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solution manual for Graphical Approach to College Algebra 4th Edition by Hornsby John Hornsby, Margaret L.Lial, Gary K.Rockswold Solution manual for Graphical Approach toto College Algebra 4th Edition by John Rockswold solution manual for Graphical Approach College Algebra 4th Edition by John Lial Hornsby, Margaret L.Lial, Gary link full link fulllink download full download: K.Rockswold download:https://getbooksolutions.com/download/solution-manual-for-graphical-approach-to-college-algebra-4th-edition-by-john-hornsby-lial-rockswold/ Full file at https://fratstock.eu https://getbooksolutions.com/download/solution-manual-for-graphical-approach-to-college-algebra-4th-edition-by-john-hornsby-lial-rock link full:https://getbooksolutions.com/download/solution-manual-for-graphical-approach-to-college-algebra-4th-edition-by-john-hornsby-lial-rockswold/ download link full download: https://getbooksolutions.com/download/solution-manual-for-graphical-approach-to-college-algebra-4th-edition-by-john-hornsby-lial-rockswold/ Chapter 2: Analysis of Graphs of Functions 2.1: Graphs of Basic Functions and Relations; Symmetry 1᎐ q, q2 1᎐ q, q2 ; 30, q2 10, 02 30, q2; 30, q2 increases 1᎐ q, 04; 30, q2 x-axis even odd 10 y-axis; origin 11 The domain can be all real numbers, therefore the function is continuous for the interval: 1᎐ q, q2 12 The domain can be all real numbers, therefore the function is continuous for the interval: 1᎐ q, q2 13 The domain can only be values where x Ն 0, therefore the function is continuous for the interval: 30, q2 14 The domain can only be values where x Յ 0, therefore The function is continuous for the interval: 1᎐ q, 04 15 The domain can be all real numbers except ᎐ 3, therefore the function is continuous for the interval: 1᎐ q, ᎐ 32; 1᎐ 3, q2 16 The domain can be all real numbers except 1, therefore the function is continuous for the interval: 1᎐ q, 12; 11, q2 17 (a) The function is increasing for the interval: 33, q2 (b) The function is decreasing for the interval: 1᎐ q, 34 (c) The function is never constant, therefore: none (d) The domain can be all real numbers, therefore the interval: 1᎐ q, q2 (e) The range can only be values where y Ն 0, therefore the interval: 30, q2 18 (a) The function is increasing for the interval: 34, q2 (b) The function is decreasing for the interval: 1᎐ q, ᎐ 14 (c) The function is constant for the interval: 3᎐ 1, 44 (d) The domain can be all real numbers, therefore the interval: 1᎐ q, q2 (e) The range can only be values where y Ն 3, therefore the interval: 33, q2 19 (a) The function is increasing for the interval: 1᎐ q, 14 (b) The function is decreasing for the interval: 34, q2 (c) The function is constant for the interval: 31, 44 (d) The domain can be all real numbers, therefore the interval: 1᎐ q, q2 (e) The range can only be values where y Յ 3, therefore the interval: 1᎐ q, 34 Full file at https://fratstock.eu 58 CHAPTER Analysis of Graphs of Functions 20 (a) The function never is increasing, therefore: none (b) The function is always decreasing, therefore the interval: 1᎐ q, q2 (c) The function is never constant, therefore: none (d) The domain can be all real numbers, therefore the interval: 1᎐ q, q2 (e) The range can be all real numbers, therefore the interval: 1᎐ q, q2 21 (a) The function never is increasing, therefore: none (b) The function is decreasing for the intervals: 1᎐ q, ᎐ 4; 33, q2 (c) The function is constant for the interval: 1᎐ 2, 32 (d) The domain can be all real numbers, therefore the interval: 1᎐ q, q2 (e) The range can only be values where y Յ 1.5 or y Ն 2, therefore the interval: 1᎐ q, 1.54 ´ 32, q2 22 (a) The function is increasing for the interval: 13, q2 (b) The function is decreasing for the interval: 1᎐ q, ᎐ 32 (c) The function is constant for the interval: 1᎐ 3, 34 (d) The domain can be all real numbers except ᎐ 3, therefore the interval: 1᎐ q, ᎐ 32 ´ 1᎐ 3, q2 (e) The range can only be values where y 1, therefore the interval: 11, q2 23 Graph f 1x2 ϭ x 5, See Figure 23 As x increases for the interval: 1᎐ q, q2 , y increases, therefore increasing 24 Graph f 1x2 ϭ ᎐ x 3, See Figure 24 As x increases for the interval: 1᎐ q, q2, y decreases, therefore decreasing 25 Graph f 1x2 ϭ x 4, See Figure 25 As x increases for the interval: 1᎐ q, 04, y decreases, therefore decreasing 26 Graph f 1x2 ϭ x 4, See Figure 26 As x increases for the interval: 30, q2 , y increases, therefore increasing 3᎐10, 10 by 3᎐10, 10 Xscl ϭ Yscl ϭ 3᎐10, 10 by ᎐10, 10 Xscl ϭ Yscl ϭ 3᎐10, 10 by 3᎐10, 10 Xscl ϭ Yscl ϭ 3᎐10, 10 by ᎐10, 104 Xscl ϭ Yscl ϭ Figure 23 Figure 24 Figure 25 Figure 26 27 Graph f 1x2 ϭ ᎐ x See Figure 27 As x increases for the interval: 1᎐ q, 04, y increases, therefore increasing 28 Graph f 1x2 ϭ ᎐ x , See Figure 28 As x increases for the interval: 30, q2, y decreases, therefore decreasing 29 Graph f 1x2 ϭ ᎐ x, See Figure 29 As x increases for the interval: 1᎐ q, q2, y decreases, therefore decreasing 30 Graph f 1x2 ϭ ᎐ 1x, See Figure 30 As x increases for the interval: 30, q2, y decreases, therefore decreasing 3᎐10, 10 by 3᎐10, 10 Xscl ϭ Yscl ϭ 3᎐10, 10 by ᎐10, 10 Xscl ϭ Yscl ϭ 3᎐10, 10 by 3᎐10, 10 Xscl ϭ Yscl ϭ 3᎐10, 10 by ᎐10, 104 Xscl ϭ Yscl ϭ Figure 27 Figure 28 Figure 29 Figure 30 Full file at https://fratstock.eu Graphs of Basic Functions and Relations; Symmetry SECTION 2.1 59 31 Graph f 1x2 ϭ Ϫ x 3, See Figure 31 As x increases for the interval: 1᎐ q, q2, y decreases, therefore decreasing 32 Graph f 1x2 ϭ x Ϫ 2x, See Figure 32 As x increases for the interval: 31, q2, y increases, therefore increasing 33 Graph f 1x2 ϭ Ϫ x See Figure 33 As x increases for the interval: 1᎐ q, 04, y increases, therefore increasing 34 Graph f 1x2 ϭ x ϩ , See Figure 34 As x increases for the interval: 1᎐ q, ᎐ 14, y decreases, therefore decreasing 3᎐10, 10 by 3᎐10, 10 Xscl ϭ Yscl ϭ 3᎐10, 10 by ᎐10, 10 Xscl ϭ Yscl ϭ ᎐10, 104 by 3᎐10, 10 Xscl ϭ Yscl ϭ 3᎐10, 10 