Chapter 2 Polynomial, Power, and Rational Functions

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1 Section. Linear and Quadratic Functions and Modeling 67 Chapter Polnomial, Power, and Rational Functions Section. Linear and Quadratic Functions and Modeling Eploration. $000 per ear. The equation will have the form v(t)=mt+b.the value of the building after 0 ear is v(0)=m(0)+b=b=0,000. The slope m is the rate of change, which is 000 (dollars per ear). So an equation for the value of the building (in dollars) as a function of the time (in ears) is v(t)= 000t+0,000.. v(0)=0,000 and v(6)= 000(6)+0,000=8,000 dollars. The equation v(t)=9,000 becomes 000t+0,000=9, t=,000 t=. ears Quick Review.. =8+.6. =.8-. -= (+),or= (, ) = (-),or= + 6 (, ) (, ) 6. (-) =(-)(-)= --+6 = (-6) =(-6)(-6)=(-8)(-6) = = (+7) = (+7)(+7) =( -)(+7)= ---7 = =( -+)=(-)(-) =(-) 0. ++=( ++)=(+)(+) =(+) Section. Eercises. Not a polnomial function because of the eponent. Polnomial of degree with leading coefficient. Polnomial of degree with leading coefficient. Polnomial of degree 0 with leading coefficient. Not a polnomial function because of cube root 6. Polnomial of degree with leading coefficient 8 7. m= so -= (-) f()= (, ) m= - 7 so -= - 7 (+) f()= (, ) 0 (6, ) (, ) 0 (, ). (+) =(+)(+)= +++9 = +6+9

2 68 Chapter Polnomial, Power, and Rational Functions 9. m= - so -6= - (+) f()= (, 6) (, ) 0 0. m= so -= (-) f()= + 0 (, 7) 6. (f) the verte is in Quadrant I, at (, ), meaning it must be either (b) or (f). Since f(0)=0, it cannot be (b): if the verte in (b) is (, ), then the intersection with the -ais occurs considerabl lower than (0, 0). It must be (f). 7. (e) the verte is at (, ) in Quadrant IV, so it must be (e). 8. (c) the verte is at (, ) in Quadrant II and the parabola opens down, so it must be (c). 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()=(-) (, ) 0. m= so -= (-0) f()= + (0, ) 0. Verticall shrink the graph of f()= b a factor of to obtain the graph of g()=, and translate this graph unit down to obtain the graph of h()= -. 0 (, 0) 0. m= so -= (-0) f()= + 0 (0, ) (, 0) 0. Translate the graph of f()= units left to obtain the graph of h()=(+), verticall shrink this graph b a factor of to obtain the graph of k()= (+), and translate this graph units down to obtain the graph of g()= (+) (a) the verte is at (, ), in Quadrant III, einating all but (a) and (d). Since f(0)=, it must be (a).. (d) the verte is at (, 7), in Quadrant III, einating all but (a) and (d). Since f(0)=, it must be (d).. (b) the verte is in Quadrant I, at (, ), meaning it must be either (b) or (f). Since f(0)=, it cannot be (f): if the verte in (f) is (, ), then the intersection with the -ais would be about (0, ). It must be (b). 0

3 Section. Linear and Quadratic Functions and Modeling 69. Verticall stretch the graph of f()= b a factor of to obtain the graph of g()=, reflect this graph across the -ais to obtain the graph of k()=, and translate this graph up units to obtain the graph of h()= h()= a b 9 = a + # b 8 7 = a b 8 Verte: a - 7 ais: = - 7, 7 8 b ;. f()=( -+)+6-=(-) +. Verte: (, ); ais: =; opens upward; does not intersect ais. For #, with an equation of the form f()=a(-h) +k, the verte is (h, k) and the ais is =h.. Verte: (, ); ais: =. Verte: (, ); ais: =. Verte: (, 7); ais: = 6. Verte: (, ); ais: = 7. f()= a + - b 7 = -- = a + a + # b 6 b Verte: a - ; ais: = 6, - 7 b 6 8. f()= a b 9 = a - # b 8 = a b 8 7 Verte: a 7 ; ais: =, 8 b 9. f()= ( -8)+ = ( - # +6)+ +6= (-) +9 Verte: (, 9); ais: = 0. f()= a - +6 b = +6- = a - a - # b b Verte: a ais: =, b ;. g()= a b = +- = a - 9 a - # b b Verte: a ais: =, b ; [, 6] b [0, 0]. g()=( -6+9)+-9=(-) +. Verte: (, ); ais: =; opens upward; does not intersect ais. [, 8] b [0, 0]. f()= ( +6)+0 = ( +6+6)+0+6= (+8) +7. Verte: ( 8, 7); ais: = 8; opens downward; intersects ais at about 6.60 and ; 7. [ 0, ] b [ 00, 00] 6. h()= ( -)+8= ( -+)+8+ = (-) +9 Verte: (, 9); ais: =; opens downward; intersects ais at and. [ 9, ] b [ 00, 0]

