Chapter 2 Polynomial, Power, and Rational Functions

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1 Section. Linear and Quadratic Functions and Modeling 9 Chapter Polynomial, Power, and Rational Functions Section. Linear and Quadratic Functions and Modeling Eploration. $000 per year.. The equation will have the form v(t)=mt+b.the value of the building after 0 year is v(0)=m(0)+b=b=0,000. The slope m is the rate of change, which is 000 (dollars per year). So an equation for the value of the building (in dollars) as a function of the time (in years) is v(t)= 000t+0,000.. v(0)=0,000 and v(6)= 000(6)+0,000=8,000 dollars. 4. The equation v(t)=9,000 becomes 000t+0,000=9, t=,000 t=. years. Quick Review.. y= m= so y-4= (-) f()= m= - 4 so y-6= - 4 (+4) f()= ( 4, 6) (, ) (, ) 0 y y (, 4) 0. y-4= - (+),ory= y. m= so y-= (-0) f()= + y (, 4) (, ) (0, ) (, 0). (+) =(+)(+)= +++9 = (-6) =(-6)(-6)=(-8)(-6) = = =( -+)=(-)(-) =(-) Section. Eercises. Not a polynomial function because of the eponent. Polynomial of degree with leading coefficient. Not a polynomial function because of cube root. (a) The verte is at (, ), in Quadrant III, einating all but (a) and (d). Since f(0)=, it must be (a).. (b) The verte is in Quadrant I, at (, 4), meaning it must be either (b) or (f). Since f(0)=, it cannot be (f): if the verte in (f) is (, 4), then the intersection with the y-ais would be about (0, ). It must be (b). 7. (e) The verte is at (, ) in Quadrant IV, so it must be (e). Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

2 40 Chapter Polynomial, Power, and Rational Functions 9. Translate the graph of f()= units right to obtain the graph of h()=(-), and translate this graph units down to obtain the graph of g()=(-) -. y 0 0 [ 4, 6] by [0, 0]. f()= ( +6)+0 = ( +6+64)+0+64= (+8) +74. Verte: ( 8, 74); ais: = 8; opens downward; intersects ais at about 6.60 and 0.60(-8 ; 74).. Translate the graph of f()= units left to obtain the graph of h()=(+), vertically shrink this graph by a factor of to obtain the graph of k()= (+), and translate this graph units down to obtain the graph of g()= (+) -. y 0 [ 0, ] by [ 00, 00] 7. f()=( +)+7 = +7- = a + 9 a b b 0 Verte: a - ais: = - opens upward; does not, b; ; intersect the -ais; vertically stretched by. For #, with an equation of the form f()=a(-h) +k, the verte is (h, k) and the ais is =h.. Verte: (, ); ais: =. Verte: (, 7); ais: = 7. f()=a + -4 b 7 = -4- =a + a + # b 6 b Verte: a - ; ais: = 6, - 7 b 6 9. f()= ( -8)+ = ( - # 4+6)+ +6= (-4) +9 Verte: (4, 9); ais: =4. g()= a b = +4- = a - 9 a - # b b [.7, ] by [,.] For #9 44, use the form y=a(-h) +k, taking the verte (h, k) from the graph or other given information. 9. h= and k=, so y=a(+) -. Now substitute =, y= to obtain =4a-, so a=: y=(+) h= and k=, so y=a(-) +. Now substitute =4, y= 7 to obtain 7=9a+, so a= : y= (-) h= and k=, so y=a(-) +. Now substitute =0, y= to obtain =a+, so a=: y=(-) Strong positive 47. Weak positive 49. (a) Verte: a ais: =, b;. f()=( -4+4)+6-4=(-) +. Verte: (, ); ais: =; opens upward; does not intersect ais. [, 4] by [0, 0] (b) Strong positive Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

3 Section. Linear and Quadratic Functions and Modeling 4. m= - 0 = 470 and b=0, so v(t)= 470t+0. At t=, v()=( 470)()+0=$940.. (a) y L The slope, m L 0.4, represents the average annual increase in hourly compensation for production workers, about $0.4 per year. (b) Setting =40 in the regression equation leads to y L $.7.. (a) [0, 00] by [0, 000] is one possibility. (b) When 07. or 7.66 either 07, units or 7, 66 units. 7. If the strip is feet wide, the area of the strip is A()=(+)(40+)-000 ft. This equals 04 ft when =. ft. 9. (a) R()=(6, )( ). (b) Many choices of Xma and Ymin are reasonable. Shown is [0, ] by [0,000, 7,000]. (b) The maimum height is 90 ft,. sec after it is shot. 6. The quadratic regression is y Plot this curve together with the curve y=40, and then find the intersection to find when the number of patent applications will reach 40,000. Note that we use y=40 because the data were given as a number of thousands. The intersection occurs at L 6., so the number of applications will reach 40,000 approimately 6 years after 980 in (a) y = 40 [, 0] by [0, 00] [, 4] by [0, 40] [0, ] by [0000, 7000] (c) The maimum revenue $6,00 is achieved when =8; that is, charging 90 cents per can. 6. (a) g L ft/sec. s 0 = 8 ft and v 0 = 9 ft/sec. So the models are height= s(t) = -6t + 9t + 8 and vertical velocity= v(t) = -t + 9. The maimum height occurs at the verte of s(t). h = - b and a = - 9 (-6) =.87, k = s(.87) =.. The maimum height of the baseball is about ft above the field. (b) The amount of time the ball is in the air is a zero of s(t). Using the quadratic formula, we obtain t = -9 ; 9-4(-6)(8) (-6) -9 ;,776 = L or 6.4. Time is not - negative, so the ball is in the air about 6.4 seconds. (c) To determine the ball s vertical velocity when it hits the ground, use v(t) = -t + 9, and solve for t = 6.4. v(6.4) = -(6.4) + 9 L -7 ft/sec when it hits the ground. 6. (a) h= 6t +80t-0. The graph is shown in the window [0, ] by [ 0, 00]. (b) y (c) On average, the children gained 0.68 pound per month. (d) [, 4] by [0, 40] (e) 9.4 lbs 69. The Identity Function f()= [ 4.7, 4.7] by [.,.] Domain: (- q, q) Range: (- q, q) Continuity: The function is continuous on its domain. Increasing decreasing behavior: Increasing for all Symmetry: Symmetric about the origin Boundedness: Not bounded Local etrema: None Horizontal asymptotes: None Vertical asymptotes: None End behavior: f() = -q and f() = q : q : -q [0, ] by [ 0, 00] Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