by ᎐10, 104 Xscl ϭ Yscl ϭ Figure 31 Figure 32 Figure 33 Figure 34 35 (a) No (b) Yes (c) No 36 (a) Yes (b) No (c) No 37 (a) Yes (b) No (c) No 38 (a) No (b) No (c) Yes 39 (a) Yes (b) Yes (c) Yes 40 (a) Yes (b) Yes (c) Yes 41 (a) No (b) No (c) Yes 42 (a) No (b) Yes (c) No 43 If f is an even function then f 1᎐ x2 ϭ f 1x2 or opposite domains have the same range See Figure 43 44 If g is an odd function then g1᎐ x2 ϭ ᎐ g1x2 or opposite domains have the opposite range See Figure 44 x –3 –2 –1 x g(x) –5 13 –3 –2 –5 0 –1 –13 ƒ(x) 21 –12 –25 –25 –12 21 Figure 43 Figure 44 45 (a) Since f 1᎐ x2 ϭ f 1x2, this is an even function and is symmetric with respect to the y-axis See Figure 45a (b) Since f 1᎐ x2 ϭ ᎐ f 1x2, this is an odd function and is symmetric with respect to the origin See Figure 45b y y 2 x Figure 45a x Figure 45b Full file at https://fratstock.eu 60 CHAPTER Analysis of Graphs of Functions 46 (a) Since this is an odd function the graph is symmetric with respect to the origin See Figure 46a (b) Since this is an even function the graph is symmetric with respect to the y-axis See Figure 46b y y 2 x Figure 46a x Figure 46b 47 Since f 1᎐ x2 ϭ f 1x2 , it is even 48 Since f 1᎐ x2 ϭ ᎐ f 1x2 , it is odd 49 If f 1x2 ϭ x Ϫ 7x ϩ 6, then f 1᎐ x2 ϭ 1᎐ x2 Ϫ 71᎐ x2 ϩ f 1᎐ x2 ϭ x Ϫ 7x ϩ Since f 1᎐ x2 ϭ f 1x2 , the function is even 50 If f 1x2 ϭ ᎐ 2x Ϫ 8x , then f 1᎐ x2 ϭ ᎐ 21᎐ x2 Ϫ 81᎐ x2 f 1᎐ x2 ϭ ᎐ 2x Ϫ 8x Since f 1᎐ x2 ϭ f 1x2 , the function is even 51 If f 1x2 ϭ x Ϫ 4x ϩ 5, then f 1᎐ x2 ϭ 1᎐ x2 Ϫ 41᎐ x2 ϩ f 1᎐ x2 ϭ x Ϫ 4x ϩ Since f 1᎐ x2 ϭ f 1x2 , the function is even 52 If f 1x2 ϭ 8, then f 1᎐ x2 ϭ Since f 1᎐ x2 ϭ f 1x2 , the function is even 53 If f 1x2 ϭ 5x , then f 1᎐ x2 ϭ 51᎐ x2 f 1᎐ x2 ϭ 5x Since f 1᎐ x2 ϭ f 1x2 , the function is even 54 If f 1x2 ϭ 2x ϩ 1, then f 1᎐ x2 ϭ 21᎐ x2 ϩ 1 f 1᎐ x2 ϭ 2x ϩ Since f 1᎐ x2 ϭ f 1x2 , the function is even 55 If f 1x2 ϭ 3x Ϫ x, then f 1᎐ x2 ϭ 31᎐ x2 Ϫ 1᎐ x2 f 1᎐ x2 ϭ ᎐ 3x ϩ x and ᎐ f 1x2 ϭ ᎐ 13x Ϫ x2 ᎐ f 1x2 ϭ ᎐ 3x ϩ x Since f 1᎐ x2 ϭ ᎐ f 1x2 , the function is odd 56 If f 1x2 ϭ ᎐ x ϩ 2x Ϫ 3x, then f 1᎐ x2 ϭ ᎐ 1᎐ x2 ϩ 21᎐ x2 Ϫ 31᎐ x2 f 1᎐ x2 ϭ x Ϫ 2x ϩ 3x and ᎐ f 1x2 ϭ ᎐ 1᎐ x ϩ 2x Ϫ 3x2 ᎐ f 1x2 ϭ x Ϫ 2x ϩ 3x Since f 1᎐ x2 ϭ ᎐ f 1x2 , the function is odd 57 If f 1x2 ϭ 3x Ϫ x ϩ 7x, then f 1᎐ x2 ϭ 31᎐ x2 Ϫ 1᎐ x2 ϩ 71᎐ x2 f 1᎐ x2 ϭ ᎐ 3x ϩ x Ϫ 7x and ᎐ f 1x2 ϭ ᎐ 13x Ϫ x ϩ 7x2 ᎐ f 1x2 ϭ ᎐ 3x ϩ x Ϫ 7x Since f 1᎐ x2 ϭ ᎐ f 1x2 , the function is odd 58 If f 1x2 ϭ x Ϫ 4x, then f 1᎐ x2 ϭ 1᎐ x2 Ϫ 41᎐ x2 f 1᎐ x2 ϭ ᎐ x ϩ 4x and ᎐ f 1x2 ϭ ᎐ 1x Ϫ 4x2 ᎐ f 1x2 ϭ ᎐ x ϩ 4x Since f 1᎐ x2 ϭ ᎐ f 1x2 , the function is odd 59 If f 1x2 ϭ 1 1 1 f 1᎐ x2 ϭ ᎐ and ᎐ f 1x2 ϭ ᎐ a b ᎐ f 1x2 ϭ ᎐ , then f 1᎐ x2 ϭ Since 21᎐ x2 2x 2x 2x 2x f 1᎐ x2 ϭ ᎐ f 1x2 , the function is odd 1 1 f 1᎐ x2 ϭ ᎐ 4x ϩ and 60 If f 1x2 ϭ 4x Ϫ , then f 1᎐ x2 ϭ 41᎐ x2 Ϫ x x 1᎐ x2 1 ᎐ f 1x2 ϭ ᎐ a4x Ϫ b ᎐ f 1x2 ϭ ᎐ 4x ϩ Since f 1᎐ x2 ϭ ᎐ f 1x2 , the function is odd x x Full file at https://fratstock.eu Graphs of Basic Functions and Relations; Symmetry SECTION 2.1 61 61 If f 1x2 ϭ ᎐ x ϩ 2x, then f 1᎐ x2 ϭ ᎐ 1᎐ x2 ϩ 21᎐ x2 f 1᎐ x2 ϭ x Ϫ 2x and ᎐ f 1x2 ϭ ᎐ 1᎐ x ϩ 2x2 ᎐ f 1x2 ϭ x Ϫ 2x Since f 1᎐ x2 ϭ ᎐ f 1x2 , the function is symmetric with respect to the origin Graph f 1x2 ϭ ᎐ x ϩ 2x The graph supports symmetry with respect to the origin 62 If f 1x2 ϭ x Ϫ 2x 3, then f 1᎐ x2 ϭ 1᎐ x2 Ϫ 21᎐ x2 f 1᎐ x2 ϭ ᎐ x ϩ 2x and ᎐ f 1x2 ϭ ᎐ 1x Ϫ 2x ᎐ f 1x2 ϭ ᎐ x ϩ 2x Since f 1᎐ x2 ϭ ᎐ f 1x2 , the function is symmetric with respect to the origin Graph f 1x2 ϭ x Ϫ 2x The graph supports symmetry with respect to the origin 63 If f 1x2 ϭ x Ϫ 2x ϩ 1, then f 1᎐ x2 ϭ 1᎐ x2 Ϫ 21᎐ x2 ϩ 1 f 1᎐ x2 ϭ x Ϫ 2x ϩ Since f 1᎐ x2 ϭ f 1x2 , the function is symmetric with respect to the y-axis Graph f 1x2 ϭ x Ϫ 2x ϩ The graph supports symmetry with respect to the y-axis 64 If f 1x2 ϭ 75 x ϩ x ϩ 1, then f 1᎐ x2 ϭ 75 1᎐ x2 ϩ 1᎐ x2 ϩ 1 f 1᎐ x2 ϭ 75 x ϩ x ϩ Since f 1᎐ x2 ϭ f 1x2 , the function is symmetric with respect to the y-axis Graph f 1x2 ϭ 75 x ϩ x ϩ The graph supports symmetry with respect to the y-axis 65 If f 1x2 ϭ x Ϫ x ϩ 3, then f 1᎐ x2 ϭ 1᎐ x2 Ϫ 1᎐ x2 ϩ f 1᎐ x2 ϭ ᎐ x ϩ x ϩ and ᎐ f 1x2 ϭ ᎐ 1x Ϫ x ϩ 32 ᎐ f 1x2 ϭ ᎐ x ϩ x Ϫ Since f 1x2 f 1᎐ x2 ᎐ f 1x2 , the function is not symmetric with respect to the y-axis or origin Graph f 1x2 ϭ x Ϫ x ϩ The graph supports no symmetry with respect to the y-axis or origin 66 If f 1x2 ϭ x Ϫ 5x ϩ 2, then f 1᎐ x2 ϭ 1᎐ x2 Ϫ 51᎐ x2 ϩ f 1᎐ x2 ϭ x ϩ 5x ϩ and ᎐ f 1x2 ϭ ᎐ 1x Ϫ 5x ϩ 22 ᎐ f 1x2 ϭ ᎐ x ϩ 5x Ϫ Since f 1x2 f 1᎐ x2 ᎐ f 1x2 , the function is not symmetric with respect to the y-axis or origin Graph f 1x2 ϭ x Ϫ 5x ϩ The graph supports no symmetry with respect to the y-axis or origin 67 If f 1x2 ϭ x Ϫ 4x 3, then f 1᎐ x2 ϭ 1᎐ x2 Ϫ 41᎐ x2 f 1᎐ x2 ϭ x ϩ 4x and ᎐ f 1x2 ϭ ᎐ 1x Ϫ 4x ᎐ f 1x2 ϭ ᎐ x ϩ 4x Since f 1x2 f 1᎐ x2 ᎐ f 1x2 , the function is not symmetric with respect to the y-axis or ori- gin Graph f 1x2 ϭ x Ϫ 4x The graph supports no symmetry with respect to the y-axis or origin 68 If f 1x2 ϭ x Ϫ 3x, then f 1᎐ x2 ϭ 1᎐ x2 Ϫ 31᎐ x2 f 1᎐ x2 ϭ ᎐ x ϩ 3x and ᎐ f 1x2 ϭ ᎐ 1x Ϫ 3x2 ᎐ f 1x2 ϭ ᎐ x ϩ 3x Since f 1᎐ x2 ϭ ᎐ f 1x2 , the function is symmetric with respect to the origin Graph f 1x2 ϭ x Ϫ 3x The graph supports symmetry with respect to the origin 69 If f 1x2 ϭ ᎐ 6, then f 1᎐ x2 ϭ ᎐ Since f 1᎐ x2 ϭ f 1x2 , the function is symmetric with respect to the y-axis Graph f 1x2 ϭ ᎐ The graph supports symmetry with respect to the y-axis 70 If f 1x2 ϭ ᎐ x ϭ x , then f 1᎐ x2 ϭ ᎐ 1᎐ x2 ϭ x f 1᎐ x2 ϭ x Since f 1᎐ x2 ϭ f 1x2 , the function is symmetric with respect to the y-axis Graph f 1x2 ϭ ᎐ x The graph supports symmetry with respect to the y-axis 71 If f 1x2 ϭ 1 1 , then f 1᎐ x2 ϭ f 1᎐ x2 ϭ ᎐ and ᎐ f 1x2 ϭ ᎐ a b ᎐ f 1x2 ϭ ᎐ Since 3 41᎐ x2 4x 4x 4x 4x f 1᎐ x2 ϭ ᎐ f 1x2 , the function is symmetric with respect to the origin Graph f 1x2 ϭ The graph 4x supports symmetry with respect to the origin 72 If f 1x2 ϭ 2x f 1x2 ϭ x, then f 1᎐ x2 ϭ 21᎐ x2 f 1᎐ x2 ϭ 2x f 1᎐ x2 ϭ x Since f 1᎐ x2 ϭ f 1x2 , the