4 70 Chapter Polnomial, Power, and Rational Functions 7. f()=( +)+7 = +7- = a + 9 a b b Verte: a - ais: = - opens upward; does not, b ; ; intersect the -ais; verticall stretched b. 9. (a) [, ] b [0, 0] (b) Strong positive 0. (a) [.7, ] b [,.] 8. g()=( -)+ = a b 77 = a - - b Verte: a ais: = opens upward; intersects, - 77 b ; ; ais at about 0.8 and.6 or ; verticall stretched b. ; 0 8 b [, 0] b [ 0, 00] For #9, use the form =a(-h) +k, taking the verte (h, k) from the graph or other given information. 9. h= and k=, so =a(+) -. Now substitute =, = to obtain =a-, so a=: =(+) h= and k= 7, so =a(-) -7. Now substitute =0, = to obtain =a-7, so a=: =(-) -7.. h= and k=, so =a(-) +. Now substitute =, = 7 to obtain 7=9a+, so a= : = (-) +.. h= and k=, so =a(+) +. Now substitute =, = to obtain =9a+, so a= : = (+) +.. h= and k=, so =a(-) +. Now substitute =0, = to obtain =a+, so a=: =(-) +.. h= and k=, so =a(+) -. Now substitute =, = 7 to obtain 7=a-, so a= : = (+) -.. Strong positive 6. Strong negative 7. Weak positive 8. No correlation [0, 90] b [0, 70] (b) Strong negative 0. m= = 70 and b=0, so v(t)= 70t+0. At t=, v()=( 70)()+0=$90.. Let be the number of dolls produced each week and be the average weekl costs. Then m=.70, and b=0, so =.70+0, or 00=.70+0: =; dolls are produced each week.. (a) L The slope, m L 0., represents the average annual increase in hourl compensation for production workers, about $0. per ear. (b) Setting =0 in the regression equation leads to L $.7.. If the length is, then the width is 0-, so A()=(0-); maimum of 6 ft when = (the dimensions are ft* ft).. (a) [0, 00] b [0, 000] is one possibilit. (b) When 07. or 7.66 either 07, units or 7, 66 units. 6. The area of the picture and the frame, if the width of the picture is ft, is A()=(+)(+) ft. This equals 08 when =, so the painting is ft* ft. 7. If the strip is feet wide, the area of the strip is A()=(+)(0+)-000 ft. This equals 0 ft when =. ft. 8. (a) R()=(800+0)(00-). (b) [0, ] b [00,000, 60,000] is one possibilit (shown). [0, ] b [00,000, 60,000] (c) The maimum income $0,000 is achieved when =0, corresponding to rent of $0 per month.

5 Section. Linear and Quadratic Functions and Modeling 7 9. (a) R()=(6, )( ). (b) Man choices of Xma and Ymin are reasonable. Shown is [0, ] b [0,000, 7,000]. [0, ] b [0,000, 7,000] (c) The maimum revenue $6,00 is achieved when =8; that is, charging 90 cents per can. 60. Total sales would be S()=(0+)(0-) thousand dollars, when additional salespeople are hired. The maimum occurs when =0 (halfwa between the two zeros, at 0 and 0). 6. (a) g L ft>sec. s 0 = 8 ft and v 0 = 9 ft>sec. So the models are height= st = -6t + 9t + 8 and vertical velocit= vt = -t + 9. The maimum height occurs at the verte of st. h = - b, and a = - 9 =.87-6 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 st. Using the quadratic formula, we obtain t = -9 ; ;,776 = L or 6.. Time is not - negative, so the ball is in the air about 6. seconds. (c) To determine the ball s vertical velocit when it hits the ground, use vt = -t + 9, and solve for t = 6.. v6. = L -7 ft>sec when it hits the ground. 6. (a) h= 6t +8t+.. (b) The graph is shown in the window [0,.] b [0, ]. The maimum height is 9. ft,. sec after it is thrown. 6. The eact answer is, or about.6 ft/sec. In addition to the guess-and-check strateg suggested, this can be found algebraicall b noting that the verte of the parabola =a +b+c has coordinate b c- (note a= 6 and c=0), and setting a = b 6 this equal to The quadratic regression is Plot this curve together with the curve =0, and then find the intersection to find when the number of patent applications will reach 0,000. Note that we use =0 because the data were given as a number of thousands. The intersection occurs at L 6., so the number of applications will reach 0,000 approimatel 6 ears after 980 in 006. [, 8.] b [0, ] 6 ft 66. (a) m= = ft (b) r 67 ft, or about 0.79 mile. (c) 7.6 ft 67. (a) [, ] b [0, 0] (b) (c) On average, the children gained 0.68 pound per month. (d) [, ] b [0, 0] [0,.] b [0, ] 6. (a) h= 6t +80t-0. The graph is shown in the window [0, ] b [ 0, 00]. (e) 9. lbs 68. (a) The linear regression is L 8.0 +,07.6, where represents the number of ears since 90. (b) 00 is 70 ears after 90, so substitute 70 into the equation to predict the median U.S. famil income in 00. = ,07.6 L $9,00. [0, ] b [ 0, 00] (b) The maimum height is 90 ft,. sec after it is shot.