4 4 Chapter Polynomial, Power, and Rational Functions 7. False. For f()= +-, the initial value is f(0)= m = The answer is E. 4 - (-) = - 6 = -. For #7 76, f()=(+) - corresponds to f()=a(-h) +k with a= and (h, k)=(, ). 7. The ais of symmetry runs vertically through the verte: =. The answer is B. 77. (a) Graphs (i), (iii), and (v) are linear functions. They can all be represented by an equation y=a+b, where a Z 0. (b) In addition to graphs (i), (iii), and (v), graphs (iv) and (vi) are also functions, the difference is that (iv) and (vi) are constant functions, represented by y=b, b Z 0. (c) (ii) is not a function because a single value (i.e., = ) results in a multiple number of y-values. In fact, there are infinitely many y-values that are valid for the equation =. 79. Answers will vary. One possibility: When using the leastsquares method, mathematicians try to minimize the residual y i -(a i +b), i.e., place the predicted y-values as close as possible to the actual y-values. If mathematicians reversed the ordered pairs, the objective would change to minimizing the residual i -(cy i +d), i.e., placing the predicted -values as close as possible to the actual -values. In order to obtain an eact inverse, the - and y-values for each y pair would have to be almost eactly the same distance from the regression line which is statistically impossible in practice. 8. (a) If a +b +c=0, then -b ; b - 4ac = by the quadratic a -b + b - 4ac formula. Thus, = and a -b - b - 4ac = and a -b + b - 4ac - b - b - 4ac + = a -b = a = -b a = -b a. (b) Similarly, # = a -b + b - 4ac ba -b - b - 4ac b a a b - (b - 4ac) 4ac c = = = 4a 4a a. 8. Multiply out f() to get -(a+b)+ab. Complete (a + b) the square to get a - a + b b +ab-. The 4 a + b verte is then (h, k) where h= and (a + b) (a - b) k=ab- = The Constant Rate of Change Theorem states that a function defined on all real numbers is a linear function if and only if it has a constant nonzero average rate of change between any two points on its graph. To prove this, suppose f() = m + b with m and b constants and m Z 0. Let and be real numbers with Z. Then the average rate of change is f( ) - f( ) = (m + b) - (m + b) = - - m - m = m( - ) = m, a nonzero constant. - - Now suppose that m and are constants, with m Z 0. Let be a real number such that Z, and let f be a function defined on all real numbers such that f() - f( ) = m. Then f() - f( ) = m( - ) = - m - m, and f() = m + (f( ) - m ). f( ) - m is a constant; call it b. Then f( ) - m = b ; so, f( ) = b + m and f() = b + m for all Z. Thus, f is a linear function. Section. Power Functions with Modeling Eploration. [.,.] by [.,.] [, ] by [, ] [ 0, 0] by [ 00, 00] The pairs (0, 0), (, ), and (, ) are common to all three graphs. The graphs are similar in that if 6 0, f(), g(), and h() 6 0 and if >0, f(), g(), and h()>0. They are different in that if 6, f(), g(), and h() : 0 at dramatically different rates, and if 7, f(), g(), and h() : q at dramatically different rates.. [.,.] by [ 0.,.] Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

5 Section. Power Functions with Modeling 4 [, ] by [, ] Continuous Decreasing on (- q, 0). Increasing on (0, q). Even. Symmetric with respect to y-ais. Bounded below, but not above Local minimum at =0. Asymptotes: None End Behavior: 4 = q, 4 = q : -q : q [, ] by [ 0, 400] The pairs (0, 0), (, ), and (, ) are common to all three graphs. The graphs are similar in that for Z 0, f(), g(), and h() 7 0. They are diffferent in that if 6, f(), g(), and h() : 0 at dramatically different rates, and if 7, f(), g(), and h() : q at dramatically different rates. Quick Review... d. q 4 7. / 9. L.7-4/ Section. Eercises. power=, constant= -. not a power function. power=, constant=c g 7. power=, constant= 9. power=, constant=k. degree=0, coefficient= 4. degree=7, coefficient= 6. degree=, coefficient=4 7. A=ks 9. I=V/R. E=mc. The weight w of an object varies directly with its mass m, with the constant of variation g.. The refractive inde n of a medium is inversely proportional to v, the velocity of light in the medium, with constant of variation c, the constant velocity of light in free space. 7. power=4, constant= Domain: (- q, q) Range: [0, q) [, ] by [, 49] 9. power= constant= 4, Domain: [0, q) Range: [0, q) Continuous Increasing on [0, q). Bounded below Neither even nor odd Local minimum at (0, 0) Asymptotes: None End Behavior: 4 = q : q [, 99] by [, 4]. Start with y = 4 and shrink vertically by Since f( )= f is even. (-)4 =. 4, [, ] by [, 9]. Start with y =, then stretch vertically by. and reflect over the -ais. Since f(-) = -.(-) =. = f(), f is odd. [, ] by [ 0, 0] Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

6 44 Chapter Polynomial, Power, and Rational Functions. Start with y = 8, then shrink vertically by Since f( )= f is even. 4 (-)8 = 4. kt (0.96 atm)(.46 L). V= so k = PV = P, T 0 K 4 8 = f(), atm - L =0.006 K atm - L a b(0 K) K At P=.4 atm, V=.4 atm =. L a.00 * 08 m b sec [, ] by [, 49] 7. (g) 9. (d) 4. (h) 4. k=, a= In the first quadrant, the function is 4. increasing and concave down. f is undefined for 6 0. c. n = c so v= = v, n. (a) [, 7] by [0, 40] (b) r L.04 # w (c).4 =.4 * 0 8 m sec [, 99] by [, 0] 4. k = -, a = 4 In the fourth quadrant, f is decreasing. and concave down. f(-) = -( (-) 4 ) = -( 4 ) = - 4/ = f(), so f is even. [, 7] by [0, 40] (d) Approimately 7.67 beats/min, which is very close to Clark s observed value. 7. (a) [ 0, 0] by [ 9, ] 47. k = In the first quadrant, f is decreasing and, a = -. concave up. f(-) = (-)- = = -f(), so f is odd. [, ] by [ 0, 0] 49. y = 8 power = -, constant = 8., (-) = - - [0.8,.] by [ 0., 9.] (b) y L 7.9 # ; yes (c) [0.8,.] by [ 0., 9.] W W (d) Approimately.76 and respectively. m m, 9. False. f( )=( ) / = ( / )= f() and so the function is odd. It is symmetric about the origin, not the y-ais. 6. f(0) = -(0) -/ = - # is undefined. 0 / = - # 0 Also, f( )= ( ) / = ( )=, f()= () / = ()=, and f()= () /.08. The answer is E. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

7 Section. Power Functions with Modeling 4 6. f()= / =( / ) = () is defined for Ú 0. The answer is B. 6. (a) [0, ] by [0, ] [0, ] by [0, ] [, ] by [, ] The graphs of f()= and h()= are similar and appear in the st and rd quadrants only. The graphs of g()= and k()= 4 are similar and appear in the st and nd quadrants only. The pair (, ) is common to all four functions. f g h k Domain Range y 0 y 0 y 0 y 0 Continuous yes yes yes yes Increasing (, 0) (, 0) Decreasing (, 0), (0, ) (0, ) (, 0), (0, ) (0, ) Symmetry w.r.t. origin w.r.t. y-ais w.r.t. origin w.r.t. y-ais Bounded not below not below Etrema none none none none Asymptotes -ais, y-ais -ais, y-ais -ais, y-ais -ais, y-ais End Behavior f() 0 g() 0 h() 0 k() 0 (b) [0, ] by [0, ] [0, ] by [0, ] [, ] by [, ] The graphs of f()= / and h()= /4 are similar and appear in the st quadrant only. The graphs of g()= / and k()= / are similar and appear in the st and rd quadrants only. The pairs (0, 0), (, ) are common to all four functions. Domain f g h k [0, ) (, ) [0, ) (, ) Range y 0 (, ) y 0 (, ) Continuous yes yes yes yes Increasing [0, ) (, ) [0, ) (, ) Decreasing Symmetry none w.r.t. origin none w.r.t. origin Bounded below not below not Etrema min at (0, 0) none min at (0, 0) none Asymptotes none none none none End Behavior f() g() g() h() k() k() Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

