Solution manual for technical calculus with analytic geometry 5th edition by kuhfittig

Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Full file at https://TestbankDirect.eu/ Chapter © Cengage Learning All rights reserved No distribution allowed without express authorization Introduction to Analytic Geometry 1.1 The Cartesian Coordinate System Let (x2 , y2 ) = (2, 4) and (x1 , y1 ) = (5, 2) From the distance formula d= (x2 − x1 )2 + (y2 − y1 )2 we get d= (2 − 5)2 + (4 − 2)2 = (−3)2 + 22 = √ 9+4= √ 13 Let (x2 , y2 ) = (−3 , 2) and (x1 , y1 ) = (5 , −4) From the distance formula 2 d= (x2 − x1 ) +(y2 − y1 ) we get d= 2 2 √ (−3 − 5) +[2 − (−4)] = (−8) +(6) = 64+36 = 10 Let (x2 , y2 ) = (−3, −6) and (x1 , y1 ) = (5, −2) Then d= (−3 − 5)2 + [−6 − (−2)]2 = (−8)2 + (−4)2 = √ 64 + 16 = √ √ 16 · = √ Let (x2 , y2 ) = − , 2) and (x1 , y1 ) = (0 , 0) √ √ 2 Then d = (− − 0) +(2 − 0) = 5+4 = √ Let (x2 , y2 ) = ( 3, 4) and (x1 , y1 ) = (0, 2) Then d= √ √ √ ( − 0)2 + (4 − 2)2 = + = d = √ √ √ √ √ ( − 2) +( − 0) = 0+5 = d = √ √ √ [1 − (−1)]2 + (− − 0)2 = + = d = [2 − (−2)] + (7 − 3)2 = d = [−9 − (−11)]2 + (−1 − 1)2 = 10 Distance from (0 , 0) to (4 , 3): √ 16 + 16 = √ √ · 16 = 22 + (−2)2 = √ 8= (4 − 0) +(3 − 0) = (6 , 0)= Distance from (4 , 3) to (6 , 0): √ √ Perimeter = + + 13 = 11 + 13 √ √ 2·4=2 √ 16 + = Distance from (0 , 0) to √ √ (6 − 4) +(0 − 3) = + = 13 Not For Sale © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Not For Sale Full file 2at https://TestbankDirect.eu/ CHAPTER INTRODUCTION TO ANALYTIC GEOMETRY 11b x/y is negative whenever x and y have opposite signs: quadrants II and IV 12 If x2 > x1 , then x2 − x1 = P1 P3 = |x2 − x1 | If x2 < x1 , then x2 − x1 = −P1 P3 and P1 P3 = |−P1 P3 | = |x2 − x1 | 13a Any point on the y-axis has coordinates of the form (0, y) (11 − 0)2 + (2 − 0)2 = 14 Distance from (11, 2) to origin: √ 125 √ Distance from (−5, 10) to origin: (−5 − + (10 − = 125 √ Distance from (−1, −11) to origin: (−1 − 0)2 + (−11 − 0)2 = 122 0)2 0)2 15 Let A = (−2, −5), B = (−4, 1) and C = (5, 4); then AB = [−2 − (−4)]2 + (−5 − 1)2 = √ √ √ 40, AC = (−2 − 5)2 + (−5 − 4)2 = 130, and BC = (−4 − 5)2 + (1 − 4)2 = 90 Since (AB)2 + (BC)2 = (AC)2 , the triangle must be a right triangle 16 Let A = (−1, −1), B = (−2, 3), and C = (6, 5) After calculating AB = √ AC = 85, we observe that (AB)2 + (BC)2 = (AC)2 √ 17, BC = √ 68, and √ 17 The points (12, 0), (−4, 8) and (−1, −13) are all 5 units from (1, −2) 18 Distance from (−2 , 10) to (3 , −2): 13 Distance from (15 , 3) to (3 , −2): 13 19 Distance from (−1, −1) to (2, 8): (−1 − 2)2 + (−1 − 8)2 = √ + 81 = √ 90 = √ √ · 10 = 10 Distance from (2, 8) to (5, 17): (5 − 2)2 + (17 − 8)2 = √ √ 90 = 10 Distance from (−1, −1) to (5, 17): 62 + 182 = √ √ 360 = 10 √ √ √ Total distance 10 = 10 + 10, the sum of the other two distances 20 Distance from (x , y) to (−1 , 2): (x + 1)2 + (y − 2)2 = (x + 1)2 + (y − 2)2 = (3)2 x + 2x + + y − 4y + = x2 + y + 2x − 4y = d= squaring both sides 21 Distance from (x, y) to y-axis: x units Distance from (x, y) to (2, 0): By assumption, (x − 2)2 + y = x (x − 2)2 + y = x2 = x y − 4x + = x − 4x + + y (x − 2)2 + (y − 0)2 = (x − 2)2 + y squaring both sides © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ © Cengage Learning All rights reserved No distribution allowed without express authorization Answer: (−1, −11) Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Full