6 7 Chapter Polnomial, Power, and Rational Functions 69. The Identit Function f()= [.7,.7] b [.,.] Domain: -q, q Range: -q, q Continuit: The function is continuous on its domain. Increasing decreasing behavior: Increasing for all Smmetr: Smmetric about the origin Boundedness: Not bounded Local etrema: None Horizontal asmptotes: None Vertical asmptotes: None End behavior: f = -q and f =q S-q 70. The Squaring Function f()= Sq [.7,.7] b [, ] Domain: -q, q Range: 0, q Continuit: The function is continuous on its domain. Increasing decreasing behavior: Increasing on 0, q, decreasing on -q, 0. Smmetr: Smmetric about the -ais Boundedness: Bounded below, but not above Local etrema: Local minimum of 0 at =0 Horizontal asmptotes: None Vertical asmptotes: None End behavior: f = f =q S-q 7. False. For f()= Sq +-, the initial value is f(0)=. 7. True. B completing the square, we can rewrite f() so that f()= a b = a - +. Since f, f()>0 for all. b 7. m = - = -. The answer is E = - 7. f()=m+b = - ( )+b = +b 7 b=- = The answer is C. For #7 76, f()=(+) - corresponds to f()=a(-h) +k with a= and (h, k)=(, ). 7. The ais of smmetr runs verticall through the verte: =. The answer is B. 76. The verte is (h, k)=(, ). The answer is E. 77. (a) Graphs (i), (iii), and (v) are linear functions. The can all be represented b an equation =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 b =b, b Z 0. (c) (ii) is not a function because a single value (i.e., = ) results in a multiple number of -values. In fact, there are infinitel man -values that are valid for the equation =. f - f (a) = = - f - f - (b) = =7 - fc - fa c - a c - ac + a (c) = = =c+a c - a c - a c - a g - g - (d) = = - g - g - (e) = = - gc - ga c + - a + (f) = c - a c - a c - a = = c - a hc - ha 7c - - 7a - (g) = c - a c - a 7c - 7a = =7 c - a kc - ka mc + b - ma + b (h) = c - a c - a mc - ma = =m c - a lc - la c - a (i) = c - a = -b a = -b c - a a = -b a c - ac + ac + a = =c +ac+a c - a 79. Answers will var. One possibilit: When using the leastsquares method, mathematicians tr to minimize the residual i -(a i +b), i.e., place the predicted -values as close as possible to the actual -values. If mathematicians reversed the ordered pairs, the objective would change to minimizing the residual i -(c 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 -values for each pair would have to be almost eactl the same distance from the regression line which is statisticall impossible in practice.

7 Section. Power Functions with Modeling (a) [0, 7] b [, 6] (b) [0, 7] b [, 6] (c) [0, ] b [0, ] (d) The median median line appears to be the better fit, because it approimates more of the data values more closel. 8. (a) If a +b +c=0, then -b ; b - ac = b the quadratic a -b + b - ac formula. Thus, = and a -b - b - ac = and a -b + b - ac - b - b - ac + = a -b = a = -b a = -b a. (b) Similarl, # = a -b + b - ac ba -b - b - ac b a a b - b - ac ac c = = = a a a. 8. f()=(-a)(-b)= -b-a+ab = +( a-b)+ab. If we use the verte form of a quadratic function, we have h= - a -a - b b a + b a + b =. The ais is =h=. 8. Multipl out f() to get -(a+b)+ab. Complete a + b the square to get a - a + b b +ab-. The a + b verte is then (h, k) where h= and a + b a - b k=ab- = and are given b the quadratic formula -b ; b - ac ; then + = - b, and the line of a a + smmetr is = - b, which is eactl equal to. a 8. The Constant Rate of Change Theorem states that a function defined on all real numbers is a linear function if and onl if it has a constant nonzero average rate of change between an 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. [.,.] b [.,.] [, ] b [, ]

