionario.net Chapter TRIGONOMETRIC FUNCTIONS Section 1.1: Angles Answers will vary 1° represents Let x = the measure of the angle If the angle is its own complement, then we have the following x + x = 90 x = 90 x = 45 A 45° angle is its own complement Let x = the measure of the angle If the angle is its own supplement, then we have the following x + x = 180 2x = 180 x = 90 A 90° angle is its own supplement 11 360 of a rotation Therefore, 45° represents 30° (a) 90° − 30° = 60° (b) 180° − 30° = 150° 45 360 = 81 of a complete rotation 60° (a) 90° − 60° = 30° (b) 180° − 60° = 120° 45° (a) 90° − 45° = 45° (b) 180° − 45° = 135° 18° (a) 90° − 18° = 72° (b) 180° − 18° = 162° 54° (a) 90° − 54° = 36° (b) 180° − 54° = 126° 10 89° (a) 90° − 89° = 1° ° (b) 180° − 89° = 91° 25 minutes x 25 = ⇒ x = ( 360 ) = 25 ( ) = 150° 60 minutes 360° 60 12 Since the minute hand is 34 the way around, the hour hand is 34 of the way between the and Thus, the hour hand is located 8.75 minutes past 12 The minute hand is 15 minutes before the 12 The smaller angle formed by the hands of the clock can be found by solving the proportion (15 + 8.75 ) minutes x = 60 minutes 360° (15 + 8.75 ) minutes 60 minutes = x x 23.75 23.75 ⇒ = ⇒x= ( 360 ) = 23.75 ( ) = 142.5° 360° 60 360 60 13 The two angles form a straight angle x + 11x = 180 ⇒ 18 x = 180 ⇒ x = 10 The measures of the two angles are ( x ) ° =  (10 )  ° = 70° and (11x ) ° = 11(10 )  ° = 110° www.elsolucionario.net Chapter 1: Trigonometric Functions 14 The two angles form a right angle y + y = 90 ⇒ y = 90 ⇒ y = 15 The two angles have measures of ( y ) ° =  (15 )  ° = 60° and ( y ) ° =  (15 )  ° = 30° 15 The two angles form a right angle ( 5k + 5) + ( 3k + 5) = 90 ⇒ 8k + 10 = 90 ⇒ 8k = 80 ⇒ k = 10 The measures of the two angles are ( 5k + ) ° = 5 (10 ) + 5 ° = ( 50 + 5) ° = 55° and ( 3k + 5) ° = 3 (10 ) + 5 ° = ( 30 + 5) ° = 35° 16 The sum of the measures of two supplementary angles is 180° (10m + ) + ( 7m + 3) = 180 ⇒ 17m + 10 = 180 ⇒ 17m = 170 ⇒ m = 10 Since (10m + ) ° = 10 (10 ) +  ° = (100 + ) ° = 107° and ( 7m + 3) ° = 7 (10 ) + 3 ° = ( 70 + 3) ° = 73°, the angle measures are 107° and 73° 17 The sum of the measures of two supplementary angles is 180° ( x − ) + ( x − 12 ) = 180 ⇒ 14 x − 16 = 180 ⇒ 14 x = 196 ⇒ x = 14 Since ( x − ) ° =  (14 ) −  ° = ( 84 − ) ° = 80° and ( x − 12 ) ° = 8 (14 ) − 12 ° = (112 − 12 ) ° = 100°, the angle measures are 80° and 100° 18 The sum of the measures of two complementary angles is 90° ( z + ) + 3z = 90 ⇒ 12 z + = 90 ⇒ 12 z = 84 ⇒ z = Since ( z + ) ° = 9 ( ) +  ° = ( 63 + ) ° = 69° and ( 3z ) ° = 3 ( ) ° = 21°, the angle measures are 69° and 21° 19 If an angle measures x degrees and two angles are complementary if their sum is 90°, then the complement of an angle of x° is (90 − x)° 20 If an angle measures x degrees and two angles are supplementary if their sum is 180°, then the supplement of an angle of x° is (180 − x)° 21 The first negative angle coterminal with x between 0° and 60° is (x − 360)° 22 The first positive angle coterminal with x between 0° and –60° is (x + 360)° 23 62° 18′ +21° 41′ 25 71°18′ − 47° 29′ = 70° 78′ − 47° 29′ 70° 78′ −47° 29′ 83° 59′ 24 23° 49′ 75° 15′ + 83° 32′ 158° 47′ www.elsolucionario.net 634 Appendix C 11 Two sets of ordered pairs, namely (1,1) and (1, −1) as well as ( 2, ) and ( 2, −4 ) , have the same xvalue paired with different y-values, so the relation is not a function domain: {0,1, 2} ; range: {−4, −1, 0,1, 4} 12 The relation is a function because for each different x-value there is exactly one y-value This correspondence can be shown as follows domain: {2,3, 4,5} ; range: {5,7,9,11} 13 The relation is a function because for each different x-value there is exactly one y-value This correspondence can be shown as follows domain: {2,5,11,17} ; range: {1, 7, 20} 14 The relation is not a function because for each different x-value there is not exactly one y-value This correspondence can