function is symmetric with respect to the y-axis Graph f 1x2 ϭ 2x The graph supports symmetry with respect to the y-axis Full file at https://fratstock.eu 62 CHAPTER Analysis of Graphs of Functions 73 (a) Functions where f 1᎐ x2 ϭ f 1x2 are even, therefore exercises: 63, 64, 69, 70, and 72 are even (b) Functions where f 1᎐ x2 ϭ ᎐ f 1x2 are odd, therefore exercises: 61, 62, 68, and 71 are odd (c) Functions where f 1x2 f 1᎐ x2 ᎐ f 1x2 are neither odd or even, therefore exercises: 65, 66, and 67 are neither odd or even 74 Answers may vary If a function f is even, then f 1x2 ϭ f 1᎐ x2 for all x in the domain Its graph is symmetric with respect to the y-axis If a function f is odd, then f 1᎐ x2 ϭ ᎐ f 1x2 for all x in the domain Its graph is symmetric with respect to the origin 2.2: Vertical and Horizontal Shifts of Graphs The equation y ϭ x shifted units upward is: y ϭ x ϩ The equation y ϭ x shifted units downward is: y ϭ x Ϫ The equation y ϭ 1x shifted units downward is: y ϭ 1x Ϫ 4 3 The equation y ϭ x shifted units upward is: y ϭ xϩ6 The equation y ϭ x shifted units to the right is: y ϭ x Ϫ The equation y ϭ x shifted units to the left is: y ϭ x ϩ The equation y ϭ x shifted units to the left is: y ϭ 1x ϩ 72 The equation y ϭ 1x shifted units to the right is: y ϭ 1x Ϫ 9 Shift the graph of f units upward to obtain the graph of g 10 Shift the graph of f units to the left to obtain the graph of g 11 The equation y ϭ x Ϫ is y ϭ x shifted units downward, therefore graph B 12 The equation y ϭ 1x Ϫ 32 is y ϭ x shifted units to the right, therefore graph C 13 The equation y ϭ 1x ϩ 32 is y ϭ x shifted units to the left, therefore graph A 14 The equation y ϭ x ϩ is y ϭ x shifted units upward, therefore graph A 15 The equation y ϭ x ϩ Ϫ is y ϭ x shifted units to the left and units downward, therefore graph B 16 The equation y ϭ x Ϫ Ϫ is y ϭ x shifted units to the right and units downward, therefore graph C 17 The equation y ϭ 1x Ϫ 32 is y ϭ x shifted units to the right, therefore graph C 18 The equation y ϭ 1x Ϫ 22 Ϫ is y ϭ x shifted units to the right and units downward, therefore graph A 19 The equation y ϭ 1x ϩ 22 Ϫ is y ϭ x shifted units to the left and units downward, therefore graph B 20 If y ϭ x Ϫ h ϩ k with h and k 0, then the graph of y ϭ x is shifted to the left ᎐ h units and ᎐ k units downward This would place the vertex or lowest point of the absolute value graph in the third quadrant 21 For the equation y ϭ x , the Domain is: 1᎐ q, q2 and the Range is: 30, q2 Shifting this units downward gives us: (a) Domain: 1᎐ q, q2 (b) Range: 3᎐ 3, q2 22 For the equation y ϭ x , the Domain is: 1᎐ q, q2 and the Range is: 30, q2 Shifting this units to the right gives us: (a) Domain: 1᎐ q, q2 (b) Range: 30, q2 Full file at https://fratstock.eu Vertical and Horizontal Shifts of Graphs SECTION 2.2 63 23 For the equation y ϭ x , the Domain is: 1᎐ q, q2 and the Range is: 30, q2 Shifting this units to the left and units downward gives us: (a) Domain: 1᎐ q, q2 (b) Range: 3᎐ 3, q2 and units downward gives us: (a) Domain: 1᎐ q, q2 (b) Range: 3᎐ 3, q2 24 For the equation y ϭ x , the Domain is: 1᎐ q, q2 and the Range is: 30, q2 Shifting this units to the right 25 For the equation y ϭ x , the Domain is: 1᎐ q, q2 and the Range is: 1᎐ q, q2 Shifting this units to the right (a) Domain: 1᎐ q, q2 gives us: (b) Range: 1᎐ q, q2 26 For the equation y ϭ x , the Domain is: 1᎐ q, q2 and the Range is: 1᎐ q, q2 Shifting this units to the right and units downward gives us: (a) Domain: 1᎐ q, q2 27 Using Y2 ϭ Y1 ϩ k and x ϭ 0, we get 19 ϭ 15 ϩ k k ϭ (b) Range: 1᎐ q, q2 28 Using Y2 ϭ Y1 ϩ k and x ϭ 0, we get ᎐ ϭ ᎐ ϩ k k ϭ ᎐ 29 From the graphs 16, 22 is a point on Y1 and 16, ᎐ 12 a point on Y2 Using Y2 ϭ Y1 ϩ k and x ϭ 6, we get ᎐ ϭ ϩ k k ϭ ᎐ 30 From the graphs 1᎐ 4, 32 is a point on Y1 and 1᎐ 4, 82 a point on Y2 Using Y2 ϭ Y1 ϩ k and x ϭ ᎐ 4, we get ϭ ϩ k k ϭ 31 The graph of y ϭ 1x Ϫ 12 is the graph of the equation y ϭ x shifted unit to the right See Figure 31 32 The graph of y ϭ 1x ϩ is the graph of the equation y ϭ 1x shifted units to the left See Figure 32 33 The graph of y ϭ x ϩ is the graph of the equation y ϭ x shifted unit upward See Figure 33 y y y 1 x –2 Figure 31 x 0 Figure 32 x Figure 33 34 The graph of y ϭ x ϩ is the graph of the equation y ϭ x shifted units to the left See Figure 34 35 The graph of y ϭ 1x Ϫ 12 is the graph of the equation y ϭ x shifted unit to the right See Figure 35 36 The graph of y ϭ x Ϫ is the graph of the equation y ϭ x shifted units downward See Figure 36 y y y –2 x –1 x –3 –3 Figure 34 Figure 35 Figure 36 x Full file at https://fratstock.eu 64 CHAPTER Analysis of Graphs of Functions 37 The graph of y ϭ 1x Ϫ Ϫ is the graph of the equation y ϭ 1x shifted units to the right and unit downward See Figure 37 38 The graph of y ϭ 1x ϩ Ϫ is the graph of the equation y ϭ 1x shifted units to the left and units downward See Figure 38 39 The graph of y ϭ 1x ϩ 22 ϩ is the graph of the equation y ϭ x shifted units to the left and units upward See Figure 39 y y y x –1 –3 x –2 x –4 Figure 37 Figure 38 Figure 39 40 The graph of y ϭ 1x Ϫ 42 Ϫ is the graph of the equation y ϭ x shifted units to the right and units downward See Figure 40 41 The graph of y ϭ x ϩ Ϫ is the graph of the equation y ϭ x shifted units to the left and units downward See Figure 41 42 The graph of y ϭ 1x ϩ 32 Ϫ is the graph of the equation y ϭ x shifted units to the left and unit downward See Figure 42 y y y x –4 Figure 40 – –4 –2 x –2 x –2 Figure 41 Figure 42 43 Since h and k are positive, the equation y ϭ 1x Ϫ h2 Ϫ k is y ϭ x shifted to the right and down, therefore: B 44 Since h and k are positive, the equation y ϭ 1x ϩ h2 Ϫ k is y ϭ x shifted to the left and down, therefore: D 45 Since h and k are positive, the equation y ϭ 1x ϩ h2 ϩ k is y ϭ x shifted to the left and up, therefore: A 46 Since h and k are positive, the equation y ϭ 1x Ϫ h2 ϩ k is y ϭ x shifted to the right and up, therefore: C 47 The