8 7 Chapter Polnomial, Power, and Rational Functions 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. The are different in that if 0 0 6, f, g, and h S 0 at dramaticall different rates, and if 0 0 7, f, g, and h Sqat dramaticall different rates.. 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. The are diffferent in that if 0 0 6, f, g, and h S 0 at dramaticall different rates, and if 0 0 7, f, g, and h Sq at dramaticall different rates. Quick Review... p... d 7 [.,.] b [ 0.,.] q [ 0, 0] b [ 00, 00] [, ] b [, ] [, ] b [ 0, 00] 6. m 7. > 8. > 9. L.7 -> 0. L 0.7 -> Section. Eercises. power=, constant= -. power=, constant=9. not a power function. power=0, constant=. power=, constant=c k 6. power=, constant= g 7. power=, constant= p 8. power=, constant= 9. power=, constant=k 0. power=, constant=m. degree=0, coefficient=. not a monomial function; negative eponent. degree=7, coefficient= 6. not a monomial function; variable in eponent. degree=, coefficient= 6. degree=, coefficient=l 7. A=ks 8. V=kr 9. I=V/R 0. V=kT. E=mc. p = gd. The weight w of an object varies directl with its mass m, with the constant of variation g.. The circumference C of a circle is proportional to its diameter D, with the constant of variation.. The refractive inde n of a medium is inversel proportional to v, the velocit of light in the medium, with constant of variation c, the constant velocit of light in free space. 6. The distance d traveled b a free-falling object dropped from rest varies directl with the square of its speed p, with the constant of variation. g

9 Section. Power Functions with Modeling 7 7. power=, constant= Domain: -q, q Range: 0, q Continuous Decreasing on -q, 0. Increasing on 0, q. Even. Smmetric with respect to -ais. Bounded below, but not above Local minimum at =0. Asmptotes: None End Behavior: =q, =q S-q Sq Discontinuous at =0 Increasing on -q, 0. Increasing on 0, q. Odd. Smmetric with respect to origin Not bounded above or below No local etrema Asmptotes at =0 and =0 End Behavior: - - = 0, - - = 0. S-q Sq [, ] b [, 9] 8. power=, constant= Domain: -q, q Range: -q, q Continuous Decreasing for all Odd. Smmetric with respect to origin Not bounded above or below No local etrema Asmptotes: None End Behavior: - =q, - = -q S-q Sq [, ] b [, ]. Start with = and shrink verticall b. Since f( )=, f is even. - = [, ] b [, 9]. Start with = and stretch verticall b. Since f - = - = - = -f, f is odd. [, ] b [ 0, 0] 9. power=, constant= Domain: 0, q Range: 0, q Continuous Increasing on 0, q Bounded below Neither even nor odd Local minimum at (0, 0) Asmptotes: None End Behavior: Sq =q [, ] b [ 0, 0]. Start with =, then stretch verticall b. and reflect over the -ais. Since f - = -. - =. = f(), f is odd. [, ] b [ 0, 0]. Start with = 6, then stretch verticall b and reflect over the -ais. Since f - = = - 6 =f(), f is even. [, 99] b [, ] 0. power=, constant= Domain: -q, 0 0, q Range: -q, 0 0, q [, ] b [ 9, ]

10 76 Chapter Polnomial, Power, and Rational Functions. Start with = 8, then shrink verticall b. Since f( )= f is even. -8 = 8 = f,. k=, a=. In the fourth quadrant, f is decreasing and concave down. f - = - - = ( )=- > =f(), so f is even. [, ] b [, 9] 6. Start with = 7, then shrink verticall b. Since 8 f( )=, f is odd. 8-7 = = -f [ 0, 0] b [ 9, ] 6. k=, a=. In the first quadrant, f is increasing and concave up. f is undefined for 6 0. [, ] b [ 0, 0] 7. (g) 8. (a) 9. (d) 0. (g). (h). (d). k=, a=. In the first quadrant, the function is increasing and concave down. f is undefined for 6 0. [, 99] b [, 0]. k=, a=. In the fourth quadrant, the function is decreasing and concave up. f - = - - = - = - > =f(), so f is even. [, 8] b [, 9] 7. k=, a=. In the first quadrant, f is decreasing and concave up. f( )= - - = - = - - = f(), so f is odd. [, ] b [ 0, 0] 8. k=, a=. In the fourth quadrant, f is increasing and concave down. f( )= = - - = -, so f is even. = -- = f [, ] b [ 9, ] 9. = 8, power=, constant=8 [ 0, 0] b [ 9, ] 0. = -, power=, constant=