file at https://TestbankDirect.eu/ 1.2 THE SLOPE 22 Let (x1 , y1 ) = (−3 , −5) and (x2 , y2 ) = (−1 , 7) Then from the midpoint formula x1 + x2 y1 + y2 −3 + (−1) −5 + , = , = (−2 , 1) 2 2 23 Let (x1 , y1 ) = (−2, 6) and (x2 , y2 ) = (2, −4) Then from the midpoint formula x1 + x2 y1 + y2 , 2 we get © Cengage Learning All rights reserved No distribution allowed without express authorization −2 + + (−4) , 2 24 x1 + x2 y1 + y2 , 2 = −3 + (−2) + , 2 = (0, 1) − ,7 = 25 Let (x1 , y1 ) = (5, 0) and (x2 , y2 ) = (9, 4) Then from the midpoint formula x1 + x2 y1 + y2 , 2 we get 5+9 0+4 , 2 26 x1 + x2 y1 + y2 , 2 = −4 + (−1) + (−7) , 2 27 The center is the midpoint: −2 + −1 + 11 , 2 = (7, 2) = − ,−2 = (2, 5) 28 Midpoint of given line segment: (2, 6) Midpoint of line segment from (2, 6) to (−2, 4): (0, 5) 1.2 The Slope Let (x2 , y2 ) = (1, 7) and (x1 , y1 ) = (2, 6) Then, by formula (1.4), m= we get m= y2 − y1 x2 − x1 7−6 = = −1 1−2 −1 Let (x2 , y2 ) = (−3 , −10) and (x1 , y1 ) = (−5, 2) Then by formula (1.4) y2 − y1 −10 − −12 m= = = = −6 x2 − x1 −3 − (−5) Let (x1 , y1 ) = (0, 2) and (x2 , y2 ) = (−4, −4) Then m= −4 − −6 = = −4 − −4 Let (x2 , y2 ) = (6 , −3) and (x1 , y1 ) = (4, 0) Then m = y2 − y1 −3 − −3 = = =− x2 − x1 6−4 2 Let (x2 , y2 ) = (7, 8) and (x1 , y1 ) = (−3, −4) Then Not For Sale m= − (−4) 8+4 12 = = = − (−3) 7+3 10 © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Not For Sale m = −4 − =− 8−0 m = 43 − (−1) 44 = = −22 −1 − −2 m = 31 − = 30 − (−1) m = −5 − −9 = (undefined) 3−3 10 m = 6−6 = =0 − (−3) 11 m = −3 − (−3) = =0 9−5 12 m − − (−2) = (undefined) 4−4 13 m = 3−2 = 12 − (−2) 14 14 (a) tan 0◦ = 0; (b) tan 30◦ = √ 3 ; INTRODUCTION TO ANALYTIC GEOMETRY (c) tan 150◦ = − √ 3 ; (d) tan 90◦ is undefined; (e) tan 45◦ = 1; (f) tan 135◦ = −1 15 See answer section of book 16 Slope of AB = 0−2 −2 = = 2; of BC = ; of AC = − −2 − (−1) −1 − (−5) 6 = = − Slope of perpendicular is given by the negative −7 − −13 13 13 reciprocal and is therefore − = (−6/13) 17 Slope of given line is −3 − = −3 −1 − (−4) −9 − Slope of line through (−4, 6) and (1, −9): = −3 − (−4) Since the lines have the same slope and pass through (−4, 6), they must coincide 18 Slope of line through (−4, 6) and (−1, −3): − 10 −4 = = −4 − −10 10 − 10 Slope of line through (6, 10) and (10, 0): = =− − 10 −4 Since the slopes are negative reciprocals, the lines are perpendicular 19 Slope of line through (−4, 6) and (6, 10): 20 Slope of line through (−4 , 2) and (−1 , 8): 2; Slope of line through (9 , 4) and (6 , −2): 2; Slope of line through (−1 , 8) and (9 , 4): −2/5; Slope of line through (−4 , 2) and (6 , −2): −2/5; −3 − −6 = = −3 − (−2) 6−0 Slope of line through (7, 6) and (9, 0): = = −3 7−9 −2 3−6 −3 Slope of line through (−2, 3) and (7, 6): = = −2 − −9 −3 − Slope of line through (0, −3) and (9, 0): = 0−9 21 Slope of line through (0, −3) and (−2, 3): © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ © Cengage Learning All rights reserved No distribution allowed without express authorization Full file 4at https://TestbankDirect.eu/ CHAPTER Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Full file at https://TestbankDirect.eu/ 1.3 THE STRAIGHT LINE Since −3 and are negative reciprocals, adjacent sides are perpendicular and opposite sides are parallel 22 Midpoint of line segment: −2 + −4 + , 2 −3 + −2 + , 2 23 Midpoint: −4 − = −2 6−3 = (3, −1) − (−1) = 5−3 Slope of line through (5, 6) and (3, −1): © Cengage Learning All rights reserved No distribution allowed without express authorization = (3, 2) Point: (6 , −4), m = 24 Let (x , y) be the other end of the diameter Since the center is the midpoint, we get y−3 and = 2; solving, x = −2, y = 25 tan θ = θ = 3.6◦ 26 tan θ = x+4 =1 rise 10.0 ft = = 0.0625 run 160 ft rise 2.5 m = ≈ 0.1316; θ = 7.5◦ run 19.0 m −1 − (−5) = = −1 −1 − −4 2+6 Slope of line through (x, 2) and (4, −6): = x−4 x−4 Since the two slopes must be equal, we have: = −1 x−4 = −x + multiplying both sides by x − 27 Slope of line through (−1, −1) and (3, −5): x = −4 −2 + 28 Slope of line segment (2 , −1) to (−3 , 2): − ; Slope of other line segment: ; so x−4 5 = (negative reciprocals); solving, x = x−4 1.3 The Straight Line Since (x1 , y1 ) = (−7, 2) and m = 1/2, we get y − = 12 (x + 7) y − y1 = m(x − x1 ) 2y − x − 2y + 11 = x+7 clearing fractions = x − 2y + + = Since (x1 , y1 ) = (0 , 3) and m = −4, we get y − = −4(x − 0) y − y1 = m(x − x1 ) 4x + y − 3 y − y1 = = m(x − x1 ) y+4 = 3(x − 3) y+4 = 3x − 3x − y − 13 = (x1 , y1 ) = (3, −4); m = Not For Sale By (1.8), y = © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Not For Sale Full file 6at https://TestbankDirect.eu/ CHAPTER y − y1 = m(x − x1 ) y−0 = − 13 (x − 0) 3y = −x x + 3y = INTRODUCTION TO ANALYTIC GEOMETRY (x1 , y1 ) = (0, 0); m = −1/3 By (1.9), x = −3 y−0 = 0(x + 4) (x1 , y1 ) = (−4, 0); m = y = x-axis First determine the slope: m = y2 −y1‘ x2 −x1 = 2−6 −3−7 = Let (x1 , y1 ) = (−3 , 2), then y − = 25 (x + 3) 5y − 10 = 2x + 2x − 5y + 16 = multiplying by First determine the slope using m = Then let (x1 , y1 ) = (−3, 4) to get y − = − 53 (x + 3) 3y − 12 = −5x − 15 5x + 3y + = 10 m = 6−3 −4−2 2y − x + 2y − 5−7 −3−1 (x1 , y1 ) = (2, 3) = −x + multiplying by = = = = (x 2y − 10 = x+3 x − 2y + 13 = 2y = choosing (x1 , y1 ) = (5, 0) y−5 13 6x + 2y multiplying by = − 12 (x − 2) −4 − = −1 9−5 y − = −1(x − 5) x+y−5 12 m = y − y1 = m(x − x1 ) = − 12 y−3 11 m = y2 − y1 − (−6) 10 = =− to get m = x2 − x1 −3 − −6 + 3) (x1 , y1 ) = (−3, 5) multiplying by = −6x + y = −3x + 52 y = mx + b m = −3, y-intercept = 52 ; see graph in answer section of book 14 Solving for y, we get y = x − Slope:1, y-intercept :−1 15 Since 2x = 3y, y = 23 x From the form y = mx + b, m = the origin and has slope 3 and b = The line passes through See graph in answer section of book 16 From y = −4x + 12, we get m = −4 and b = 12 © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ © Cengage Learning All rights reserved No distribution allowed without express authorization The line y = = 0x + has slope y − y1 = m(x − x1 ) Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Full file at https://TestbankDirect.eu/ 1.3 THE STRAIGHT LINE 17 2y − = y = 0x + 72 y = mx + b m = 0, y-intercept = 27 ; see graph in answer section of book 18 Solving for y: y = 14 x − 19 2x − 3y = −3y 2 3x = − © Cengage Learning All rights reserved No distribution allowed without express authorization = = −4x − y 6x 3x = + y = + in both cases, so that the lines are parallel y − 2x = = − 12 x − y = −6y From the form y = mx + b, m = 20 2x + 4y + y-intercept :− 32 4x − 6y + = −2x + y 4; slope: y = 2x + slope = slope Answer: The lines are perpendicular = − 12 21 3x − 4y = −4y 3y − 4x = 3y = 4x + = −3x + 