be shown as follows domain: {1, 2,3} ; range: {10,15, 20, 25} 15 This graph represents a function If you pass a vertical line through the graph, one xvalue corresponds to only one y-value domain: ( −∞, ∞ ) ; range: ( −∞, ∞ ) 16 This graph represents a function If you pass a vertical line through the graph, one xvalue corresponds to only one y-value domain: ( −∞, ∞ ) ; range: ( −∞, 4] domain: ( −∞, ∞ ) ; range: ( −∞, ∞ ) 19 This graph does not represent a function If you pass a vertical line through the graph, there are places where one value of x corresponds to two values of y domain: [ −4, 4] ; range: [ −3,3] 17 This graph does not represent a function If you pass a vertical line through the graph, there are places where one value of x corresponds to two values of y domain: [ 3, ∞ ) ; range: ( −∞, ∞ ) 18 This graph represents a function If you pass a vertical line through the graph, one xvalue corresponds to only one y-value 20 This graph represents a function If you pass a vertical line through the graph, one xvalue corresponds to only one y-value domain: [ −2, 2] ; range: [ 0, 4] www.elsolucionario.net Appendix C: Functions 635 21 y = x represents a function since y is always found by squaring x Thus, each value of x corresponds to just one value of y x can be any real number Since the square of any real number is not negative, the range would be zero or greater domain: ( −∞, ∞ ) ; range: [ 0, ∞ ) 22 y = x represents a function since y is always found by cubing x Thus, each value of x corresponds to just one value of y x can be any real number Since the cube of any real number could be negative, positive, or zero, the range would be any real number domain: ( −∞, ∞ ) ; range: ( −∞, ∞ ) 23 The ordered pairs (1,1) and (1, −1) both satisfy x = y This equation does not represent a function Because x is equal to the sixth power of y, the values of x are nonnegative Any real number can be raised to the sixth power, so the range of the relation is all real numbers domain: [ 0, ∞ ) range: ( −∞, ∞ ) 24 The ordered pairs (1,1) and (1, −1) both satisfy x = y This equation does not represent a function Because x is equal to the fourth power of y, the values of x are nonnegative Any real number can be raised to the fourth power, so the range of the relation is all real numbers domain: [ 0, ∞ ) range: ( −∞, ∞ ) 25 y = x − represents a function since y is found by multiplying x by and subtracting Each value of x corresponds to just one value of y x can be any real number, so the domain is all real numbers Since y is twice x, less 6, y also may be any real number, and so the range is also all real numbers domain: ( −∞, ∞ ) ; range: ( −∞, ∞ ) www.elsolucionario.net 636 Appendix C 26 y = −6 x + represents a function since y is found by multiplying x by −6 and adding Each value of x corresponds to just one value of y x can be any real number, so the domain is all real numbers Since y is −6 times x, plus 8, y also may be any real number, and so the range is also all real numbers domain: ( −∞, ∞ ) ; range: ( −∞, ∞ ) 27 For any choice of x in the domain of y = x , there is exactly one corresponding value of y, so this equation defines a function Since the quantity under the square root cannot be negative, we have x ≥ Because the radical is nonnegative, the range is also zero or greater domain: [ 0, ∞ ) ; range: [ 0, ∞ ) 28 For any choice of x in the domain of y = − x, there is exactly one corresponding value of y, so this equation defines a function Since the quantity under the square root cannot be negative, we have x ≥ The outcome of the radical is nonnegative, when you change the sign (by multiplying by −1 ), the range becomes nonpositive Thus the range is zero or less domain: [ 0, ∞ ) ; range: ( −∞,0] 29 For any choice of x in the domain of y = x + 2, there is exactly one corresponding value of y, so this equation defines a function Since the quantity under the square root cannot be