equation y ϭ f 1x2 ϩ is y ϭ f 1x2 shifted up units or add to the y-coordinate of each point as follows: 1᎐ 3, ᎐ 22 1᎐ 3, 02; 1᎐ 1, 42 1᎐ 1, 62; 15, 02 15, 22 See Figure 47 Full file at https://fratstock.eu Vertical and Horizontal Shifts of Graphs SECTION 2.2 65 48 The equation y ϭ f 1x2 Ϫ is y ϭ f 1x2 shifted down units or subtract from the y-coordinate of each point as follows: 1᎐ 3, ᎐ 22 1᎐ 3, ᎐ 42; 1᎐ 1, 42 1᎐ 1, 22; 15, 02 15, ᎐ 22 See Figure 48 y y (– 1, 6) (–1, 2) (5, 2) x x (– 3, 0) 0 (5, –2) (–3, –4) Figure 47 Figure 48 49 The equation y ϭ f 1x ϩ 22 is y ϭ f 1x2 shifted left units or subtract from the x-coordinate of each point as follows: 1᎐ 3, ᎐ 22 1᎐ 5, ᎐ 22; 1᎐ 1, 42 1᎐ 3, 42; 15, 02 13, 02 See Figure 49 50 The equation y ϭ f 1x Ϫ 22 is y ϭ f 1x2 shifted right units or add to the x-coordinate of each point as follows: 1᎐ 3, ᎐ 22 1᎐ 1, ᎐ 22; 1᎐ 1, 42 11, 42; 15, 02 17, 02 y See Figure 50 y (– 3, 4) (1, 4) (7, 0) (3, 0) x x 0 (– 5, – 2) Figure 49 (–1, –2) Figure 50 51 The graph is the basic function y ϭ x translated units to the left and units up, therefore the new equation is: y ϭ 1x ϩ 42 ϩ The equation is now increasing for the interval: (a) 3᎐ 4, q2 and decreasing for the interval: (b) 1᎐ q, ᎐ 44 52 The graph is the basic function y ϭ 1x translated units to the left, therefore the new equation is: y ϭ 1x ϩ The equation is now increasing for the interval: (a) 3᎐ 5, q2 and does not decrease, therefore: (b) none 53 The graph is the basic function y ϭ x translated units down, therefore the new equation is: y ϭ x Ϫ The equation is now increasing for the interval: (a) 1᎐ q, q2 and does not decrease, therefore: (b) none 54 The graph is the basic function y ϭ x translated 10 units to the left, therefore the new equation is: y ϭ x ϩ 10 The equation is now increasing for the interval: (a) 3᎐ 10, q2 and decreasing for the interval: (b) 1᎐ q, ᎐ 10 55 The graph is the basic function y ϭ 1x translated units to the right and unit up, therefore the new equation is: y ϭ 1x Ϫ ϩ The equation is now increasing for the interval: (a) 32, q2 and does not decrease, therefore: (b) none 56 The graph is the basic function y ϭ x translated units to the right and units down, therefore the new equation is: y ϭ 1x Ϫ 22 Ϫ The equation is now increasing for the interval: (a) 32, q2 and decreasing for the interval: (b) 1᎐ q, 24 Full file at https://fratstock.eu 66 CHAPTER Analysis of Graphs of Functions 57 (a) f 1x2 ϭ 0: 53, 46 (b) f 1x2 0: for the intervals 1᎐ q, 32 ´ 14, q2 (c) f 1x2 0: for the interval 13, 42 58 (a) f 1x2 ϭ 0: 126 (b) f 1x2 0: for the interval 12, q2 (c) f 1x2 0: for the interval 1᎐ q, 122 59 (a) f 1x2 ϭ 0: 5᎐ 4, 56 (b) f 1x2 Ն 0: for the intervals 1᎐ q, ᎐ 4 ´ 35, q2 (c) f 1x2 Յ 0: for the interval 3᎐ 4, 54 60 (a) f 1x2 ϭ 0: never, therefore: л (b) f 1x2 Ն 0: for the interval 31, q2 (c) f 1x2 Յ 0: never, therefore: л 61 The translation is units to the left and unit up, therefore the new equation is: y ϭ x ϩ ϩ The form y ϭ x Ϫ h ϩ k will equal y ϭ x ϩ ϩ when: h ϭ ᎐ and k ϭ 62 The equation y ϭ x has a Domain: 1᎐ q, q2 and a Range: 30, q2 After the translation the Domain is still: 1᎐ q, q2 , but now the Range is: 338, q2 , a positive or upward shift of 38 units Therefore, the horizontal shift can be any number of units, but the vertical shift is up 38 This makes h any real number and k ϭ 38 63 (a) Since corresponds to 1998, our equation using exact years would be: y ϭ 895.51x Ϫ 19982 ϩ 14,709 (b) y ϭ 895.512006 Ϫ 19982 ϩ 14,709 y ϭ 7164 ϩ 14,709 y ϭ $21,873 64 (a) Since corresponds to 1998, our equation using exact years would be: y ϭ 299.81x Ϫ 19982 ϩ 5249.1 (b) y ϭ 299.812007 Ϫ 19982 ϩ 5249.1 y ϭ 2698.2 ϩ 5249.1 y ϭ $ 7947.3 billion 65 (a) Enter the year in L1 and enter tuition and fees in L The year 1991 corresponds to x ϭ and so on The regression equation is: y Ϸ 190x ϩ 2071.3 (b) Since x ϭ corresponds to 1991, the equation when the exact year is entered is: y Ϸ 1901x Ϫ 19912 ϩ 2071.3 (c) y Ϸ 19012008 Ϫ 19912 ϩ 2071.3 y Ϸ 3230 ϩ 2071.3 y Ϸ $5300 66 (a) Enter the year in L1 and enter the percent of women in the workforce in L The year 1965 corresponds to x ϭ and so on The regression equation is: y Ϸ 6162x ϩ 40.6167 (b) Since x ϭ corresponds to 1965, the equation when the exact year is entered is: y Ϸ 61621x Ϫ 19652 ϩ 40.6167 (c) y Ϸ 616212010 Ϫ 19652 ϩ 40.6167 y Ϸ 27.729 ϩ 40.6167 y Ϸ 68.3% 67 See Figure 67 Ϫ 1᎐ 22 mϭ ϭ2 3Ϫ1 69 Using slope-intercept form yields: y1 Ϫ ϭ 21x Ϫ 32 y1 Ϫ ϭ 2x Ϫ y1 ϭ 2x Ϫ 68 m ϭ 70 11, ᎐ ϩ 62 and 13, ϩ 62 11, 42 and 13, 82 8Ϫ4 71 m ϭ mϭ ϭ2 3Ϫ1 Full file at https://fratstock.eu 108 CHAPTER Analysis of Graphs of Functions x 90 (a) If x ϭ 4s then s ϭ x x x2 (b) If y ϭ s2 and s ϭ , then y1x2 ϭ a b y1x2 ϭ 4 16 162 36 (c) Sine x is perimeter and x ϭ 6, y162 ϭ ϭ ϭ ϭ 2.25 16 16 x (d) Show that the point 16, 2.252 is on the graph y ϭ A square with perimeter will have area 2.25 square units 16 13 13 91 (a) A12x2 ϭ 12x2 ϭ 14x A12x2 ϭ 13 x 4 13 13 (b) A1x2 ϭ 1162 ϭ 12562 A1x2 ϭ 6413 square units 4 13 (c) On the graph of y ϭ x , locate the point where x ϭ 16 to find y Ϸ 110.85, an approximation for 64 13 92 (a) If A1r2 ϭ π r and r1t2 ϭ 2t then 1A ؠr21t2 ϭ A3r 1t2 ϭ A32t4 ϭ π 12t2 ϭ π t (b) 1A ؠr21t2 is a composite function that expresses the area of the circular region covered by the pollutants as a function of time t (in hours) (c) Since t ϭ is A.M., noon would be t ϭ 1A ؠr2142 ϭ π 142 ϭ 64 π mi (d) Graph y1 ϭ π x and show that for x ϭ 4, y Ϸ 201 (an approximation for 64 π) 93 (a) The function h is the addition of functions f and g x 1999 h(x) 76 2000 82 2001 79 2002 89 2003 103 (b) The function h is the addition of functions f and g Therefore h1x2 ϭ f 1x2 ϩ g1x2 94 (a) The domain is the years See Table Therefore: D ϭ 51998, 1999, 2000, 2001, 20026 x 1998 h(x) 94.2 1999 95.3 2000 99.3 2001 106.4 2002 93.2 (b) The function h computes the value of animals produced in the U.S and sold in billions of dollars 95 (a) f ϩ g2119702 ϭ 32.4 ϩ 17.6 