11 Section. Power Functions with Modeling 77 kt 0.96 atm.6 L. V=, so k = PV = P T 0 K atm L =0.006 K atm L a b0 K K At P=. atm, V=. atm =. L.6 L. V=kPT, so k = V = PT 0.96 atm0 K L =0.0 atm K L At T=8 K, V= a 0.0 (0.96 atm) atm K b (8 K)=.87 L a.00 * 08 m b c sec. n = c, so v= = =. * 0 8 m v n. sec. P = kv, so k = P v = w =. * 0-0 mph Wind Speed (mph) Power (W) Since P = kv is a cubic, power will increase significantl with onl a small increase in wind speed.. (a) 7. (a) [0.8,.] b [ 0., 9.] (b) L 7.9 # -.987; es (c) [0.8,.] b [ 0., 9.] W W (d) Approimatel.76 and 0.697, respectivel. m 8. True, because f( )=( ) -/ =[( ) ] -/ =( ) -/ = -/ =f(). 9. False. f( )=( ) / = ( / )= f() and so the function is odd. It is smmetric about the origin, not the -ais. 60. f = -> = = > = =. The answer is A. 6. f0 = -0 -> = - # is undefined. 0 = - # > 0 Also, f( )= ( ) / = ( )=, f()= () / = ()=, and f()= () /.08. The answer is E. m 6. f( )=( ) / =[( ) ] / =( ) / = / =f() The function is even. The answer is B. [, 7] b [0, 0] (b) r L.0 # w (c) [, 7] b [0, 0] (d) Approimatel 7.67 beats/min, which is ver close to Clark s observed value. 6. Given that n is an integer, n : If n is odd, then f - = - n = - n = -f and so f() is odd. If n is even, then f( )=( ) n = n =f() and so f() is even. 6. f()= / =( / ) = is defined for 0. The answer is B. 6. Answers will var. In general, however, students will find n n even: f = k # m>n = k # m, so f is undefined for <0. n m even, n odd: f = k # m>n = k # m ; f - m odd, n odd: = k # n - m = k # n m = f, so f is even. f = k # m>n = k # n m ; f - = k # n - m = -k # n m = -k # m>n = -f, so f is odd.

12 78 Chapter Polnomial, Power, and Rational Functions 6. (a) [0, ] b [0, ] [0, ] b [0, ] [, ] b [, ] The graphs of f()= and h()= are similar and appear in the st and rd quadrants onl. The graphs of g()= and k()= are similar and appear in the st and nd quadrants onl. The pair (, ) is common to all four functions. f g h k Domain Range Continuous es es es es Increasing (, 0) (, 0) Decreasing (, 0), (0, ) (0, ) (, 0), (0, ) (0, ) Smmetr w.r.t. origin w.r.t. -ais w.r.t. origin w.r.t. -ais Bounded not below not below Etrema none none none none Asmptotes -ais, -ais -ais, -ais -ais, -ais -ais, -ais End Behavior f() 0 g() 0 h() 0 k() 0 (b) [0, ] b [0, ] [0, ] b [0, ] [, ] b [, ] The graphs of f()= / and h()= / are similar and appear in the st quadrant onl. The graphs of g()= / and k()= / are similar and appear in the st and rd quadrants onl. The pairs (0, 0), (, ) are common to all four functions. f g h k Domain [0, ) (, ) [0, ) (, ) Range 0 (, ) 0 (, ) Continuous es es es es Increasing [0, ) (, ) [0, ) (, ) Decreasing Smmetr 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 Asmptotes none none none none End behavior f() g() g() h() k() k()