negative, we have x + ≥ ⇒ x ≥ −2 ⇒ x ≥ −42 or x ≥ − 21 Because the radical is nonnegative, the range is also zero or greater domain:  − 12 , ∞ ) ; range: [ 0, ∞ ) 30 For any choice of x in the domain of y = − 2x , there is exactly one corresponding value of y, so this equation defines a function Since the quantity under the square root cannot be negative, we have − x ≥ ⇒ −2 x ≥ −9 ⇒ x ≤ −−92 or x ≤ 92 Because the radical is nonnegative, the range is also zero or greater domain: ( −∞, 92  ; range: [ 0, ∞ ) www.elsolucionario.net Appendix C: Functions 637 31 Given any value in the domain of y = x 2−9 , we find y by subtracting 9, then dividing into This process produces one value of y for each value of x in the domain, so this equation is a function The domain includes all real numbers except those that make the denominator equal to zero, namely x = Values of y can be negative or positive, but never zero Therefore, the range will be all real numbers except zero domain: ( −∞,9 ) ∪ ( 9, ∞ ) ; range: ( −∞,0 ) ∪ ( 0, ∞ ) 32 Given any value in the domain of y = −7 , x −16 we find y by subtracting 16, then dividing into −7 This process produces one value of y for each value of x in the domain, so this equation is a function The domain includes all real numbers except those that make the denominator equal to zero, namely x = 16 Values of y can be negative or positive, but never zero Therefore, the range will be all real numbers except zero domain: ( −∞,16 ) ∪ (16, ∞ ) ; range: ( −∞,0 ) ∪ ( 0, ∞ ) 33 B 34 Answers will vary An example is: The cost of gasoline depends on the number of gallons used; so cost is a function of number of gallons 35 f ( x ) = −3x + f ( ) = −3 ⋅ + = + = 36 f ( x ) = −3 x + 40 g ( x ) = − x + x + g ( k ) = − k + 4k + 41 f ( −3) = −3 ( −3) + = + = 13 37 g ( x ) = − x2 + 4x + f ( − x ) = −3 ( − x ) + = x + 42 g ( −2 ) = − ( −2 ) + ( −2 ) + g ( x ) = − x2 + x + g ( x ) = − x2 + 4x + g (−x) = − (−x) + (−x) +1 = −4 + ( −8 ) + = −11 38 f ( x ) = −3 x + = − x2 − 4x + 43 f ( x ) = −3 x + f ( a + ) = −3 ( a + ) + g (10 ) = −102 + ⋅10 + = −3a − 12 + = −3a − = −100 + 40 + = −59 39 f ( x ) = −3x + f ( p ) = −3 p + 44 f ( x ) = −3 x + f ( m − ) = −3 ( m − ) + = −6m + + = −6m + 13 www.elsolucionario.net 638 Appendix C 45 (a) f ( 2) = (b) f ( −1) = 46 (a) f (2) = (b) f ( −1) = 11 (c) 47 (a) f (2) = (b) f ( −1) = −3 50 (a) 48 (a) f ( ) = −3 (b) f ( −1) = 51 (a) f ( ) = 11 (b) f ( ) = 49 (a) (c) f ( −2 ) = (b) f ( ) = f (1) = (d) f ( ) = f ( −2 ) = (b) f ( ) = f (1) = (d) f ( ) = (c) Since f ( −2 ) = 0, a = −2 (d) Since f ( ) = f ( ) = f (8 ) = 10, x = 2, 7,8 (e) We have (8, f (8 ) ) = (8,10 ) and (10, f (10 ) ) = (10,0 ) Using the distance formula we have d= 52 (a) (10 − 8) + ( − 10 ) = 22 + ( −10 ) = + 100 = 104 2 f ( ) = 1.2 (b) Since f ( 5) = −2.4, x = (c) Since f ( ) = 3.6, the graph of f intersects the y-axis at ( 0,3.6 ) (d) Since f ( 3) = 0, the graph of f intersects the x-axis at ( 3,0 ) [ 4, ∞ ) [ −1, 4] (b) (b) (c) ( −∞,1] [1, 4] 55 (a) ( −∞, 4] (b) 53 (a) (c) 54 (a) (c) none ( −∞, −1] 56 (a) none (b) ( −∞, ∞ ) (b) ( −∞, −3) (c) none [ 4, ∞ ) [ 4, ∞ ) 57 (a) none (b) ( −∞, −2] ; [ 3, ∞ ) (c) ( −2,3) 58 (a) ( 3, ∞ ) ( −3,3] (c) 59 (a) yes (b) [ 0, 24] (c) When t = 8, y = 1200 from the graph At A.M., approximately 1200 megawatts is being used (d) at 17 hr or P.M.; at A.M (e) f (12 ) = 2000; At 12 noon, electricity use is 2000 megawatts (f) increasing from A.M to P.M.; decreasing from midnight to A.M and from P.M to midnight 60 (a) At t = 8, y = 24 from the graph Therefore, there are 24 units of the drug in the bloodstream at hours (b) Level increases between and hours and decreases between and 12 hours (c) The coordinates of the highest point are (2, 64) Therefore, at hours, the level of the drug in the bloodstream reaches its greatest value of 64 units (d) After the peak, y = 16 at t = 10 10 hours – hours = hours after the peak additional hours are required for