8 46 Chapter Polynomial, Power, and Rational Functions 67. Our new table looks like: Table.0 (revised) Average Distances and Orbit Periods for the Si Innermost Planets Average Distance Period of Planet from Sun (Au) Orbit (yrs) Mercury Venus Earth Mars..88 Jupiter.0.86 Saturn Source: Shupe, Dorr, Payne, Hunsiker, et al., National Geographic Atlas of the World (rev. 6th ed.). Washington, DC: National Geographic Society, 99, plate 6. Using these new data, we find a power function model of: y L #.0 L.. Since y represents years, we set y = T and since represents distance, we set = a then, y =. Q T = a / Q (T) = (a / ) Q T = a. 69. If f is even, so (f() 0). f() = f() = f(-), f(-), Since g()= =g( ), g is also even. f() = f(-) If g is even, g()=g( ), so g( )= =g()= f(-) f(). Since = f( )=f(), and f is even. f(-) f(), If f is odd, f() = -f(), so f() Z 0. f() = - f(), Since g()= = g(), g is also odd. f() = - f() If g is odd, g()=g( ), so g( )= = g()= - f(-) f(). Since = - f( )= f(), and f is odd. f(-) f(), 7. (a) The force F acting on an object varies jointly as the mass m of the object and the acceleration a of the object. (b) The kinetic energy KE of an object varies jointly as the mass m of the object and the square of the velocity v of the object. (c) The force of gravity F acting on two objects varies jointly as their masses m and m and inversely as the square of the distance r between their centers, with the constant of variation G, the universal gravitational constant. Section. Polynomial Functions of Higher Degree with Modeling Eploration. (a) = q, (b) (c) (d). (a) : q (- ) = -q, : q : q : q : q = q, : -q [, ] by [, ] [, ] by [, ] : -q [, ] by [, ] (-0. 7 ) = -q, [, ] by [, ] (- 4 ) = -q, [, ] by [, ] = -q : -q = -q : -q : -q (- ) = q (-0. 7 ) = q (- 4 ) = -q Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

9 Section. Polynomial Functions of Higher Degree with Modeling 47 (b) : q = q, : -q = q (c) : q [, ] by [, ] 6 = q, : -q 6 = q [, ] by [, ] If a n 7 0 and n>0, f() = q and : q f() = - q. If a n <0 and n>0, : -q f() = -q and f() = q. : q : -q Eploration. y= It is an eact fit, which we epect with only 4 data points! (d) : q [, ] by [, ] (-0. ) = -q, : -q (-0. ) = -q [, 0] by [, ] [, ] by [, ]. y= It is an eact fit, eactly what we epect with only data points!. (a) : q (-0. ) = -q, : -q (-0. ) = q [., 8.] by [ 8, ] (b) (c) : q : q [, ] by [, ] (- ) = -q, [, ] by [, ] 4 = q, : -q : -q 4 = q (- ) = -q Quick Review.. (-4)(+). (-)(-). (-)(-) 7. =0, = 9. = 6, =, =. Section. Eercises. Start with y=, shift to the right by units, and then stretch vertically by. y-intercept: (0, 4). 0 y 0 (d). = q,. = -q : q [, ] by [, ] : -q Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

10 48 Chapter Polynomial, Power, and Rational Functions. Start with y=, shift to the left by unit, vertically shrink by reflect over the -ais, and then vertically, shift up units. y-intercept: a0,. b. One possibility: y [ 0, 0] by [ 000, 000] For #7 4, when one end of a polynomial function s graph curves up into Quadrant I or II, this indicates a it at q. And when an end curves down into Quadrant III or IV, this indicates a it at - q. 7.. Start with y= 4, shift to the left units, vertically stretch by, reflect over the -ais, and vertically shift down units. y-intercept: (0, ). 40 y 9. : q : -q [, ] by [ 8, ] f() = q f() = -q 7. Local maimum: (0.79,.9), zeros: =0 and.6. The general shape of f is like y= 4, but near the origin, f behaves a lot like its other term,. f is neither even nor odd.. : q : -q [ 8, 0] by [ 0, 00] f() = - q f() = q [, ] by [, ] 9. Cubic function, positive leading coefficient. The answer is (c).. Higher than cubic, positive leading coefficient. The answer is (a).. One possibility:. : q : -q [, ] by [ 4, 6] f() = q f() = q [ 00, 00] by [ 000, 000] : q : -q [, ] by [ 0, 0] f() = q f() = q Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

11 Section. Polynomial Functions of Higher Degree with Modeling 49 For # 8, the end behavior of a polynomial is governed by the highest-degree term.. f() = q, f() = q : q : -q 7. f() = - q, f() = q : q : -q 9. (a); There are zeros: they are.,, and... (c); There are zeros: approimately 0.7 (actually /), 0., and. For #, factor or apply the quadratic formula.. 4 and. / and / For #6 8, factor out, then factor or apply the quadratic formula. 7. 0, /, and 9. Degree ; zeros: =0 (multiplicity, graph crosses -ais), = (multiplicity, graph is tangent). y 0 4. Degree ; zeros: = (multiplicity, graph crosses -ais), = (multiplicity, graph is tangent). 0 y Zeros: 4.90, 0.4,,. [ 6, ] by [, ] 49. 0, 6, and 6. Algebraically factor out first..,, and. Graphically. [ 0, ] by [ 00, 0] For # 6, the minimal polynomials are given; any constant (or any other polynomial) can be multiplied by the answer given to give another answer.. f()=(-)(+4)(-6) = f()= ( - )( + ) (-4) =( -)(-4)= y= [ 4, 4] by [ 0, 0] 9. y= Zeros:.4, 0.74,.67 [, ] by [ 0, 0] 4. Zeros:.47,.46,.94 [0, 0] by [, 4] 6. f()= has an odd-degree leading term, which means that in its end behavior it will go toward q at one end and toward q at the other. Thus the graph must cross the -ais at least once. That is to say, f() takes on both positive and negative values, and thus by the Intermediate Value Theorem, f()=0 for some. 6. (a) [, ] by [ 0, 0] [0, 60] by [ 0, 0] Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