ϭ 50.0 (b) The function f ϩ g21x2 computes the total SO2 emissions from burning coal and oil during year x (c) Add functions f and g x 1860 (f ϩ g)(x) 2.4 1900 12.8 1940 26.5 1970 50.0 2000 78.0 96 (a) The function h is the addition of functions f and g x 1990 h(x) 32 2000 35.5 2010 39 2020 42.5 2030 46 (b) The function h is the addition of functions f and g Therefore h1x2 ϭ f 1x2 ϩ g1x2 97 (a) The function h is the subtraction of function f from g Therefore h1x2 ϭ g 1x2 Ϫ f 1x2 (b) h119962 ϭ g119962 Ϫ f 119962 ϭ 841 Ϫ 694 ϭ 147 h120062 ϭ g120062 Ϫ f 120062 ϭ 1165 Ϫ 1012 ϭ 153 153 Ϫ 147 ϭ ϭ 2006 Ϫ 1996 10 Now using point slope form: y Ϫ 147 ϭ 61x Ϫ 19962 y ϭ 61x Ϫ 19962 ϩ 147 (c) Using the points 11996, 1472 and 12006, 1532 from part b, the slope is: m ϭ Full file at https://fratstock.eu Operations and Composition 98 (a) Graph h1x2 ϭ 19001x Ϫ 19822 ϩ 619 32001x Ϫ 19822 ϩ 1586 SECTION 2.6 109 , in the window 31982, 19944 by 30, 14 See Figure 98a Approximately 59% of the people who contracted AIDS during this time period died (b) Divide number of deaths by number of cases for each year The results compare favorably with the graph See Figure 98b 3᎐10, 10 by ᎐10, 10 Xscl ϭ Yscl ϭ Year 1982 Ratio 39 1984 51 Figure 98a 1986 59 1988 58 1990 61 1992 60 1994 61 Figure 98b Reviewing Basic Concepts (Sections 2.4—2.6) 1 1 1 (a) ` x ϩ ` ϭ x ϩ ϭ x ϭ x ϭ or x ϩ ϭ ᎐ x ϭ ᎐ x ϭ ᎐ 12 2 2 Therefore, the solution set is: 5᎐ 12, 46 (b) ` 1 1 x ϩ ` x ϩ x x or x ϩ ᎐ x ᎐ x ᎐ 12 2 2 Therefore, the solution interval is: 1᎐ q, ᎐ 122 h 14, q2 (c) ` 1 x ϩ ` Յ ᎐ Յ x ϩ Յ ᎐ Յ x Յ ᎐ 12 Յ x Յ 2 Therefore, the solution interval is: 3᎐ 12, 44 For the graph of y ϭ f 1x2 , we reflect the graph of y ϭ f 1x2 across the x-axis for all points for which y Where y Ն 0, the graph remains unchanged See Figure y (–3, 2) (3, 2) Figure x Full file at https://fratstock.eu 110 CHAPTER Analysis of Graphs of Functions (a) The range of values for RL Ϫ 26.75 Յ 1.42 is: ᎐ 1.42 Յ RL Ϫ 26.75 Յ 1.42 25.33 Յ RL Յ 28.17 The range of values for RE Ϫ 38.75 Յ 2.17 is: ᎐ 2.17 Յ RE Ϫ 38.75 Յ 2.17 36.58 Յ RE Յ 40.92 (b) If TL ϭ 2251RL then the range for TL is: 225125.33 Յ TL Յ 28.172 ϭ 5699.25 Յ TL Յ 6338.25 If TE ϭ 2251RE then the range for TL is: 225136.58 Յ TE Յ 40.922 ϭ 8230.5 Յ TE Յ 9207 (a) f 1᎐ 32 ϭ 21᎐ 32 ϩ ϭ ᎐ (a) See Figure 5a (b) f 102 ϭ 102 ϩ ϭ (c) f 122 ϭ 122 ϩ ϭ (b) Graph y1 ϭ 1᎐ x 2 * 1x Յ 02 ϩ 1x Ϫ 42 * 1x 02 in the window 3᎐ 10, 104 by 3᎐ 10, 104 See Figure 5b y 3᎐10, 10 by 3᎐10, 10 30, 1000 by 3᎐20, 100 Xscl ϭ 100 Yscl ϭ 10 Figure 5b Figure Xscl ϭ –2 Yscl ϭ x –4 Figure 5a (a) f ϩ g21x2 ϭ 1᎐ 3x Ϫ 42 ϩ 1x 2 ϭ x Ϫ 3x Ϫ Therefore, f ϩ g2112 ϭ 112 Ϫ 3112 Ϫ ϭ ᎐ (b) f Ϫ g21x2 ϭ 1᎐ 3x Ϫ 42 Ϫ 1x 2 ϭ ᎐ x Ϫ 3x Ϫ Therefore, f Ϫ g2132 ϭ ᎐ 132 Ϫ 3132 Ϫ ϭ ᎐ 22 (c) f g21x2 ϭ 1᎐ 3x Ϫ 421x 2 ϭ ᎐ 3x Ϫ 4x Therefore, f g21᎐22 ϭ ᎐ 31᎐ 22 Ϫ 41᎐ 22 ϭ 24 Ϫ 16 ϭ ᎐ 31᎐ 32 Ϫ f f ᎐ 3x Ϫ Therefore, a b 1᎐ 32 ϭ ϭ (d) a b 1x2 ϭ 2 g g x 1᎐ 32 (e) f ؠg21x2 ϭ f 3g1x2 ϭ ᎐ 31x 2 Ϫ 1 f ؠg21x2 ϭ ᎐ 3x Ϫ (f) 1g ؠf 21x2 ϭ g3 f 1x2 ϭ 1᎐ 3x Ϫ 42 1g ؠf 21x2 ϭ 9x ϩ 24x ϩ 16 One of many possible solutions for f ؠg21x2 ϭ h1x2 is: f 1x2 ϭ x and g1x2 ϭ x ϩ Then f ؠg21x2 ϭ f 3g1x2 ϭ 1x ϩ 22 ᎐ 21x ϩ h2 ϩ 31x ϩ h2 Ϫ Ϫ 1᎐ 2x ϩ 3x Ϫ 52 ᎐ 21x ϩ 2xh ϩ h2 ϩ 3x ϩ 3h Ϫ ϩ 2x Ϫ 3x ϩ ϭ ϭ h h ᎐ 2x Ϫ 4xh Ϫ 2h2 ϩ 3x ϩ 3h Ϫ ϩ 2x Ϫ 3x ϩ ᎐ 4xh Ϫ 2h2 ϩ 3h ϭ ϭ ᎐ 4x Ϫ 2h ϩ h h (a) At 4% simple interest the equation for interest earned is: y1 ϭ 04 x (b) If he invested x dollars in the first account, then he invested x ϩ 500 in the second account The equation for the amount of interest earned on this account is: y2 ϭ 0251 x ϩ 5002 y2 ϭ 025x ϩ 12.5 (c) It represents the total interest earned in both accounts for year (d) Graph y1 ϩ y2 ϭ 04x ϩ 1.025x ϩ 12.52 y1 ϩ y2 ϭ 04x ϩ 025x ϩ 12.5 in the window 0, 10004 by 30, 1004 See Figure An input value of x ϭ 250, results in $28.75 earned interest (e) At x ϭ 250, y1 ϩ y2 ϭ 0412502 ϩ 02512502 ϩ 12.5 ϭ 10 ϩ 6.25 ϩ 12.5 ϭ $28.75 10 If the radius is r, then the height is 2r and the equation is S ϭ π r2r ϩ 12r2 ϭ π r2r ϩ 4r ϭ π r25r S ϭ π r 15 Full file at https://fratstock.eu Chapter Review SECTION 2.R 111 Chapter Review Exercises The graphs for exercises 1–10 can be found in the “Function Capsule” boxes located in section 2.1 in the text True Both f 1x2 ϭ x and f 1x2 ϭ x have the interval: 30, q2 as the range True Both f 1x2 ϭ x and f 1x2 ϭ x increase on the interval: 30, q2 False The function f 1x2 ϭ 1x has the domain: 30, q2 and f 1x2 ϭ x the domain: 1᎐ q, q2 False The function f 1x2 ϭ x increases on its entire domain True The function f 1x2 ϭ x has a domain and range of: 1᎐ q, q2 False The function f 1x2 ϭ 1x is not defined on 1᎐ q, 02 , so certainly cannot be continuous True All of the functions show increases on the interval: 30, q2 True Both f 1x2 ϭ x and f 1x2 ϭ x have graphs that are symmetric with respect to the origin True Both f 1x2 ϭ x and f 1x2 ϭ x have graphs that are symmetric with respect to the y-axis 10 True No graphs are symmetric with respect to the x-axis 11 Only values where x Ն can be input for x, therefore the domain of f 1x2 ϭ 1x is: 30, q2 12 Only positive solutions are possible in absolute value functions, therefore the range of f 1x2 ϭ 1x is: 0, q2 13 All solutions are possible in cube root functions, therefore the range of f 1x2 ϭ x is: 1᎐ q, q2 14 All values can be input for x, therefore the domain of f 1x2 ϭ x is: 1᎐ q, q2 15 The function f 1x2 ϭ x increases for all inputs for x, therefore the interval is: 1᎐ q, q2 16 The function f 1x2 ϭ x increases for all inputs where x Ն 0, therefore the interval is: 30, q2 17 The equation y ϭ x is the equation y ϭ 1x Only values where x Ն can be input for x, therefore the domain of y ϭ 1x is: 30, q2 18 The equation y ϭ x is the equation y ϭ 1x Square root functions have both positive and negative solutions and all solution are possible, therefore the range of y ϭ 1x is: 1᎐ q, q2 19 The graph of f 1x2 ϭ 1x ϩ 32 Ϫ is the graph y ϭ x shifted units to the left and