the level to drop to 16 units www.elsolucionario.net Appendix D GRAPHING TECHNIQUES (a) B; y = ( x − 7)2 is a shift of y = x , units to the right (b) E; y = x − is a shift of y = x , units downward (c) F; y = x is a vertical stretch of y = x , by a factor of (d) A; y = ( x + 7)2 is a shift of y = x , units to the left (e) D; y = x + is a shift of y = x , units upward (f) C; y = x is a vertical shrink of y = x (a) E; y = x is a vertical stretch of y = x , by a factor of (b) C; y = − x is a reflection of y = x , over the x-axis (c) D; y = ( − x ) is a reflection of y = x , over the y-axis (d) A; y = ( x + ) is a shift of y = x , units to the left (e) B; y = x + is a shift of y = x , units up (a) B; y = x + is a shift of y = x , units upward (b) A; y = x − is a shift of y = x , units downward (c) G; y = ( x + 2)2 is a shift of y = x , units to the left (d) C; y = ( x − 2)2 is a shift of y = x , units to the right (e) F; y = x is a vertical stretch of y = x , by a factor of (f) D; y = − x is a reflection of y = x , across the x-axis (g) H; y = ( x − 2) + is a shift of y = x , units to the right and unit upward (h) E; y = ( x + 2)2 + is a shift of y = x , units to the left and unit upward y = 3x x y = x2 y = 3x −2 12 −1 0 1 12 639 www.elsolucionario.net 640 Appendix D y = x2 x y = x2 y = x2 −2 16 −1 0 1 4 16 x y = x2 y = 13 x −3 −2 4 −1 1 0 1 4 3 x y = x2 y = 23 x −3 −2 −1 0 1 3 y = 13 x y = 23 x www.elsolucionario.net Appendix D: Graphing Techniques 641 y = − 12 x x y = x2 y = − 12 x −3 − 92 −2 −2 −1 − 12 0 1 − 12 −2 − 92 x y = x2 y = −3 x −3 −27 −2 −12 −1 −3 0 1 −3 −12 −27 x −x y = −x y = −x 0 0 −1 1 −4 4 −9 −16 16 y = −3 x 10 y = − x www.elsolucionario.net 642 Appendix D 11 y = −2 x x −2x y = −2 x 0 − 12 1 −2 − 92 −8 16 12 (a) y = f ( x + ) is a horizontal translation of f, units to the left The point that corresponds to (8,12 ) on this translated function would be (8 − 4,12 ) = ( 4,12 ) (b) y = f ( x ) + is a vertical translation of f, units up The point that corresponds to (8,12 ) on this translated function would be (8,12 + ) = ( 8,16 ) 13 (a) y = f ( x ) is a vertical shrinking of f, by a factor of this translated function would be (8, 14 ⋅ 12 ) = (8,3) The point that corresponds to (8,12 ) on (b) y = f ( x ) is a vertical stretching of f, by a factor of The point that corresponds to (8,12 ) on this translated function would be (8, ⋅ 12 ) = (8, 48 ) 14 (a) The point that corresponds to (8,12 ) when reflected across the x-axis would be (8, −12 ) (b) The point that corresponds to (8,12 ) when reflected across the y-axis would be ( −8,12 ) 15 (a) The point that is symmetric to (5, –3) with respect to the x-axis is (5, 3) (b) The point that is symmetric to (5, –3) with respect to the y-axis is (–5, –3) (c) The point that is symmetric to (5, –3) with respect to the origin is (–5, 3) 16 (a) The point that is symmetric to (– 6, 1) with respect to the x-axis is (– 6, –1) (b) The point that is symmetric to (– 6, 1) with respect to the y-axis is (6, 1) (c) The point that is symmetric to (– 6, 1) with respect to the origin is (6, –1) www.elsolucionario.net Appendix D: Graphing Techniques 643 17 (a) The point that is symmetric to (– 4, –2) with respect to the x-axis is (– 4, 2) (b) The point that is symmetric to (– 4, –2) with respect to the y-axis is (4, –2) (c) The point that is symmetric to (– 4, –2) with respect to the origin is (4, 2) 18 (a) The point that is symmetric to (–8, 0) with respect to the x-axis is (–8, 0), since this point lies on the x-axis (b) The point that is symmetric to the point (–8, 0) with respect to the y-axis is (8, 0) (c) The point that is symmetric to the point (–8, 0) with respect to the origin is (8, 0) 19 y = x + Replace x with –x to obtain y = (− x)2 + = x + The result is the same as the original equation, so the graph is symmetric with respect to the y-axis Since y is a function of x, the graph cannot be symmetric with respect to the x-axis Replace