12 0 Chapter Polynomial, Power, and Rational Functions (b) y= (c) [0, 60] by [ 0, 0] (d) y() 6.9 ft (e) Using the quadratic equation to solve 0= (0.6-00), we find two answers: =67.74 mph and = mph. Clearly the negative value is etraneous. 6. (a) [0.9999,.000] by [ *0 7, *0 7 ] 79. A maimum and minimum are not visible in the standard window, but can be seen on the window [0., 0.4] by [.9,.]. [0, 0.8] by [0,.0] (b) 0.9 cm from the center of the artery 67. The volume is V()=(0-)(-); use any with 0< 0.99 or.644 <. [0., 0.4] by [.9,.0] 8. The graph of y=( -) (shown on the window [, ] by [, ]) increases, then decreases, then increases; the graph of y= only increases. Therefore, this graph cannot be obtained from the graph of y= by the transformations studied in Chapter (translations, reflections, and stretching/shrinking). Since the right side includes only these transformations, there can be no solution. [0, ] by [0, 00] 69. True. Because f is continuous and f()=() -() -= <0 while f()=() -() -=>0, the Intermediate Value Theorem assures us that the graph of f crosses the -ais (f()=0) somewhere between = and =. 7. When =0, f()=(-) +=( ) +=. The answer is C. 7. The graph indicates three zeros, each of multiplicity : =, =0, and =. The end behavior indicates a negative leading coefficient. So f()= (+)(-), and the answer is B. 7. The first view shows the end behavior of the function but obscures the fact that there are two local maima and a local minimum (and 4 -ais intersections) between and 4. These are visible in the second view, but missing is the minimum near =7 and the -ais intersection near =9. The second view suggests a degree 4 polynomial rather than degree. 77. The eact behavior near = is hard to see. A zoomed-- in view around the point (, 0) suggests that the graph just touches the -ais at 0 without actually crossing it that is, (, 0) is a local maimum. One possible window is [0.9999,.000] by [ 0 7, 0 7 ]. [, ] by [, ] 8. (a) Substituting =, y=7, we find that 7=(-)+7, so Q is on line L, and also f()= =7, so Q is on the graph of f(). (b) Window [.8,.] by [6, 8]. Calculator output will not show the detail seen here. [.8,.] by [6, 8] (c) The line L also crosses the graph of f() at (, ). [, ] by [, ] Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

13 Section.4 Real Zeros of Polynomial Functions 8. (a) Label the points of the diagram as shown, adding the horizontal segment FH. Therefore, ECB is similar (in the geometric sense) to HGB, and also ABC is similar to AFH. Therefore: HG 8 D - u AF or = and also AB = FH EC = BG BC, D, BC, y or = D - u Then = y - 8 y D. y. A E y H D u F 8 D u B G C D 8 (b) Equation (a) says Multiply both sides by = - 8 y. y: 8y = y - 8. Subtract y from both sides and factor: y(8 - ) = -8. Divide both sides by 8 - : -8 y = Factor out - from numerator and denominator: y = - 8. (c) Applying the Pythagorean Theorem to EBC and ABC, we have +D =0 and y +D =0, which combine to give D =400- =900-y, or y - =00. Substituting y=8/(-8), we get a 8 - =00, so that - 8 b 64 - =00, or 64 - (-8) ( - 8) =00(-8). Epanding this gives ,000 = This is equivalent to ,000 = 0. (d) The two solutions are.9446 and.78. Based on the figure, must be between 8 and 0 for this problem, so.78. Then D= 0-6. ft. Section.4 Real Zeros of Polynomial Functions Quick Review ( - 4) = ( - ) = ( + )( - ) 7. 4( + - ) = 4( + )( - ) 9. ( + ) - ( + ) = ( + ) - ( + ) = ( + )( - ) = ( + )( + )( - ) Section.4 Eercises f()=(-) +; f() = ( + + 4)( + ) - ; f() + = f() = ( )( + - ) - + 8; f() + - = f() - = = = = The remainder is f()=.. The remainder is f( )= 4. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

14 Chapter Polynomial, Power, and Rational Functions 7. The remainder is f()=. 9. Yes: is a zero of the second polynomial.. No: When =, the second polynomial evaluates to 0.. Yes: - is a zero of the second polynomial.. From the graph it appears that (+) and (-) are factors f()=(+)(-)(-7) 7. (+)(-)(-4)= ( - )a - ba - b = ( - )( - )( - ) 9 = Since f( 4)=f()=f()=0, it must be that (+4), (-), and (-) are factors of f.so f()=k(+4)(-)(-) for some constant k. Since f(0)=80, we must have k=. So f()=(+4)(-)(-). ;. Possible rational zeros: or, ;, ;, ;, ;6,,, ; is a zero. 6 ;, ;, ;9. Possible rational zeros: or,, 9, ;, ;,, ; ; 9 ; is a zero., Since all numbers in the last line are Ú 0, is an upper bound for the zeros of f Since all values in the last line are Ú 0, is an upper bound for the zeros of f() Since the values in the last line alternate signs, - is a lower bound for the zeros of f() Since the values in the last line alternate signs, 0 is a lower bound for the zeros of f(). 4. By the Upper and Lower Bound Tests, is a lower bound and is an upper bound. No zeros outside window Synthetic division shows that the Upper and Lower Bound Tests were not met. There are zeros not shown (appro..00 and.00), because and are not bounds for zeros of f() , , For #49 6, determine the rational zeros using a grapher (and the Rational Zeros Test as necessary). Use synthetic division to reduce the function to a quadratic polynomial, which can be solved with the quadratic formula (or otherwise). The first two are done in detail; for the rest, we show only the synthetic division step(s). 49. Possible rational zeros: ;, ;, ;, ;6, or ;, ; ;, ;, ;, ;6, ; The only rational zero is, ;.. Synthetic division (below) leaves -4, so the irrational zeros are ;. / Rational: -; irrational: ; Rational: and 4; irrational: ; Rational: - and 4; irrational: none Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