unit downward See Figure 19 20 The graph of f 1x2 ϭ ᎐ x ϩ is the graph y ϭ x reflected across the x-axis, vertically shrunk by a factor of , and shifted unit upward See Figure 20 21 The graph of f 1x2 ϭ 1x ϩ 12 Ϫ is the graph y ϭ x shifted unit to the left and units downward See Figure 21 y y y –2 x x –1 x –2 Figure 19 Figure 20 Figure 21 Full file at https://fratstock.eu 112 CHAPTER Analysis of Graphs of Functions 22 The graph of f 1x2 ϭ ᎐ 2x ϩ is the graph y ϭ x reflected across the x-axis, vertically stretched by a factor of 2, and shifted units upward See Figure 22 23 The graph of f 1x2 ϭ ᎐ x ϩ is the graph y ϭ x reflected across the x-axis and shifted units upward See Figure 23 24 The graph of f 1x2 ϭ 1x Ϫ 32 is the graph y ϭ x shifted units to the right See Figure 24 y y y 2 x x x –2 Figure 22 25 The graph of f 1x2 ϭ Figure 23 Figure 24 x is the graph y ϭ 1x horizontally stretched by a factor of See Figure 25 B2 26 The graph of f 1x2 ϭ x Ϫ ϩ is the graph y ϭ 1x shifted units to the right and unit upward See Figure 26 3 27 The graph of f 1x2 ϭ 21 x is the graph y ϭ x vertically stretched by a factor of See Figure 27 y y y 1 x Figure 25 x Figure 26 x Figure 27 3 28 The graph of f 1x2 ϭ x Ϫ is the graph y ϭ x shifted units downward See Figure 28 29 The graph of f 1x2 ϭ x Ϫ ϩ is the graph y ϭ x shifted units right and unit upward See Figure 29 30 The graph of f 1x2 ϭ ᎐ 2x ϩ is the graph y ϭ x horizontally shrunk by a factor of , shifted a b 132 or 2 units to the left, and reflected across the y-axis See Figure 30 Full file at https://fratstock.eu Chapter Review SECTION 2.R y y 113 y 3 –1 x x x –2 Figure 28 Figure 29 Figure 30 31 (a) From the graph, the function is continuous for the intervals: 1᎐ q, ᎐ 22, 3᎐ 2, 14, and 11, q2 (b) From the graph, the function is increasing for the interval: 3᎐ 2, 14 (c) From the graph, the function is decreasing for the interval: 1᎐ q, ᎐ 22 (d) From the graph, the function is constant for the interval: 11, q2 (e) From the graph, all values can be input for x, therefore the domain is: 1᎐ q, q2 (f) From the graph, the possible values of y or the range is: 5᎐ 26 h 3᎐ 1, 14 h 12, q2 32 x ϭ y Ϫ y ϭ x ϩ y ϭ 1x ϩ and y ϭ ᎐ 1x ϩ 33 From the graph, the relation is symmetric with respect to the x-axis, y-axis, and origin The relation is not a function since some inputs x have two outputs y 34 If F1x2 ϭ x Ϫ 6, then F1᎐ x2 ϭ 1᎐ x2 Ϫ F1᎐ x2 ϭ ᎐ x Ϫ and ᎐ F1x2 ϭ ᎐ 1x Ϫ 62 ᎐F1x2 ϭ ᎐ x ϩ Since F1x2 F1᎐ x2 ᎐ F1x2 , the function has no symmetry and is neither an even nor an odd function 35 If f 1x2 ϭ x ϩ , then f 1᎐ x2 ϭ 1᎐ x2 ϩ f 1᎐ x2 ϭ x ϩ and ᎐ f 1x2 ϭ ᎐ x Ϫ Since f 1᎐ x2 ϭ f 1x2 , the function is symmetric with respect to the y-axis and is an even function 36 If f 1x2 ϭ 1x Ϫ 5, then f 1᎐ x2 ϭ 11᎐ x2 Ϫ and ᎐ f 1x2 ϭ ᎐ 1x Ϫ Since f 1x2 f 1᎐ x2 ᎐ f 1x2 , the func- tion has no symmetry and is neither an even nor an odd function 37 If y ϭ x Ϫ then y ϭ Ϯ1x Ϫ Since f 1x2 ϭ ᎐ 1x Ϫ is the reflection of f 1x2 ϭ 1x Ϫ across the x-axis, the relation has symmetry with respect to the x-axis Also, one x input can produce two y outputs The relation is not a function 38 If f 1x2 ϭ 3x ϩ 2x ϩ 1, then f 1᎐ x2 ϭ 31᎐ x2 ϩ 21᎐ x2 ϩ 1 f 1᎐ x2 ϭ 3x ϩ 2x ϩ and ᎐ f 1x2 ϭ ᎐ 3x Ϫ 2x Ϫ Since f 1᎐ x2 ϭ f 1x2 , the function is symmetric with respect to the y-axis and is an even function 39 True, a graph that is symmetrical with respect to the x-axis means that for every 1x, y2 there is also 1x, ᎐ y2 , which is not a function 40 True, since an even function and one that is symmetric with respect to the y-axis both contain the points 1x, y2 and 1᎐ x, y2 41 True, since an odd function and one that is symmetric with respect to the origin both contain the points 1x, y2 and 1᎐ x, ᎐ y2 Full file at https://fratstock.eu 114 CHAPTER Analysis of Graphs of Functions 42 False, for an even function, if 1a, b2 is on the graph, then 1᎐ a, b2 is on the graph and not 1a, ᎐ b2 For example, f 1x2 ϭ x is even, and 12, 42 is on the graph, but 12, ᎐ 42 is not 43 False, for an odd function, if 1a, b2 is on the graph, then 1᎐ a, ᎐ b2 is on the graph and not 1᎐ a, b2 For example, f 1x2 ϭ x is odd, and 12, 82 is on the graph, but 1᎐ 2, 82 is not 44 True, if 1x, 02 is on the graph of f 1x2 ϭ 0, then 1᎐ x, 02 is on the graph 45 The graph of y ϭ ᎐ 31x ϩ 42 Ϫ is the graph of y ϭ x shifted units to the left, vertically stretched by a factor of 3, reflected across the x-axis, and shifted units downward 46 The equation y ϭ 1x reflected across the y-axis is: y ϭ 1᎐ x, then reflected across the x-axis is: y ϭ ᎐ 1᎐ x, 2 now vertically shrunk by a factor of is: y ϭ ᎐ 1᎐ x , and finally shifted units upward is: y ϭ ᎐ 1᎐ x ϩ 3 47 Shift the function f upward units See Figure 47 48 Shift the function f to the right units See Figure 48 49 Shift the function f to the left units and downward units See Figure 49 y y y (8, 3) (4, 1) (–4, 3) –4 (10, 0) x –3 x (6, –2) x (–2, 0) (–7, –2) (5, –2) –4 –4 (1, –4) Figure 47 Figure 48 Figure 49 50 For values where f 1x2 the graph remains the same For values where f 1x2 reflect the graph across the x-axis See Figure 50 51 Horizontally shrink the function f by a factor of See Figure 51 52 Horizontally stretch the function f by a factor of See Figure 52 y y y 2 (4, 2) (0, 0) (–4, 0) x (8, 0) (–1, 0) (2, 0) (16, 0) (0, 0) x (–8, 0) x –2 (1, –2) Figure 50 Figure 51 (8, –2) Figure 52 53 The function is shifted upward units, therefore the domain remains the same: 3᎐ 3, 44 and the range is increased by and is: 32, 94 Full file at https://fratstock.eu Chapter Review SECTION 2.R 115 54 The function is shifted left 10 units, therefore the domain is decreased by 10 and is: 3᎐ 13, ᎐ 64 ; and the function is stretched vertically by a factor of 5, therefore the range is multiplied by and is: 3᎐ 10, 254 55 The function is horizontally shrunk by a factor of , therefore the domain is divided by and is: c ᎐ , d ; and 2 the function is reflected across the x-axis, therefore the