4x y = − y = 43 x + The lines are neither parallel nor perpendicular 22 7x − 10y = y = 10 x 10 slope = Answer: neither 23 x + 3y = = y = slope = y − 3x − = y = 3x + = −x + 3y The y−4 − y = − 13 x slopes are − 13 + and 3, respectively Since the slopes are negative reciprocals, the lines are perpendicular 24 2x + 5y = 6x + 15y − 25 x = = + y = − 25 x + Slope is − in each case; the lines are parallel y 25 3x − 5y −5y y = 9x − 15y = −15y = −9x + y = −9 −15 x + −15 x − 15 in both cases; = −3x + = 5x − y = From the form y = mx + b, the slope m is 26 4y − 3x + = 4y = 3x − = 4x y 6x − 8y + − 15 −8y = so the lines are parallel = −6x − y = y = −6 −1 −8 x + −8 4x + Not For Sale slope = Answer: the lines are parallel slope = © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Not For Sale Full file 8at https://TestbankDirect.eu/ CHAPTER 27 2y − 3x 2y y = 6y + 4x = 3x + = 2x 6y +3 y The respective slopes are = INTRODUCTION TO ANALYTIC GEOMETRY = −4x + = −4 x+ − 23 x − 56 y = and − 23 Since the slopes are negative reciprocals, the lines are perpendicular x − 4y − = 2y + 8x − −4y = −x + 1 y = x− The respective slopes are −8x + 7 y = −4x + and −4 Since the slopes are negative reciprocals, the lines are 2y = = perpendicular 29 3y − 2x − 12 = 3y = 2x + 12 2x + 3y − = 3y = −2x + 3x y = +4 y = The lines are neither parallel nor pependicular 30 3x + 4y − = 6x − 8y − = − 23 x + −3x + −8y = −6x + 3 3 y = − x+1 y = x− 4 The lines are neither parallel nor perpendicular 4y = 3x + 4y = y = − 34 x y − y1 = m(x − x1 ) point-slope form y−1 = − 34 (x point: (−2, 1) 4y − = −3x − 3x + 4y + = 31 (given line) + slope = − 34 + 2) 32 The given line can be written y = − x + So the slope is − Slope of perpendicular: 4 y−1 = 3y − = (x + 2) 4x + 4x − 3y + 11 = y − y1 = m(x − x1 ) 33 To find the coordinates of the point of intersection, solve the equations simultaneously: 2x − 4y = 3x 5x + 4y = = adding x = From the second equation, 3(1) + 4y = 4, and y = So the point of intersection is (1, 14 ) From the equation 5x + 7y + = 0, we get 7y = −5x − y = − 75 x − slope= −5/7 © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ © Cengage Learning All rights reserved No distribution allowed without express authorization 28 Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Full file at https://TestbankDirect.eu/ 1.3 THE STRAIGHT LINE Thus (x1 , y1 ) = (1, 14 ) and m = − 57 The desired line is y − = − 57 (x − 1) To clear fractions, we multiply both sides by 28: 28y − = −20(x − 1) 28y − 20x + 28y − 27 = −20x + 20 = 34 The first two lines are perpendicular: y = 25 x + and y = −3x − © Cengage Learning All rights reserved No distribution allowed without express authorization 35 See graph in answer section of book 36 F = 3x, slope = 3, passing through the origin 37 From F = kx, we get = k · 12 Thus k = and F = 6x 38 y-intercept : initial value; t-intercept : the year the value becomes zero 39 F 212 32 = mC + b = m(100) + b F = 212, C = 100 = m(0) + b F = 32, C = b = 212 32 = m(100) + 32 second equation substituting into first equation 180 = 100 Solution: F = 95 C + 32 m = 40 C = 1800 + 500t 41 R = aT + b 51 = a · 100 + b R = 51, T = 100 54 = a · 400 + b R = 54, T = 400 −3 = −300a subtracting −3 a = = 0.01 −300 From the first equation, 51 = a · 100 + b, we get 51 = (0.01)(100) + b (a = 0.01) b = 50 So the formula R = aT + b becomes R = 0.01T + 