15 Section.4 Real Zeros of Polynomial Functions / The supply and demand graphs are shown on the window [0, 0] by [0, 00]. They intersect when p=$6.7, at which point the supply and demand equal.7. [0, 0] by [0, 00] 9. Using the Remainder Theorem, the remainder is (-) 40 - = (a) Lower bound: Upper bound: The Upper and Lower Bound Tests are met, so all real zeros of f lie on the interval [, 4]. (b) Potential rational zeros: Factors of 8 ;, ;, ;9, ;8 : Factors of ; A graph shows that is most promising, so we verify with synthetic division: Use the Remainder Theorem: f(-) = 0 Z 0 f(-) = 9 Z 0 f() = 7 Z 0 f( 8)=,960,040 f(8)=,78,40 f( 9)=,97 f(9)=9,89 Since all possible rational roots besides yield nonzero function values, there are no other rational roots. 6. The statement f()=0 means that = is a zero of f(). and that is an -intercept of the graph of f(). And it follows that - is a factor of f() and thus that the remainder when f() is divided by - is zero. So the answer is A. 67. f()=(+)( +-)- yields a remainder of when divided by either + or +-, from which it follows that + is not a factor of f() and that f() is not evenly divisible by +. The answer is B. 69. (a) The volume of a sphere is V = 4 In this case, the pr. radius of the buoy is, so the buoy s volume is weight (b) Total weight = volume # unit volume = volume # density. In this case, the density of the buoy is so, the weight W b of the buoy is 4 d, W b = 4p # 4 d = dp. (c) The weight of the displaced water is W H volume # O = density. We know from geometry that the volume of a spherical cap is V = p so, 6 (r + h ) h, p pd W. 6 (r + ) 6 (r + H ) # O = d = (d) Setting the two weights equal, we have: W b = W H O pd = pd 6 (r + ) = (r + ) 0 = (6 - + ) - 0 = = - +. Solving graphically, we find that L 0.67 m, the depth that the buoy will sink. 7. (a) Shown is one possible view, on the window [0, 600] by [0, 00]. 4 p. [, 4] by [, 49] (c) f() = ( - ) ( ) (d) From our graph, we find that one irrational zero of is L.04. (e) f() L ( - )( -.04)( ) 6. False. -a is a factor if and only if f(a)=0. So (+) is a factor if and only if f( )=0. [0, 600] by [0, 00] (b) The maimum population, after 00 days, is 460 turkeys. (c) P=0 when t L. about days after release. (d) Answers will vary. One possibility: After the population increases to a certain point, they begin to compete for food and eventually die of starvation. 7. (a) sign changes in f(), sign change in f( )= + ++; 0 or positive zeros, negative zero. (b) No positive zeros, or negative zeros. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

16 4 Chapter Polynomial, Power, and Rational Functions (c) positive zero, no negative zeros. (d) positive zero, negative zero Divide numerator and = denominator by. - Write coefficients of zero of divisor - dividend. line for products line for sums - Quotient, remainder 4 8 Copy into the first quotient position. Multiply # = and add this to - Multiply - # and add this to. = - 4 Multiply # and add this to The last line tells 4 =. 8. us a - ba b = (a) g()=f(), so the zeros of f and the zeros of g are identical. If the coefficients of a polynomial are rational, we may multiply that polynomial by the least common multiple (LCM) of the denominators of the coefficients to obtain a polynomial, with integer coefficients, that has the same zeros as the original. (b) The zeros of f() are the same as the zeros of 6f() = Possible rational ;, ;, ;7, ; zeros: or ;, ;, ;, ;6, ;, ;, ;7, ;, ;, ;, ; 7 ; 7 The actual zeros are, ; 6, ; 7, ;, ;, 6. 7/, /, and. (c) The zeros of f() are the same as the zeros of f() = Possible rational zeros: ;, ;, ;, ;, ;6, ;0, ;, ;0, or ;, ;, ;, ;4, ;6, ; ;, ;, ;, ;, ;6, ;0, ;, ;0, ;, ;, ;, ;, ; 0, ;, ;, ;, ; 4, ; 4, ; 4, ; 4, ; 6, ; 6, ;, ;. There are no rational zeros. 79. (a) Approimate zeros:.6,.07, 0.90,.9 (b) f() L g() = ( +.6)( +.07)( )( -.9) (c) Graphically: Graph the original function and the approimate factorization on a variety of windows and observe their similarity. Numerically: Compute f(c) and g(c) for several values of c. Section. Comple Zeros and the Fundamental Theorem of Algebra Eploration. f(i) = (i) - i(i) + = = 0; f(-i) = (-i) - i(-i) + = = 0; no.. g(i) = i - i + ( + i) = - - i + + i = 0; g( - i) = ( - i) - ( - i) + ( + i) = -i + i = 0; no.. The Comple Conjugate Zeros Theorem does not necessarily hold true for a polynomial function with comple coefficients. Quick Review.. ( - i) + (- + i) = ( - ) + (- + )i = + i. ( + i)( - i) = ( - i) + i( - i) = - i + 6i - 4i = 7 + 4i. (-)(+) ; - 4()() 7. = = ; -9 = ; 9 i ;, ; 9. ;, ;, or ;, ;, ;, ; Section. Eercises. (-i)(+i)= -(i) = +9. The factored form shows the zeros to be = ; i. The absence of real zeros means that the graph has no -intercepts.. (-)(-)(+i)(-i) =( -+)( +4) = The factored form shows the zeros to be = (multiplicity ) and = ;i. The real zero = is the -intercept of the graph. In # 6, any constant multiple of the given polynomial is also an answer.. (-i)(+i)= + 7. (-)(-i)(+i)=(-)( +9) = (-)(-)(-i)(+i) =(-)(-)( +) = (-)(--i)(-+i) =(-)( -6+)= (-) (+) = (-) (--i)(-+i) =(-) ( -6+0) =( -4+4)( -6+0) = Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

17 Section. Comple Zeros and the Fundamental Theorem of Algebra In #7 0, note that the graph crosses the -ais at oddmultiplicity zeros, and kisses (touches but does not cross) the -ais where the multiplicity is even. 7. (b) 9. (d) In # 6, the number of comple zeros is the same as the degree of the polynomial; the number of real zeros can be determined from a graph. The latter always differs from the former by an even number (when the coefficients of the polynomial are real).. comple zeros; none real.. comple zeros; real.. 4 comple zeros; real. In #7, look for real zeros using a graph (and perhaps the Rational Zeros Test). Use synthetic division to factor the polynomial into one or more linear factors and a quadratic factor. Then use the quadratic formula to find comple zeros. 7. Inspection of the graph reveals that = is the only real zero. Dividing f() by - leaves ++ (below). The quadratic formula gives the remaining zeros of f() Zeros: =, f() = - ; 9 = ( - )c - a = ( - )( i)( i) 4 9. Zeros: =_(graphically) and = - ; i (applying the quadratic formula to ++6) f()=(-)(+) c - a - + ibd = ( - )( + )( + + i) ( + - i) 4 7. Zeros: = - and = (graphically) and = ; i (applying the quadratic formula to =6( -+)). 7/ ib d i c - a c - a - - ibd ib d / f()=(+7)(-)[-(-i)] [-(+i)] =(+7)(-)(-+i)(--i) In # 6, since the polynomials coefficients are real, for the given zero z=a+bi, the comple conjugate z =a-bi must also be a zero. Divide f() by -z and -z to reduce to a quadratic.. First divide f() by -(+i) (synthetically). Then divide the result, +( +i) -+(-i),by -(-i). This leaves the polynominal -. Zeros: = ;, = ; i +i 6 6 +i -i 6 +i -i 0 -i +i -i -i 0 +i 0 0 f()= ( - )( + )[ - ( - i)][ - ( + i)] = ( - )( + )( - + i)( - - i). First divide f() by -(-i).then divide the result, +( -i) -+6+4i, by -(+i). This leaves -. Zeros: = ;, = ; i - i i i i i 0 + i - - i i + i i 0-0 f()= ( - )( + ) [-(-i)] [-(+i)] = ( - )( + ) (-+i)(--i) For #7 4, find real zeros graphically, then use synthetic division to find the quadratic factors. Only the synthetic divison step is shown. 7. f()=(-)( ++) 9. f()=(-)( ++) f()=(-)(+4)( +) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