13 Section. Polnomial Functions of Higher Degree with Modeling π π π π a π b The graphs look like those shown in Figure. on page 9. f = p looks like the red graph in Figure.(a) because k=>0 and a = p 7. f = >p looks like the blue graph in Figure.(a) because k=>0 and 0<a=/ p<. f = -p looks like the green graph in Figure.(a) because k=<0 and a = -p 6 0. f = - p looks like the red graph in Figure.(b) because k= <0 and a = p 7. f = - >p looks like the blue graph in Figure.(b) because k= <0 and a = -p 6 0. f = - -p looks like the green graph in Figure.(b) because k= <0 and a = -p 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 (rs) Mercur Venus Earth Mars..88 Jupiter.0.86 Saturn a π b Source: Shupe, Dorr, Pane, Hunsiker, et al., National Geographic Atlas of the World (rev. 6th ed.). Washington, DC: National Geographic Societ, 99, plate 6. Using these new data, we find a power function model of: L #.0 L.. Since represents ears, we set = T and since represents distance, we set = a then, =. T = a > T = a > T = a. 68. Using the free-fall equations in Section., we know that st = - and that vt = -gt + v. If gt + v 0 t + s 0 0 t=0 is the time at which the object is dropped, then v. So d = s 0 - s = s 0 - a - gt + s 0 b = 0 = 0 gt and p = 0 v 0 = 0 -gt 0. Solving d = for t, we have gt d d. Then p = ` -g. A g ` = dg t = = dg A g B g The results of Eample 6 approimate this formula. 69. If f is even, f = f -, so =, f Z 0. 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 70. Let g = -a and f = a. Then g = > a = >f. Eercise 69 shows that g = >f is even if and onl if f is even, and g = >f is odd if and onl if f is odd. Therefore, g = -a is even if and onl if f = a is even, and that g = -a is odd if and onl if f = a is odd. 7. (a) The force F acting on an object varies jointl as the mass m of the object and the acceleration a of the object. (b) The kinetic energ KE of an object varies jointl as the mass m of the object and the square of the velocit v of the object. (c) The force of gravit F acting on two objects varies jointl as their masses m and m and inversel as the square of the distance r between their centers, with the constant of variation G, the universal gravitational constant. Section. Polnomial Functions of Higher Degree with Modeling Eploration. (a) =q, Sq S-q [, ] b [, ] = -q

14 80 Chapter Polnomial, Power, and Rational Functions (b) - = -q, Sq S-q - =q (d) -0. = -q, Sq S-q -0. = -q (c) =q, Sq [, ] b [, ] S-q = -q. (a) -0. = -q, Sq [, ] b [, ] S-q -0. =q (d) = -q, Sq [, ] b [, ] S-q =q (b) - = -q, Sq [, ] b [, ] S-q - = -q. (a) - = -q, Sq [, ] b [, ] S-q - = -q (c) =q, Sq [, ] b [, ] S-q =q (b) 0.6 =q, Sq [, ] b [, ] S-q 0.6 =q (d). =q, Sq [, ] b [, ] S-q. = -q (c) 6 =q, Sq [, ] b [, ] S-q 6 =q If a n 7 0 and n>0, f =qand Sq f = -q. If a n 6 0 and n>0, S-q f = -q and f =q. Sq [, ] b [, ] S-q [, ] b [, ]

15 Section. Polnomial Functions of Higher Degree with Modeling 8 Eploration. = It is an eact fit, which we epect with onl data points! 00 [, 0] b [, ]. = It is an eact fit, eactl what we epect with onl data points!. Start with =, shift to the left b unit, verticall shrink b, reflect over the -ais, and then verticall shift up units. -intercept: a 0, b [., 8.] b [ 8, ] Quick Review.. (-)(+). (-7)(-). (-)(-). (-)(-). (-)(-) 6. (-)(-) 7. =0, = 8. =0, =, = 9. = 6, =, =. 0. = 6, =, = Section. Eercises. Start with =, shift to the right b units, and then stretch verticall b. -intercept: (0, ). Start with =, shift to the right b units, verticall shrink b, and verticall shift up unit. -intercept: (0, 7) Start with =, shift to the left units, verticall stretch b, reflect over the -ais, and verticall shift down units. -intercept: (0, ) 0. Start with =, shift to the left b units, and then reflect over the -ais. -intercept: (0, )

16 8 Chapter Polnomial, Power, and Rational Functions 6. Start with =, shift to the right unit, verticall stretch b, and verticall shift down units. -intercept: (0, ). One possibilit: [ 0, 0] b [ 000, 000]. One possibilit: 7. local maimum: (0.79,.9), zeros: =0 and.6. The general shape of f is like =, but near the origin, f behaves a lot like its other term,. f is neither even nor odd. [ 0, 0] b [ 000, 000] 6. One possibilit: [, ] b [, ] 8. local maimum at (0, 0) and local minima at (.,.) and (.,.), zeros: =0,.8,.8. f behaves a lot like = ecept in the interval [.8,.8], where it behaves more like its second building block term,. [ 00, 00] b [ 000, 000] For #7, when one end of a polnomial 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. [, ] b [, ] 9. Cubic function, positive leading coefficient. The answer is (c). 0. Cubic function, negative leading coefficient. The answer is (b).. Higher than cubic, positive leading coefficient. The answer is (a).. Higher than cubic, negative leading coefficient. The answer is (d).. One possibilit: 8. Sq S-q Sq S-q [, ] b [ 8, ] f =q f = -q [, ] b [, ] f = - q f = q [ 00, 00] b [ 000, 000]