50 42 P = kx; let P = 187.2 lb and x = 3.0 ft Then 187.2 = k(3.0) and k = 187.2 = 62.4 3.0 Not For Sale So the relationship is P = 62.4x © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Not For Sale Full file 10at https://TestbankDirect.eu/ CHAPTER 1.4 INTRODUCTION TO ANALYTIC GEOMETRY Curve Sketching In the answers below, the intercepts are given first, followed by symmetry, asymptotes, and extent Intercepts If x = 0, then y = −9 If y = 0, then x2 = x2 − = solving for x x = ±3 x = and x = −3 Symmetry If x is replaced by −x, we get y = (−x)2 − 9, which reduces to the given equation y = x2 − The graph is therefore symmetric with respect to the y-axis There is no other type of symmetry Asymptotes Since the equation is not in the form of a quotient with a variable in the denominator, there are no asymptotes Extent y is defined for all x Graph y x y = 1; y-axis; none; all x Intercepts If x = 0, then y = If y = 0, then = − x2 = x = ±1 x solving for x x = and x = −1 © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ © Cengage Learning All rights reserved No distribution allowed without express authorization y = 2, x = 12 ; none; none; all x Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Full file at https://TestbankDirect.eu/ 1.4 CURVE SKETCHING 11 Symmetry If x is replaced by −x, we get y = − (−x)2 , which reduces to the given equation y = − x2 The graph is therefore symmetric with respect to the y-axis There is no other type of symmetry Asymptotes Since the equation is not in the form of a quotient with a variable in the denominator, there are no asymptotes Extent y is defined for all x Graph © Cengage Learning All rights reserved No distribution allowed without express authorization y x √ y = 5, x = ± 5; y-axis; none; all x Intercepts If x = 0, then y = 0, and if y = 0, then x = So the only intercept is the origin Symmetry If we replace x by −x, we get y = −x, which does not reduce to the given equation So there is no symmetry with respect to the y-axis If y is replaced by −y, we get (−y)2 = x, which reduces to y = x, the given equation It follows that the graph is symmetric with respect to the x-axis To check for symmetry with respect to the origin, we replace x by −x and y by −y: (−y)2 = −x The resulting equation, y = −x, does not reduce to the given equation So there is no symmetry with respect to the origin Asymptotes Since the equation is not in the form of a fraction with a variable in the denominator, there are no asymptotes Extent Solving the equation for y in terms of x, we get √ y = ± x Not For Sale Note that to avoid imaginary values, x cannot be negative It follows that the extent is x ≥ © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Not For Sale Full file 12at https://TestbankDirect.eu/ CHAPTER Graph INTRODUCTION TO ANALYTIC GEOMETRY y x Intercepts If x = 0, then y = ±1 If y = 0, then x = −1 Symmetry If we replace x by −x we get y = −x + 1, which does not reduce to the given equation So there is no symmetry with respect to the y-axis If y is replaced by −y, we get (−y)2 = x + 1, which reduces to y = x + 1, the given equation It follows that the graph is symmetric with respect to the x-axis The graph is not symmetric with respect to the origin Asymptotes Since the equation is not in the form of a fraction with a variable in the denominator, there are no asymptotes √ Extent Solving the equation for y, we get y = ± x + To avoid imaginary values, we must have x + ≥ or x ≥ −1 Therefore the extent is x ≥ −1 Graph y x √ 10 y = ± 2; x = 2; x-axis; none; x ≤ © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ © Cengage Learning All rights reserved No distribution allowed without express authorization origin; x-axis; none; x ≥ Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Full file at https://TestbankDirect.eu/ 1.4 CURVE SKETCHING 13 11 Intercepts If x = 0, then y = (0 − 3)(0 + 5) = −15 If y = 0, then = (x − 3)(x + 5) x−3=0 x=3 x+5=0 x = −5 Symmetry If x is replaced by −x, we get y = (−x − 3)(−x + 5), which does not reduce to the given equation So there is no symmetry with respect to the y-axis Similarly, there is no © Cengage Learning All rights reserved No distribution allowed without express authorization other type of symmetry Asymptotes Since the equation is not in the form of a quotient with a variable in the denominator, there are no asymptotes Extent y is defined for all x Graph • y x (0,-15) 12 y = −24; x = −6, 4; none; none; all x 13 Intercepts If x = 0, then y = If y = 0, then = x(x + 3)(x − 2) x = 0, −3, Symmetry If x is replaced by −x, we get y = −x(−x + 3)(−x − 2), which does not reduce to the given equation So the graph is not symmetric with respect to the y-axis There is no other type of symmetry Asymptotes Since the equation is not in the form of a quotient with a variable in the denom- Not For Sale inator, there are no asymptotes Extent y is defined for all x © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Not For Sale Full file 14at https://TestbankDirect.eu/ CHAPTER Graph INTRODUCTION TO ANALYTIC GEOMETRY y x 15 Intercepts If x = 0, y = 0; if y = 0, then x(x − 1)(x − 2)2 = x = 0, 1, Symmetry If x is replaced by −x, we get y = −x(−x − 1)(−x − 2)2 , which does not reduce to the given equation So there is no symmetry with respect to the y-axis Similarly, there is no other type of symmetry Asymptotes None (the equation does not have the form of a fraction) Extent y is defined for all x Graph y x 16 y = 0; x = −2, 0, 3; none; none; all x © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ © Cengage Learning All rights reserved No distribution allowed without express authorization 14 y = 0; x = 0,1, 4; none; none; all x Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Full file at https://TestbankDirect.eu/ 1.4 CURVE SKETCHING 15 17 Intercepts If x = 0, then y = If y = 0, then = x(x − 1)2 (x − 2) x = 0, 1, Symmetry If x is replaced with −x, we get y = −x(−x − 1)2 (−x − 2), which does not reduce to the given equation Therefore there is no symmetry with respect to the y-axis There is no other type of symmetry Asymptotes None (the equation does not have the form of a fraction) Extent y is defined for all x Graph © Cengage Learning All rights reserved No distribution allowed without express authorization y x 18 y = 0; x = −2, 0, 3; none; none; all x y x 19 Intercepts If x = 0, y = 1; if y = 0, we have 0= x+2 y= −x + This equation has no solution Symmetry Replacing x by −x, we get which does not reduce to the given equation So there is no symmetry with respect to the y-axis Similarly, there is no other type of symmetry Asymptotes Setting the denominator equal to 0, we get x + = or x = −2 It follows that x = −2 is a vertical asymptote Also, as