18 6 Chapter Polynomial, Power, and Rational Functions p 4. Solve for h: (h -h 4 )(6.)= ()(0), so that p h -h =60. Of the three solutions (found graphically), only h.776 ft makes sense in this setting. 4. Yes: (+)( +)= + ++ is one such polynomial. Other eamples can be obtained by multiplying any other quadratic with no real zeros by (+). 47. No: if all coefficients are real, -i and +i must also be zeros, giving zeros for a degree 4 polynomial. 49. f() must have the form a(-)(+)(-+i) (--i); since f(0)=a( )()( +i)( -i) = a=0,we know that a=. Multiplied out, this gives f()= (a) The model is D 0.080t +0.96t -.6t Synthetic division shows that f(i)=0 (the remainder), and at the same time gives f() (-i) = + -i=h(), so f()=(-i)( +-i). i - i -4i - i i -i 0 6. From graphing (or the Rational Zeros Test), we epect = to be a zero of f()= -8. Indeed, f()=8-8=0.so,= is a zero of f(). Using synthetic division we obtain: f()=(-)( ++4). We then apply the quadratic formula to find that the cube roots of -8 are, - + i, and - - i. Section.6 Graphs of Rational Functions (b) Sally walks toward the detector, turns and walks away (or walks backward), then walks toward the detector again. (c) The model changes direction at t.8 sec (D. m) and t.64 sec (when D.6 m).. False. Comple, nonreal solutions always come in conjugate pairs, so that if -i is a zero, then +i must also be a zero.. Both the sum and the product of two comple conjugates are real numbers, and the absolute value of a comple number is always real. The square of a comple number, on the other hand, need not be real. The answer is E. 7. Because the comple, non-real zeros of a real-coefficient polynomial always come in conjugate pairs, a polynomial of degree can have either 0,, or 4 non-real zeros. The answer is C. 9. (a) Power Real Part Imaginary Part (b) [, 9] by [0, ] ( + i) 7 = 8-8i ( + i) 8 = 6 ( + i) 9 = 6 + 6i ( + i) 0 = i (c) Reconcile as needed. 6. f(i)=i -i(i) +i(i)+= i+i-+=0. One can also take the last number of the bottom row from synthetic division. Eploration. g() = -. h() = - -. k() = [, 7] by [, ] [, 9] by [, ] [ 8, ] by [, ] Quick Review.6. f()=(-)(+) Q = or =. g() = ( + )( - ) Q = ;. h() = ( - )( + + ) Q = Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

19 Section.6 Graphs of Rational Functions Quotient:, Remainder: 7 Quotient:, Remainder: Section.6 Eercises. The domain of f()=/(+) is all real numbers Z -. The graph suggests that f() has a vertical asymptote at =. 7. Translate left units, reflect across -ais, vertically stretch by 7, translate up units. Asymptotes: =, y=. y Translate left 4 units, vertically stretch by, translate down units. Asymptotes: = 4, y=. y 8 [ 4.7, 4.7] by [ 4, 4] As approaches from the left, the values of f() decrease without bound. As approaches from the right, the values of f() increase without bound. That is, f() = q : -- and. The domain of f()= /( -4) is all real numbers Z -,. The graph suggests that f() has vertical asymptotes at = and =. [ 4.7, 4.7] by [, ] As approaches from the left, the values of f() decrease without bound, and as approaches from the right, the values of f() increase without bound. As approaches from the left, the values of f() increase without bound, and as approaches from the right, the values of f() decrease without bound. That is, f() = -q, - : - f() = -q. + :. Translate right units. Asymptotes: =, y=0. y f() = q. + : - f() = q, + : - f() = q, - : and : - : q : - + : -q f() = q f() = 0 f() = q f() = 9. The graph of f()=( -)/( +) suggests that there are no vertical asymptotes and that the horizontal asymptote is y=. [ 4.7, 4.7] by [ 4, 4] 6 The domain of f() is all real numbers, so there are indeed no vertical asymptotes. Using polynomial long division, we find that - f()=. + = When the value of is large, the denominator + is a large positive number, and 7/( +) is a small positive number, getting closer to zero as increases. Therefore, f() = f() =, so y= is indeed a : q : -q horizontal asymptote.. The graph of f()=(+)/( -) suggests that there are vertical asymptotes at =0 and =, with f() = q, f() = -q, f() = -q, and :0 :0 : Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

20 8 Chapter Polynomial, Power, and Rational Functions f() = q, and that the horizontal asymptote is : + 9. Intercept: a0, b. Asymptotes: =, y=-4. y=0. [ 4.7, 4.7] by [, ] The domain of f()=(+)/( -)= (+)/[(-)] is all real numbers 0,, so there are indeed vertical asymptotes at =0 and =. Rewriting one rational epression as two, we find that f() = + - = = When the value of is large, both terms get close to zero. Therefore, f() = f() = 0, : q : -q so y=0 is indeed a horizontal asymptote.. Intercepts: a0, and (, 0). Asymptotes: =, b =, and y=0.. (d); Xmin=, Xma=8, Xscl=, and Ymin=, Yma=, Yscl=.. (a); Xmin=, Xma=, Xscl=, and Ymin=, Yma=0, Yscl=.. (e); Xmin=, Xma=8, Xscl=, and Ymin=, Yma=, Yscl=. 7. For f()=/( --), the numerator is never zero, and so f() never equals zero and the graph has no - intercepts. Because f(0)= /, the y-intercept is /. The denominator factors as -- =(-)(+), so there are vertical asymptotes at = and =/. And because the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y=0. The graph supports this information and allows us to conclude that : - and [ 0, 0] by [ 0, 0] f() = q, f() = -q, - + : - f() = q. + :(/) f() = -q, - :(/) The graph also shows a local maimum of 6/ at =/4. [ 4, 6] by [, ]. No intercepts. Asymptotes: =, =0, =, and y=0. [ 4.7, 4.7] by [.,.] Intercept: a0, - b [ 4.7, 4.7] by [ 0, 0] 7. Intercepts: (0, ), (.8, 0), and (0.78, 0). Asymptotes: =, =, and y=. [, ] by [ 4, 6] Domain: ( q, ) ª a -, ª a b, q b Range: a - q, - 6 ª (0, q) b Continuity: All Z -, Increasing on ( q, ) and a -, 4 b Decreasing on c and a 4, b, q b Not symmetric. Unbounded. Local maimum at a 4, - 6 b Horizontal asymptote: y=0 Vertical asymptotes: = and =/ End behavior: f() = f() = 0 : -q : q Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