x with –x and y with –y to obtain − y = ( − x )2 + ⇒ − y = x + ⇒ y = − x − The result is not the same as the original equation, so the graph is not symmetric with respect to the origin Therefore, the graph is symmetric with respect to the y-axis only 20 y = x − Replace x with –x to obtain y = 2(− x) − = x − The result is the same as the original equation, so the graph is symmetric with respect to the y-axis Since y is a function of x, the graph cannot be symmetric with respect to the x-axis Replace x with –x and y with –y to obtain – y = 2( − x )4 − − y = 2x4 −1 y = −2 x + The result is not the same as the original equation, so the graph is not symmetric with respect to the origin Therefore, the graph is symmetric with respect to the y-axis only 21 x + y = 10 Replace x with –x to obtain the following ( − x ) + y = 10 x + y = 10 The result is the same as the original equation, so the graph is symmetric with respect to the y-axis Replace y with –y to obtain the following x + ( − y )2 = 10 x + y = 10 The result is the same as the original equation, so the graph is symmetric with respect to the x-axis Since the graph is symmetric with respect to the x-axis and y-axis, it is also symmetric with respect to the origin www.elsolucionario.net 644 Appendix D 22 y = −5 x2 Replace x with –x to obtain −5 y2 = (− x)2 −5 y2 = x The result is the same as the original equation, so the graph is symmetric with respect to the y-axis Replace y with –y to obtain −5 (− y ) = x −5 y = x The result is the same as the original equation, so the graph is symmetric with respect to the x-axis Since the graph is symmetric with respect to the x-axis and y-axis, it is also symmetric with respect to the origin Therefore, the graph is symmetric with respect to the x-axis, the y-axis, and the origin 24 y = x3 − x Replace x with –x to obtain 23 y = −3x3 Replace x with –x to obtain y = −3(− x)3 y = −3(− x3 ) y = 3x3 The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis Replace y with –y to obtain − y = −3 x y = 3x3 The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis Replace x with –x and y with –y to obtain − y = −3(− x)3 − y = −3(− x3 ) − y = x3 y = −3 x The result is the same as the original equation, so the graph is symmetric with respect to the origin Therefore, the graph is symmetric with respect to the origin only 25 y = x − x + Replace x with –x to obtain y = ( − x )3 − ( − x ) y = ( − x ) − (− x ) + y = − x3 + x The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis Replace y with –y to obtain y = x + x + The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis Since y is a function of x, the graph cannot be symmetric with respect to the x-axis Replace x with –x and y with –y to obtain − y = x3 − x y = − x3 + x The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis Replace x with –x and y with –y to obtain − y = ( − x )3 − ( − x ) − y = − x3 + x y = x3 − x The result is the same as the original equation, so the graph is symmetric with respect to the origin Therefore, the graph is symmetric with respect to the origin only − y = (− x) − (− x) + − y = x2 + x + y = − x − x − The result is not the same as the original equation, so the graph is not symmetric with respect to the origin Therefore, the graph has none of the listed symmetries www.elsolucionario.net Appendix D: Graphing Techniques 645 26 y = x + 12 Replace x with –x to obtain the following y = ( − x ) + 12 y = − x + 12 The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis Since y is a function of x, the graph cannot be symmetric with respect to the x-axis Replace x with –x and y with –y to obtain the following − y = ( − x ) + 12 y = x − 12 The result is not the same as the original equation, so the graph is not symmetric with respect to the origin Therefore, the graph has none of the listed symmetries 27 y = x − This graph may be obtained by translating the graph of y = x , unit downward 28 y = x + This graph may