x gets large, y approaches So the Not For Sale x-axis is a horizontal asymptote Extent To avoid division by 0, x cannot be equal to −2 So the extent is all x except x = −2 © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Not For Sale Full file 16at https://TestbankDirect.eu/ CHAPTER Graph INTRODUCTION TO ANALYTIC GEOMETRY y x 21 Intercepts If x = 0, then y = If y = 0, then = (x − 1)2 This equation has no solution Symmetry Replacing x by −x, we get y= (−x − 1)2 which does not reduce to the given equation There are no other types of symmetry Asymptotes Setting the denominator equal to gives (x − 1)2 = or x = It follows that x = is a vertical asymptote Also, as x gets large, y approaches So the x-axis is a horizontal asymptote Extent To avoid division by 0, x cannot be equal to So the extent is the set of all x except x = Graph y x © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ © Cengage Learning All rights reserved No distribution allowed without express authorization 20 y = −1; none; x = 3; y = 0; x = Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Full file at https://TestbankDirect.eu/ 1.4 CURVE SKETCHING 17 22 y = − 14 ; none; none; x = −2; y = 0; x = −2 y x −6 −4 −2 23 Intercepts If x = 0, then y = If y = 0, then © Cengage Learning All rights reserved No distribution allowed without express authorization 0= x2 x−1 The only solution is x = Symmetry Replacing x by −x yields y= (−x)2 x2 = −x − −x − which is not the same as the given equation So the graph is not symmetric with respect to the y-axis Replacing y by −y, we have −y = x2 x−1 which does not reduce to the given equation So the graph is not symmetric with respect to the x-axis Similarly, there is no symmetry with respect to the origin Asymptotes Setting the denominator equal to 0, we get x − = 0, or x = So x = is a vertical asymptote There are no horizontal asymptotes (Observation: for very large x the in the denominator becomes insignificant So the graph x2 gets ever closer to y = = x; the line y = x is a slant asymptote.) x Extent To avoid division by 0, x cannot be equal to So the extent is all x except x = Graph y x 24 y = 0; x = 0; none; x = −2; y = 1; x = −2 Not For Sale © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Not For Sale Full file 18at https://TestbankDirect.eu/ CHAPTER INTRODUCTION TO ANALYTIC GEOMETRY 25 Intercepts If x = 0, then y = −1/2 If y = 0, then 0= x+1 (x − 1)(x + 2) The only solution is x = −1 Symmetry Replacing x by −x yields y= −x + (−x − 1)(−x + 2) which is not the same as the given equation There are no types of symmetry Asymptotes Setting the denominator equal to 0, we get (x − 1)(x + 2) = So x = horizontal asymptote Extent To avoid division by 0, the extent is all x except x = and x = −2 Graph y x 26 y = 0, x = 1, 0; none; x = −1, 2, y = 1; x = −1, 27 Intercepts If x = 0, then y = If y = 0, then x2 − x2 − x2 − x = = multiplying by x2 − = ±2 solution Symmetry Replacing x by −x reduces to the given equation So there is symmetry with respect to the y-axis There is no other type of symmetry Asymptotes Vertical: setting the denominator equal to 0, we have x2 − = or x = ±1 Horizontal: dividing numerator and denominator by x2 , the equation becomes y= 1− 1− x2 x2 © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ © Cengage Learning All rights reserved No distribution allowed without