21 Section.6 Graphs of Rational Functions 9 9. For h()=(-)/( --), the numerator is zero when =, so the -intercept of the graph is. Because h(0)=/, the y-intercept is /. The denominator factors as --=(+)(-4), so there are vertical asymptotes at = and =4. And because the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y=0. The graph supports this information and allows us to conclude that : - h() = -q, h() = q, - + : - and h() = q. :4 + The graph shows no local etrema. [.87,.87] by [.,.] Intercepts: a0, b, (, 0) Domain: ( q, ) ª (, 4) ª (4, q) Range: ( q, q) Continuity: All Z -, 4 Decreasing on ( q, ), (, 4), and (4, q) Not symmetric. Unbounded. No local etrema. Horizontal asymptote: y=0 Vertical asymptotes: = and =4 End behavior: h() = h() = 0 : -q : q 4. For f()=( +-)/( -9), the numerator factors as +-=(+)(-), so the -intercepts of the graph are and. Because f(0)=/9, the y-intercept is /9. The denominator factors as -9=(+)(-), so there are vertical asymptotes at = and =. And because the degree of the numerator equals the degree of the denominator with a ratio of leading terms that equals, the horizontal asymptote is y=. The graph supports this information and allows us to conclude that f() = q, f() = -q, - + : - : - and f() = q. + : h() = -q, - :4 f() = -q, - : The graph also shows a local maimum of about 0.60 at about = Intercepts: (-, 0), (, 0), a0, 9 b Domain: ( q, ) ª (, ) ª (, q) Range: ( q, 0.60] ª (, q) Continuity: All Z -, Increasing on ( q, ) and (, 0.67] Decreasing on [ 0.67, ) and (, q) Not symmetric. Unbounded. Local maimum at about ( 0.67, 0.60) Horizontal asymptote: y= Vertical asymptotes: = and = End behavior: f() = f() =. : -q : q 4. For h()=( +-)/(+), the numerator factors as +-=(+)(-), so the -intercepts of the graph are and. Because h(0)= /, the y-intercept is /. The denominator is zero when =, so there is a vertical asymptote at =. Using long division, we rewrite h() as h() = + - = - + +, so the end-behavior asymptote of h() is y=. The graph supports this information and allows us to conclude that h() = q and h() = -q. - + : - : - The graph shows no local etrema. [ 9.4, 9.4] by [, ] Intercepts: (, 0), (, 0), a0, - b Domain: ( q, ) ª (, q) Range: ( q, q) Continuity: All Z - Increasing on ( q, ) and (, q) Not symmetric. Unbounded. No local etrema. Horizontal asymptote: None Vertical asymptote: = Slant asymptote: y= End behavior: h() = -q and h() = q. : -q : q 4. Divide -- by - to show that f() = = The end-behavior asymptote of f() is y=+. (a) [ 9.4, 9.4] by [, ] [ 0, 0] by [ 0, 0] Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

22 60 Chapter Polynomial, Power, and Rational Functions (b) 47. Divide - + by + to show that - + f()= = The end-behavior asymptote of f() is y= -+6. (a) (b) [ 40, 40] by [ 40, 40] [ 0, 0] by [ 0, 60] [ 0, 0] by [ 00, 00] 49. Divide 4 -+ by - to show that f()= = The end-behavior asymptote of f() is y= (a) Intercept: a0, 4 b Domain: ( q, q) Range: [0.77, 4.7] Continuity: ( q, q) Increasing on [ 0.4,.44] Decreasing on ( q, 0.4], [.44, q) Not symmetric. Bounded. Local maimum at (.44, 4.7); local minimum at ( 0.4, 0.77) Horizontal asymptote: y= No vertical asymptotes. End behavior: f()= f()=. For h()=( -)/(-), the numerator factors as -=(-)( ++), so the -intercept of the graph is. The y-intercept is h(0)=/. The denominator is zero when =, so the vertical asymptote is =. Because we can rewrite h() as - h()= = , we know that the end-behavior asymptote is y= ++4. The graph supports this information and allows us to conclude that h()= q, h()=q. : - [, ] by [, ] : -q : + : q The graph also shows a local maimum of about 0.86 at about =0.44, a local minimum of about 0.44 at about = 0.84, and another local minimum of about.970 at about =.94. [, ] by [ 00, 00] (b) [ 0, 0] by [ 0, 0] [ 0, 0] by [ 000, 000]. For f()=( -+4)/( -4+), the numerator is never zero, and so f() never equals zero and the graph has no -intercepts. Because f(0)=4/, the y-intercept is 4/. The denominator is never zero, and so there are no vertical asymptotes. And because the degree of the numerator equals the degree of the denominator with a ratio of leading terms that equals, the horizontal asymptote is y=.the graph supports this information. The graph also shows a local maimum of about 4.7 at about =.44 and a local minimum of about 0.77 at about = 0.4. Intercepts: (, 0), a0, b Domain: ( q, ) ª (, q) Range: ( q, q) Continuity: All real Z Increasing on [ 0.84, 0.44], [.94, q) Decreasing on ( q, 0.84], [0.44, ), (,.94] Not symmetric. Unbounded. Local maimum at (0.44, 0.86); local minimum at ( 0.84, 0.44) and (.94,.970) No horizontal asymptote. End-behavior asymptote: y= ++4 Vertical asymptote: = End behavior: h()= h()=q : -q : q Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

23 Section.6 Graphs of Rational Functions 6. f()=( - +-)/(-) has only one -intercept, and we can use the graph to show that it is about.7. The y-intercept is f(0)=. The denominator is zero when =/, so the vertical asymptote is =/. Because we can rewrite f() as f()= - = ( - ), we know that the end-behavior asymptote is y= - + The graph supports this information 4 8. and allows us to conclude that f()=q, f()= q. :/ - :/ + The graph also shows a local minimum of about 0.90 at about = For f()=( -)/(+), the numerator factors as -=(-)( ), and since the second factor is never zero (as can be verified by Descartes Rule of Signs or by graphing), the -intercept of the graph is. Because f(0)= /, the y-intercept is /. So the intercepts are (, 0) and (0, /). The denominator is zero when =, so = is a vertical asymptote. Divide - by + to show that - f()= = The end-behavior asymptote of f() is y= [ 0, 0] by [ 00, 400] [, ] by [ 0, 0] Intercepts: (.7, 0), (0, ) Domain: All Range: ( q, q) Continuity: All Z Increasing on [ 0.84, 0.), (0., q) Decreasing on ( q, 0.84] Not symmetric. Unbounded. Local minimum at ( 0.84, 0.90) No horizontal asymptote. End-behavior asymptote: y= Vertical asymptote: = End behavior: f()= f()=q : -q : q 7. For h()=( 4 +)/(+), the numerator is never zero, and so h() never equals zero and the graph has no - intercepts. Because h(0)=, the y-intercept is. So the one intercept is the point (0, ). The denominator is zero when =, so = is a vertical asymptote. Divide 4 + by + to show that 4 + h()= = The end-behavior asymptote of h() is y= [, ] by [ 0, 0] 6. h()=( -+)/( -) has only one -intercept, and we can use the graph to show that it is about.476. Because h(0)=, the y-intercept is. So the intercepts are (.476, 0) and (0, ). The denominator is zero when =, so = is a vertical asymptote. Divide -+ by - to show that h()= = The end-behavior asymptote of h() is y=, a horizontal line. [, ] by [, ] 6. False. If the denominator is never zero, there will be no vertical asymptote. For eample, f()=/( +) is a rational function and has no vertical asymptotes. 6. The ecluded values are those for which +=0, namely 0 and. The answer is E. 67. Since +=0 when =, there is a vertical asymptote. And because /(+)=-+/(+), the end behavior is characterized by the slant asymptote y=-. The answer is D. 69. (a) No: the domain of f is (- q, ) h (, q) ; the domain of g is all real numbers. (b) No: while it is not defined at, it does not tend toward ; q on either side. (c) Most grapher viewing windows do not reveal that f is undefined at. (d) Almost but not quite; they are equal for all. 7. (a) The volume is f() = k/, where is pressure and k is a constant. f() is a quotient of polynomials and hence is rational, but f() = k # -, so is a power function with constant of variation k and pressure -. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