be obtained by translating the graph of y = x , units upward 29 y = x + This graph may be obtained by translating the graph of y = x , units upward 30 y = x − This graph may be obtained by translating the graph of y = x , units downward www.elsolucionario.net 646 Appendix D 31 y = ( x − 1) This graph may be obtained by translating the graph of y = x , units to the right 32 y = ( x − ) This graph may be obtained by translating the graph of y = x , units to the right 33 y = ( x + ) This graph may be obtained by translating the graph of y = x , units to the left 34 y = ( x + 3) This graph may be obtained by translating the graph of y = x , units to the left 35 y = ( x + 3)2 − This graph may be obtained by translating the graph of y = x , units to the left and units down www.elsolucionario.net Appendix D: Graphing Techniques 647 36 y = ( x − 5)3 − This graph can be obtained by translating the graph of y = x , units to the right and units down 37 y = x − This graph may be obtained by translating the graph of y = x , unit down It is then stretched vertically by a factor of 38 y = 23 ( x − 2)2 This graph may be obtained by translating the graph of y = x , units to the right It is then shrunk vertically by a factor of 39 f ( x ) = 2( x − 2)2 − This graph may be obtained by translating the graph of y = x , units to the right and units down It is then stretched vertically by a factor of 40 f ( x ) = −3( x − 2)2 + This graph may be obtained by translating the graph of y = x , units to the right and unit up It is then stretched vertically by a factor of and reflected over the x-axis www.elsolucionario.net 648 Appendix D 41 It is the graph of f ( x ) = x translated unit to the left, reflected across the x-axis, and translated units up The equation is y = − x + + 42 It is the graph of g ( x ) = x translated units to the left, reflected across the x-axis, and translated units up The equation is y = − x + + 43 It is the graph of g ( x ) = x translated units to the left, stretched vertically by a factor of 2, and translated units down The equation is y = x + − 44 It is the graph of f ( x ) = x translated units to the right, shrunken vertically by a factor of translated unit down The equation is y = 1, and x − − 45 f(x) = 2x + Translate the graph of f ( x ) up units to obtain the graph of t ( x ) = (2 x + 5) + = x + Now translate the graph of t(x) = 2x + left units to obtain the graph of g ( x ) = 2( x + 3) + = x + + = x + 13 (Note that if the original graph is first translated to the left units and then up units, the final result will be the same.) 46 f(x) = – x Translate the graph of f ( x ) down units to obtain the graph of t ( x ) = (3 − x ) − = − x + Now translate the graph of t ( x ) = − x + right units to obtain the graph of g ( x ) = −( x − 3) + = − x + + = − x + (Note that if the original graph is first translated to the right units and then down units, the final result will be the same.) 47 Answers will vary There are four possibilities for the constant, c i) c > c > The graph of F ( x ) is stretched vertically by a factor of c ii) c > c < The graph of F ( x ) is shrunk vertically by a factor of c iii) c < iv) c < c > The graph of F ( x ) is stretched vertically by a factor of −c and reflected over the x-axis c < The graph of F ( x ) is shrunk vertically by a factor of −c and reflected over the x-axis 48 The graph of y = F ( x + h ) represents a horizontal shift of the graph of y = F ( x ) If h > 0, it is a shift to the left h units If h < 0, it is a shift to the left −h units (h is negative) The graph of y = F ( x ) + h is not the same as the graph of y = F ( x + h ) The graph of y = F ( x ) + h represents a vertical shift of the graph of y = F ( x ) www.elsolucionario.net ... Exercises Chapter 6: Test Chapter 6: Quantitative Reasoning 341 358 373 384 395 406 410 APPLICATIONS OF TRIGONOMETRY AND VECTORS 7.1 Oblique Triangles and the Law of Sines 7.2 The Ambiguous Case and the