17 Section. Polnomial Functions of Higher Degree with Modeling Sq S-q Sq S-q Sq S-q Sq S-q [ 8, 0] b [ 0, 00] f = - q f = q [ 0, 0] b [ 00, 0] f =q f = - q [, ] b [, 6] f =q f =q [, 6] b [ 00, ] f =q f =q Sq S-q [, ] b [ 0, 90] f = - q f = - q For # 8, the end behavior of a polnomial is governed b the highest-degree term.. f =q, f = q Sq S-q 6. f = - q, f = q Sq S-q 7. f = - q, f = q Sq S-q 8. f = - q, f = - q Sq S-q 9. (a); There are zeros: the are.,, and.. 0. (b); There are zeros: the are 0., approimatel 0.9 (actuall /7), and.. (c); There are zeros: approimatel 0.7 (actuall /), 0., and.. (d); There are zeros:, 0., and. For #, factor or appl the quadratic formula.. and. and /. / and / For #6 8, factor out, then factor or appl the quadratic formula. 6. 0,, and 7. 0, /, and 8. 0,, and 9. Degree ; zeros: =0 (multiplicit, graph crosses -ais), = (multiplicit, graph is tangent) 0 6 Sq S-q [, ] b [ 0, 0] f =q f =q

18 8 Chapter Polnomial, Power, and Rational Functions 0. Degree ; zeros: =0 (multiplicit, graph crosses -ais), = (multiplicit, graph crosses -ais). Zeros:.7,.6,.9 [, ] b [ 0, 0] 6. Zeros:.,. Degree ; zeros: = (multiplicit, graph crosses -ais), = (multiplicit, graph is tangent) 0 [, ] b [ 0, 90] 7. Zeros:.90, 0.,,. 0. Degree 6; zeros: = (multiplicit, graph is tangent), = (multiplicit, graph is tangent) [ 6, ] b [, ] 8. Zeros:.98, 0.6,.,.77,.6 00,000. Zeros:., 0.7,.67 0 [, ] b [ 00, 00] 9. 0, 6, and 6. Algebraicall factor out first. 0.,, and 0. Graphicall. Cubic equations can be solved algebraicall, but methods of doing so are more complicated than the quadratic formula. [, ] b [ 0, 0]. Zeros:.7, 0.6,.7 [, ] b [ 800, 800].,, and. Graphicall. [, ] b [ 8, ] [ 0, ] b [ 00, 0]

19 Section. Polnomial Functions of Higher Degree with Modeling 8. 6,, and 8. Graphicall. [ 0, ] b [ 00, 00] For # 6, the minimal polnomials are given; an constant (or an other polnomial) can be multiplied b the answer given to give another answer.. f()=(-)(+)(-6) = f()=(+)(-)(+)= f()= - + (-) =( -)(-)= f()=(-) = (-)[(-) -]= = must cross the -ais at least once. That is to sa, f() takes on both positive and negative values, and thus b the Intermediate Value Theorem, f()=0 for some. 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 sa, f() takes on both positive and negative values, and thus b the Intermediate Value Theorem, f()=0 for some. 6. (a) [0, 60] b [ 0, 0] (b) = (c) [, ] b [ 0, 0] 8. = [, 8] b [, 0] 9. = [0, 60] b [ 0, 0] (d) () 6.9 ft (e) Using the quadratic equation to solve 0= (0.6-00), we find two answers: =67.7 mph and = mph. Clearl the negative value is etraneous. 6. (a) P()=R()-C() is positive if 9.7<<.7 (appro.), so the need between 0 and customers. (b) P()=60000 when =00.9 or =9.7. Either 0 or 9 customers gives a profit slightl over $60,000; 00 or 0 customers both ield slightl less than $60, (a) [0, 0] b [, ] 60. = [0, 0.8] b [0,.0] (b) 0.9 cm from the center of the arter 66. (a) The height of the bo will be, the width will be -, and the length 60-. (b) An value of between approimatel 0.0 and inches. [, ] b [, ] 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 [0, 8] b [0, 00]