range is opposite of the original and is: 3᎐ 5, 24 56 The function is shifted right unit, therefore the domain is increased by and is: 3᎐ 2, 54 ; and the function is also shifted upward units, therefore the range is increased by and is: 1, 84 57 We reflect the graph of y ϭ f 1x2 across the x-axis for all points for which y Where y Ն 0, the graph remains unchanged See Figure 57 58 We reflect the graph of y ϭ f 1x2 across the x-axis for all points for which y Where y Ն 0, the graph remains unchanged See Figure 58 y (–2, 2) y y (0, 2) (0, 2) (0, 2) x (–2, 0) (2, 0) Figure 57 x (2, 0) Figure 58 x Figure 60 59 Since the range is 526, y Ն 0, so the graph remains unchanged 60 Since the range is 5᎐ 26, y 0, so we reflect the graph across the x-axis See Figure 60 61 4x ϩ ϭ 12 4x ϩ ϭ 12 4x ϭ x ϭ therefore the solution set is: e ᎐ 15 or 4x ϩ ϭ ᎐ 12 4x ϭ ᎐ 15 x ϭ ᎐ , 4 15 , f 4 62 ᎐ 2x Ϫ ϩ ϭ 1 ᎐ 2x Ϫ ϭ ᎐ Since an absolute value equation can not have a solution less than zero, the solution set is: л 63 5x ϩ ϭ x ϩ 11 5x ϩ ϭ x ϩ 11 4x ϭ x ϭ or 5x ϩ ϭ ᎐ 1x ϩ 112 6x ϭ ᎐ 14 xϭ᎐ 14 7 ϭ ᎐ , therefore the solution set is: e ᎐ , f 3 64 2x ϩ ϭ 2x ϩ ϭ 2x ϭ x ϭ or 2x ϩ ϭ ᎐ 2x ϭ ᎐ 12 x ϭ ᎐ 6, therefore the solution set is: 5᎐ 6, 16 65 2x ϩ Յ ᎐ Յ 2x ϩ Յ ᎐ 12 Յ 2x Յ ᎐ Յ x Յ , therefore the interval is: 3᎐ 6, 14 66 2x ϩ Ն 2x ϩ Ն 2x Ն x Ն or 2x ϩ Յ ᎐ 2x Յ ᎐ 12 x Յ ᎐ 6, therefore the solution is the interval: 1᎐ q, ᎐ 64 h 31, q2 Full file at https://fratstock.eu 116 CHAPTER Analysis of Graphs of Functions 67 5x Ϫ 12 5x Ϫ 12 5x 12 x therefore the solution is the interval: a᎐ q, 12 12 or 5x Ϫ 12 5x ϭ 12 x , 5 12 12 b h a , qb or e x x 5 12 68 Since an absolute value equation can not have a solution less than zero, the solution set is: л 69 3x Ϫ ϩ ϭ 21 3x Ϫ ϭ 20 3x Ϫ ϭ 10 3x Ϫ ϭ 10 3x ϭ 11 x ϭ 3x Ϫ ϭ ᎐ 10 3x ϭ ᎐ x ϭ ᎐ , therefore the solution set is: e ᎐ 3, 11 or 11 f 70 2x ϩ ϭ ᎐ 3x ϩ 1 2x ϩ ϭ ᎐ 3x ϩ 1 5x ϭ x ϭ or 2x ϩ ϭ ᎐ 1᎐ 3x ϩ 12 ᎐ x ϭ ᎐ x ϭ , therefore the solution set is: 50, 26 71 The x-coordinates of the points of intersection of the graphs are ᎐ and Thus, 5᎐ 6, 16 is the solution set of y1 ϭ y2 The graph of y1 lies on or below the graph of y2 between ᎐ and 1, so the solution set of y1 Յ y2 is 3᎐ 6, 14 The graph of y1 lies above the graph of y2 everywhere else, so the solution set of y1 Ն y2 is 1᎐ q, ᎐ 64 h 31, q2 72 Graph y1 ϭ x ϩ ϩ x Ϫ and y2 ϭ See Figure 72 The intersections are x ϭ ᎐ and x ϭ , therefore the solution set is: 5᎐ 3, 56 Check: 1᎐ 32 ϩ ϩ 1᎐ 32 Ϫ ϭ ᎐ ϩ ᎐ ϭ ϩ ϭ 8 ϭ and 152 ϩ ϩ 152 Ϫ ϭ ϩ ϭ ϩ ϭ 8 ϭ y 500 Gallons ᎐10, 104 by 3᎐ 4, 16 Xscl ϭ Yscl ϭ y (30, 500) (0, 500) 400 300 200 100 (10, 0) 10 20 30 x x Minutes Figure 72 Figure 74 Figure 75 73 Initially, the car is at home After traveling 30 mph for hr, the car is 30 mi away from home During the second hour the car travels 20 mph until it is 50 mi away During the third hour the car travels toward home at 30 mph until it is 20 mi away During the fourth hour the car travels away from home at 40 mph until it is 60 mi away from home During the last hour, the car travels 60 mi at 60 mph until it arrives home 74 See Figure 74 75 See Figure 75 76 See Figure 76 77 Graph y1 ϭ 13x ϩ 12 * 1x 22 ϩ 1᎐ x ϩ 42 * 1x Ն 22 in the window 3᎐ 10, 104 by 3᎐ 10, 104 See Figure 77 78 See Figure 78 Full file at https://fratstock.eu Chapter Review SECTION 2.R y 3᎐10, 10 by 3᎐10, 10 Xscl ϭ Yscl ϭ 3᎐5, by 3᎐5, Xscl ϭ 100 Yscl ϭ 10 x Figure 76 Figure 77 Figure 78 79 From the graphs f ϩ g2112 ϭ ϩ ϭ 80 From the graphs f Ϫ g2102 ϭ Ϫ ϭ ᎐ 81 From the graphs f g21᎐ 12 ϭ 102132 ϭ f 82 From the graphs a b 122 ϭ g 83 From the graphs f ؠg2122 ϭ f 3g122 ϭ f 122 ϭ 84 From the graphs 1g ؠf 2122 ϭ g3 f 122 ϭ g 132 ϭ 85 From the graphs 1g ؠf 21᎐ 42 ϭ g f 1᎐ 42 ϭ g 122 ϭ 86 From the graphs f ؠg21᎐ 22 ϭ f 3g1᎐ 22 ϭ f 122 ϭ 87 From the table f ϩ g2112 ϭ ϩ ϭ 88 From the table f Ϫ g2132 ϭ Ϫ ϭ 89 From the table f g21᎐ 12 ϭ 1321᎐ 22 ϭ ᎐ f 90 From the table a b 102 ϭ , which is undefined g 91 From the tables 1g ؠf 21᎐ 22 ϭ g f 1᎐ 22 ϭ g 112 ϭ 92 From the graphs f ؠg2132 ϭ f 3g132 ϭ f 1᎐ 22 ϭ 21x ϩ h2 ϩ Ϫ 12x ϩ 92 2x ϩ 2h ϩ Ϫ 2x Ϫ 2h ϭ ϭ ϭ2 h h h 1x ϩ h2 Ϫ 51x ϩ h2 ϩ Ϫ 1x Ϫ 5x ϩ 32 x ϩ 2xh ϩ h2 Ϫ 5x Ϫ 5h ϩ Ϫ x ϩ 5x Ϫ 94 ϭ ϭ h h 2xh ϩ h2 Ϫ 5h ϭ 2x ϩ h Ϫ h 95 One of many possible solutions for f ؠg21x2 ϭ h1x2 is: f 1x2 ϭ x and g1x2 ϭ x Ϫ 3x Then 93 f ؠg21x2 ϭ f 3g1x2 ϭ 1x Ϫ 3x2 96 One of many possible solutions for f ؠg21x2 ϭ h1x2 is: f 1x2 ϭ f ؠg21x2 ϭ f 3g1x2 ϭ xϪ5 and g1x2 ϭ x Ϫ Then x 4 π r , then a inch increase would be: V1r2 ϭ π 1r ϩ 42 3, and the volume gained would be: 3 4 V1r2 ϭ π 1r ϩ 42 Ϫ π r 3 97 If V1r2 ϭ 117 Full file at https://fratstock.eu 118 CHAPTER Analysis of Graphs of Functions d 98 (a) Since h ϭ d, r ϭ , and the formula for the volume of a can is: V ϭ π r 2h, the function is: d πd3 V1d2 ϭ π a b d V1d2 ϭ d (b) Since h ϭ d, r ϭ , c ϭ 2π r , and the formula for the surface area of a can is: A ϭ 2πrh ϩ 2πr2, the d d πd2 3πd2 S1d2 ϭ function is: S1d2 ϭ π a b d ϩ π a b S1d2 ϭ π d ϩ 2 2 99 The function for changing yards to inches is: f 1x2 ϭ 36x and the function for changing miles to yards is: g1x2 ϭ 1760x The composition of this which would change miles into inches is: f 3g 1x2 ϭ 36317601x2 1ƒ ؠg21x2 ϭ 63,360 x 100 If x ϭ width, then length ϭ 2x A formula for Perimeter can now be written as: P ϭ x ϩ 2x ϩ x ϩ 2x and the function is: P1x2 ϭ 6x This is a linear function Chapter Test (a) D, only values where x Ն can be input into a square root function (b) D, only values where y Ն can be the range of a square root function (c) C, all values can be input for x in a squaring function (d) B, only values where y Ն can be the range of f 1x2 ϭ x ϩ (e) C, all values can be input for x in a cube root function (f) C, all values can be the range of a cube root function (g) C, all values can be input for x in an absolute value function (h) D, only values where y Ն can be the range to an absolute value function (i) D, if x ϭ y then y ϭ 1x and only values where x