24 6 Chapter Polynomial, Power, and Rational Functions (b) If f()=k a, where a is a negative integer, then the power function f is also a rational function. (c) V = k so k=(.9)(0.866)=.494. P,.494 If P=0., then V= 4. L Horizontal asymptotes: y= and y=. Intercepts: a0, - b, a, 0b h() = μ [, ] by [, ] 7. Horizontal asymptotes: y=. Intercepts: a0, 4 b, a, 0b f() = μ [ 0, 0] by [, ] Ú Ú The graph of f is the graph of y = shifted horizontally -d/c units, stretched vertically by a factor of bc - ad /c, reflected across the -ais if and only if bc-ad<0, and then shifted vertically by a/c. Section.7 Solving Equations in One Variable Quick Review.7. The denominator is +-=(-)(+4),so the new numerator is (+4)= +8.. The LCD is the LCM of, 8, and 6, namely = = - 6. The LCD is (+)(-) = ( - ) ( + )( - ) - ( + ) ( + )( - ) = ( + )( - ) = ( + )( - ) 7. For --=0: a=, b=, and c=. = -b ; b - 4ac a = -(-) ; (-) - 4()(-) () ; 9 - (-8) = = ; For +-=0: a=, b=, and c=. = -b ; b - 4ac a = - ; () - 4()(-) () = = Section.7 Eercises -. Algebraically: + + = (-)+(+)= += = = Numerically: For =, = = = - ; 4 - (-4) = 6 - ; 7 6. Algebraically: + = 4 +=4 ( Z 0) +-4=0 (-)(+7)=0 -=0 or +7=0 = or = 7 Numerically: For =, +=+=7 and 4 = 4 = 7. For = 7, += 7+= and 4 = 4-7 = -. = - ; 7 - ; 8 6 Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

25 Section.7 Solving Equations in One Variable 6 4. Algebraically: + - = - (-)+4= ( Z ) -+4= +-=0 (+4)(-)=0 +4=0 or -=0 = 4 or = but = is etraneous. Numerically: For = 4, 4 4(-4) 6 + = -4 + = = - 7 and - = -4 - = Algebraically: + 0 = 7 +0=7 ( Z 0) -7+0=0 (-)(-)=0 -=0 or -=0 = or = Graphically: The graph of f() = + 0 suggests - 7 that the -intercepts are and. [ 9.4, 9.4] by [, ] Then the solutions are = and =. 9. Algebraically: + =7 +=7 ( Z 0) -7+=0 (-)(-4)=0 -=0 or -4=0 = or =4 Graphically: The graph of f()=+ -7 suggests that the -intercepts are and 4. [ 9.4, 9.4] by [, ] Then the solutions are = and =4.. Algebraically: - + = + [and +=(+)] ( +)-= ( Z 0, ) +-=0 (-)(+)=0 -=0 or +=0 = or = but = is etraneous. Graphically: The graph of f()= suggests that the -intercept is There is a hole at. =. [ 4.7, 4.7] by [ 4, 4] Then the solution is =.. Algebraically: = [and +-0=(+)(-)] (-)+(+)=7 ( Z -, ) --=0 (+)(-)=0 +=0 or -=0 = - or = but = is etraneous. Graphically: The graph of f()= suggests that the -intercept is There is a hole at =.. [ 4.4, 4.4] by [, 9] Then the solution is = -.. Algebraically: = 0 [and +=(+)] (-)(+)-+=0 ( Z 0, ) -=0 (-)=0 =0 or -=0 =0 or = but =0 is etraneous. Graphically: The graph of - f()= - suggests that the intercept is. The -ais hides a hole at =0. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

26 64 Chapter Polynomial, Power, and Rational Functions [ 6.4,.4] by [ 0, ] Then the solution is =. 7. Algebraically: = - [and +=(+)] +6=(-)(+) ( Z -, 0) +6= ++6 +=0 (+)=0 =0 or +=0 =0 or = but both solutions are etraneous. No real solutions. Graphically: The graph of f() = suggests that there are no -intercepts. There is a hole at =, and the -ais hides a hole at =0. [ 4.7, 4.7] by [, ] Then there are no real solutions. 9. There is no -intercept at =. That is the etraneous solution.. Neither possible solution corresponds to an -intercept of the graph. Both are etraneous.. += - +(-)=(-) ( Z ) -+=- -6+7=0 = -(-6) ; (-6) - 4()(7) () = 6 ; 8 = ; = + L 4.44 or = - L = =0 ( Z ) (-) =0 -=0 = = [and +-4=(+4)(-)] 4(-)+(+4)= ( Z -4, ) 4 ++=0 The discriminant is b -4ac= -4(4)()= 79<0. There are no real solutions = 8 +=8 ( Z 0) Using a graphing calculator to find the -intercepts of f()= -8+ yields the solutions L -.00, L 0.66, and L.49.. (a) The total amount of solution is (+) ml; of this, the amount of acid is plus 60% of the original amount, or +0.6(). (b) y= (c) C()= Multiply both sides by + = , then rearrange to get 0.7 = 8.7, so that L 69. ml (a) C() = (b) A profit is realized if C() 6.7, or Then , so that hats per week. (c) They must have.7-(000+.)>000 or 0.6>4000: 60 hats per week.. (a) If is the length, then 8/ is the width. P() = + a 8 64 b = + (b) The graph of P()=+64/ has a minimum when.49, so that the rectangle is square. Then P(.49)=(.49)+64/ ft. 7. (a) Since V= r h, the height here is V/( r ). And since in general, S= r + rh= r +V/r, here S()= +000/ (0. L=00 cm ). (b) Solving +000/=900 graphically by finding the zeros of f()= +000/-900 yields two solutions: either. cm, in which case h 6.88 cm, or.7 cm, in which case h. cm. 9. (a) R = + R R R =. +.=R+.R. R() = +. (b).=r+.r.r =. - R For R=.7, 6. ohms. 4. (a) Drain A can drain /4.7 of the pool per hour, while drain B can drain /t of the pool per hour. Together, they can drain a fraction Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.