express authorization and x = −2 are the vertical asymptotes As x gets large, y approaches 0, so the x-axis is a Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Full file at https://TestbankDirect.eu/ 1.4 CURVE SKETCHING 19 As x gets large, y approaches So y = is a horizontal asymptote Extent All x except x = ±1 (to avoid division by 0) Graph y © Cengage Learning All rights reserved No distribution allowed without express authorization x 28 y = 14 , x = ±1; y-axis; x = ±2, y = 1; x = ±2 3y x −3 −2 −1 −1 −2 −3 29 Intercepts If x = 0, then y = −4 −1 = 4, or y = ±2 If y = 0, then 0= x2 − x2 − which is possible only if x2 − = 0, or x = ±2 Symmetry The even powers on x and y tell us that if x is replaced by −x and y is replaced by −y, the resulting equation will reduce to the given equation The graph is therefore symmetric with respect to both axes and the origin Asymptotes Vertical: setting the denominator equal to 0, we get x2 − = or x = ±1 Horizontal: dividing numerator and denominator by x2 , we get y2 = 1− 1− x2 x2 Not For Sale The right side approaches as x gets large Thus y approaches 1, so that y = ±1 are the horizontal asymptotes © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Not For Sale Full file 20at https://TestbankDirect.eu/ CHAPTER INTRODUCTION TO ANALYTIC GEOMETRY Extent From y=± x2 − x2 − we conclude that x2 − (x − 2)(x + 2) = ≥ x2 − (x − 1)(x + 1) Since the signs change only at x = 2, −2, 1, and −1, we need to use arbitrary “test val- 1 √ 31 Intercepts If x = 0, y = (−3)(5) = −15, or y = ± 15 j, which is a pure imaginary number If y = 0, (x − 3)(x + 5) = x = 3, −5 Symmetry Replacing y by −y, we get (−y)2 = (x − 3)(x + 5), which reduces to the given equation Hence the graph is symmetric with respect to the x-axis Asymptotes None (no fractions) Extent From y = ± (x − 3)(x + 5), we conclude that (x − 3)(x + 5) ≥ If x ≥ 3, (x − 3)(x + 5) ≥ If x ≤ −5, (x − 3)(x + 5) ≥ 0, since both factors are negative (or zero) If −5 < x < 3, (x − 3)(x + 5) < [For example, if x = 0, we get (−3)(5) = −15.] These observations are summarized in the following chart © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ © Cengage Learning All rights reserved No distribution allowed without express authorization ues”between these points The results are summarized in the following chart test (x − 2)(x + 2) values x − x − x + x + (x − 1)(x + 1) x>2 + + + + + Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig Full file at https://TestbankDirect.eu/ 1.4 CURVE SKETCHING 21 test values x−3 x+5 (x − 3)(x + 5) x>3 + + + −5 < x < − + − x < −5 −6 − − + Extent: x ≤ −5, x ≥ © Cengage Learning All rights reserved No distribution allowed without express authorization Graph • (-5,0) y (3,0) • x 32 y = 0, x = 0; x-axis; x = −2, y = ±1; x < −2, x ≥ 33 Intercepts If x = 0, y = 0; if y = 0, x = Symmetry Replacing y by −y leaves the equation unchanged So there is symmetry with respect to the x-axis There is no other type of symmetry Asymptotes Vertical: setting the denominator equal to 0, we get (x − 3)(x − 2) = or x = 3, Horizontal: as x gets large, y approaches (x-axis) Extent From y=± we conclude that x (x − 3)(x − 2) x ≥ (x − 3)(x − 3) Since signs change only at x = 0, and 3, we need to use “test values”between these points The results are summarized in the following chart test x values x x − x − (x − 3)(x − 2) x