Ն can be input into a square root function (j) C, all values can be the range in this function (a) This is f 1x2 shifted units upward See Figure 2a (b) This is f 1x2 shifted units to the left See Figure 2b (c) This is f 1x2 reflected across the x-axis See Figure 2c (d) This is f 1x2 reflected across the y-axis See Figure 2d (e) This is f 1x2 vertically stretched by a factor of See Figure 2e (f) We reflect the graph of y ϭ f 1x2 across the x-axis for all points for which y Where y Ն 0, the graph remains unchanged See Figure 2f Full file at https://fratstock.eu Chapter Test y SECTION 2.T 119 y y (1, 3) (0, 2) (4, 2) (–2, 0) x x (1, –1) x (2, 0) (–1, –3) Figure 2a y Figure 2b Figure 2c y y (1, 3) (0, 0) (–4, 0) (4, 0) (0, 0) x (0, 0) (4, 0) x (0, 0) x (4, 0) (–1, –3) (1, –6) Figure 2d Figure 2e Figure 2f (a) Since y ϭ f 12x2 is y ϭ f 1x2 horizontally shrunk by a factor of , the point 1᎐ 2, 42 on y ϭ f 1x2 becomes the point 1᎐ 1, 42 on the graph of y ϭ f 12x2 (b) Since y ϭ f a xb is y ϭ f 1x2 horizontally stretched by a factor of , the point 1᎐ 2, 42 on y ϭ f 1x2 becomes the point 1᎐ 4, 42 on the graph of y ϭ f a xb (a) The graph of f 1x2 ϭ ᎐ 1x Ϫ 22 ϩ is the basic graph f 1x2 ϭ x reflected across the x-axis, shifted units to the right, and shifted units upward See Figure 4a (b) The graph of f 1x2 ϭ ᎐21᎐ x is the basic graph f 1x2 ϭ 1x reflected across the y-axis and vertically stretched by a factor of See Figure 4b y y 4 x Figure 4a –4 Figure 4b x Full file at https://fratstock.eu 120 CHAPTER Analysis of Graphs of Functions (a) If the graph is symmetric with respect to the y-axis, then 1x, y2 1᎐ x, y2, therefore 13, 62 1᎐ 3, 62 (b) If the graph is symmetric with respect to the x-axis, then 1x, y2 1᎐ x, ᎐ y2, therefore 13, 62 1᎐ 3, ᎐ 62 (c) See Figure We give an actual screen here The drawing should resemble it ᎐ 4, 44 by 30, Xscl ϭ Yscl ϭ y (3, 2) x Figure Figure (a) Shift the graph of y ϭ x to the left units, vertically stretch by a factor of 4, and shift units downward (b) Graph y ϭ x reflected across the x-axis, vertically shrunk by a factor of , shifted units to the right, and shifted up units See Figure From the graph the domain is: 1᎐ q, q2; and the range is: 1᎐ q, 24 (a) From the graph, the function is increasing for the interval: 1᎐ q, ᎐ 32 (b) From the graph, the function is decreasing for the interval: 14, q2 (c) From the graph, the function is constant for the interval: 3᎐ 3, 44 (d) From the graph, the function is continuous for the intervals: 1᎐ q, ᎐ 32, 3᎐ 3, 44, 14, q2 (e) From the graph, the domain is: 1᎐ q, q2 (f) From the graph, the range is: 1᎐ q, 22 (a) 4x ϩ ϭ 4x ϩ ϭ 4x ϭ ᎐ x ϭ ᎐ or 4x ϩ ϭ ᎐ 4x ϭ ᎐ 12 x ϭ ᎐ , therefore the solution set is: 5᎐ 3, ᎐ 16 From the graph, the x-coordinates of the points of intersection of the graphs of Y1 and Y2 are ᎐ and ᎐ See Figure (b) 4x ϩ ᎐ 4x ϩ ᎐ 12 4x ᎐ ᎐ x ᎐ 1, therefore the solution is: 1᎐ 3, ᎐ 12 From the graph, See Figure 8, the graphs of Y1 lies below the graph of Y2 for x-values between ᎐ and ᎐ (c) 4x ϩ 4x ϩ 4x ᎐ x ᎐ or 4x ϩ ᎐ 4x ᎐ 12 x ᎐ , therefore the solution is: 1᎐ q, ᎐ 32 h 1᎐ 1, q2 From the graph, See Figure 8, the graph of Y1 lies above the graph of Y2 for x-values less than ᎐ or for x-values greater than ᎐ 3᎐10, 10 by ᎐10, 10 Xscl ϭ Yscl ϭ Figure Full file at https://fratstock.eu Chapter Test SECTION 2.T 121 (a) f Ϫ g21x2 ϭ 2x Ϫ 3x ϩ Ϫ 1᎐ 2x ϩ 12 1 f Ϫ g21x2 ϭ 2x Ϫ x ϩ f 2x Ϫ 3x ϩ (b) a b 1x2 ϭ g ᎐ 2x ϩ (c) The domain can be all values for x, except any that make g1x2 ϭ Therefore ᎐ 2x ϩ 1 ᎐ 2x ᎐1 1 or the interval: a᎐ q b h a , qb 2 x (d) f ؠg21x2 ϭ f 3g1x2 ϭ 21᎐ 2x ϩ 12 Ϫ 31᎐ 2x ϩ 12 ϩ ϭ 214x Ϫ 4x ϩ 12 ϩ 6x Ϫ ϩ ϭ 8x Ϫ 8x ϩ ϩ 6x Ϫ ϩ ϭ 8x Ϫ 2x ϩ 21x ϩ h2 Ϫ 31x ϩ h2 ϩ Ϫ 12x Ϫ 3x ϩ 22 21x ϩ 2xh ϩ h2 Ϫ 3x Ϫ 3h ϩ Ϫ 2x ϩ 3x Ϫ ϭ ϭ h h 2x ϩ 4xh ϩ 2h2 Ϫ 3x Ϫ 3h ϩ Ϫ 2x ϩ 3x Ϫ 4xh ϩ 2h2 Ϫ 3h ϭ ϭ 4x ϩ 2h Ϫ h h 10 (a) See Figure 10a (e) (b) Graph y1 ϭ 1᎐ x ϩ 32 * 1x Յ 12 ϩ 1 x ϩ 22 * 1x 12 in the window 3᎐ 4.7, 4.74 by 3᎐ 5.1, 5.14 See Figure 10b y 3᎐ 4.7, 4.7 by 3᎐5.1, 5.1 Xscl ϭ Yscl ϭ 30, 10 by 30, Xscl ϭ Yscl ϭ Figure 10b Figure 11 –2 x Figure 10a 11 (a) See Figure 11 (b) Set x ϭ 5.5, then $2.75 is the cost of a 5.5-minute call See the display at the bottom of the screen 12 (a) With an initial set-up cost of $3300 and a production cost of $4.50 the function is: C1x2 ϭ 3300 ϩ 4.50 x (b) With a selling price of $10.50 the revenue function is: R1x2 ϭ 10.50 x (c) P1x2 ϭ R1x2 Ϫ C1x2 P1x2 ϭ 10.50x Ϫ 13300 ϩ 4.50x2 P1x2 ϭ 6x Ϫ 3300 (d) To make a profit P1x2 0, therefore 6x Ϫ 3300 6x 3300 x 550 Tyler needs to sell 551before he earns a profit (e) Graph y1 ϭ 6x Ϫ 3300, See Figure 12 The first integer x-value for which P1x2 is 551 30, 1000 by 3᎐ 4000, 4000 Xscl ϭ 50 Yscl ϭ 500 Figure 12 Full file at https://fratstock.eu 122 CHAPTER Analysis of Graphs of Functions Chapter Project Since the front is moving at 40 mph for hr the front moved 160 miles with each unit representing 100 miles This is a shift of 1.6 units south or downward The function f 1x2 ϭ x shifted 1.6 units downward is: 20 x Ϫ 1.6 20 (a) Because the front has moved 250 south and 210 miles east, graph the shifted equation: yϭ 1x Ϫ 2.12 Ϫ 2.5 Plot the point 15.5, ᎐.82 for Columbus, Ohio Here we see that the front has 20 reached the city See Figure 2a yϭ (b) Because the front has moved 250 south and 210 miles east, graph the shifted equation: 1x Ϫ 2.12 Ϫ 2.5 Plot the point 11.9, ᎐ 4.32 for Memphis, Tennessee Here we see that the front has 20 not reached the city See Figure 2b yϭ (c) Because the front has moved 250 south and 210 miles east, graph the shifted equation: 1x Ϫ 2.12 Ϫ 2.5 Plot the point 14.2, ᎐2.32 for Louisville, Kentucky Although the graph is 20 difficult to read, repeated zooming will show that the front has not reached the city See Figure 2c yϭ 3᎐10, 10 by ᎐10, 10 Xscl ϭ Yscl ϭ ᎐10, 10 by ᎐10, 10 Xscl ϭ Yscl ϭ 3᎐10, 10 by 3᎐10, 10 Xscl ϭ Yscl ϭ Figure 2a Figure 2b Figure 2c
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