20 86 Chapter Polnomial, Power, and Rational Functions 67. The volume is V()=(0-)(-); use an with 0< 0.99 or.6 <. [0, ] b [0, 00] 68. Determine where the function is positive: 0<<.. (The side lengths of the rectangle are and 6 units.) [0.9999,.000] b [ *0 7, *0 7 ] 78. This also has a maimum near = but this time a window such as [0.6,.] b [ 0., 0.] reveals that the graph actuall rises above the -ais and has a maimum at (0.999, 0.0). [0, ] b [0,,000] 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 =. 70. False. If a>0, the graph of f()=(+a) is obtained b translating the graph of f()= to the left b a units. Translation to the right corresponds to a<0. 7. When =0, f()=(-) +=( ) +=. The answer is C. 7. In f()=(-) (+) (+) 7, the factor - occurs twice. So = is a zero of multiplicit, and the answer is B. 7. The graph indicates three zeros, each of multiplicit : =, =0, and =. The end behavior indicates a negative leading coefficient. So f()= (+)(-), and the answer is B. 7. The graph indicates four zeros: = (multiplicit ), =0 (multiplicit ) and = (multiplicit ). The end behavior indicates a positive leading coefficient. So f()=(+) (-), and the answer is A. 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 -ais intersections) between and. 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 polnomial rather than degree. 76. The end behavior is visible in the first window, but not the details of the behavior near =. The second view shows those details, but at the loss of the end behavior. 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 actuall crossing it that is, (, 0) is a local maimum. One possible window is [0.9999,.000] b [ 0 7, 0 7 ]. [0.6,.] b [ 0., 0.] 79. A maimum and minimum are not visible in the standard window, but can be seen on the window [0., 0.] b [.9,.]. [0., 0.] b [.9,.0] 80. A maimum and minimum are not visible in the standard window, but can be seen on the window [0.9,.0] b [ 6.000,.999]. [0.9,.0] b [ 6.000,.999] 8. The graph of =( -) (shown on the window [, ] b [, ]) increases, then decreases, then increases; the graph of = onl increases. Therefore, this graph cannot be obtained from the graph of = b the transformations studied in Chapter (translations, reflections, and stretching/shrinking). Since the right side includes onl these transformations, there can be no solution. [, ] b [, ]

21 Section. Polnomial Functions of Higher Degree with Modeling The graph of = has a flat bottom, while the graph of = + -- (shown on [, ] b [ 8, ]) is bump. Therefore, this graph cannot be obtained from the graph of = through the transformations of Chapter. Since the right side includes onl these transformations, there can be no solution. (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 or = D - u. Then. = - 8 D A E [, ] b [ 8, ] 8. (a) Substituting =, =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,.] b [6, 8]. Calculator output will not show the detail seen here. [.8,.] b [6, 8] (c) The line L also crosses the graph of f() at (, ). [, ] b [, ] 8. (a) Note that f(a)=a n and f( a)= a n - ; m= -a n - a n -a n - = = =a n-. -a - a -a (b) First observe that f( 0 )=(a /(n-) ) n =a n/(n-). Using point slope form: -a n/(n-) =a n- (-a /(n-) ). (c) With n= and a=, this equation becomes - / = (- / ), or =9 - + =9-6. So =. F D u H 8 D u B G C D 8 (b) Equation (a) sas Multipl both sides b = - 8. : 8 = - 8. Subtract from both sides and factor: 8 - = -8. Divide both sides b 8 - : = -8 Factor out - from numerator and denominator: =. - 8 (c) Appling the Pthagorean Theorem to EBC and ABC, we have +D =0 and +D =0, which combine to give D =00- =900-, or - =00. Substituting =8/(-8), 8 we get a - =00, so that - 8 b 6 - =00, or 6 - (-8) - 8 =00(-8). Epanding this gives ,000 = This is equivalent to ,000 = 0. (d) The two solutions are.96 and.78. Based on the figure, must be between 8 and 0 for this problem, so.78. Then D= 0-6. ft. 86. (a) Regardless of the value of b, f -b = - b, f =+q, f = -q, and the graph of Sq S-q f has a -intercept of. If 0 b 0, the graph of f is strictl increasing. If 0 b 0 7, f has one local maimum and one local minimum. If 0 b 0 is large, the graph of f appears to have a double root at 0 and a single root at -b, because f = + b = + b for large. (b) Answers will var. (c) [, ] b [ 0, 0] 8. (a) Label the points of the diagram as shown, adding the horizontal segment FH. Therefore, ECB is similar [ 8.8, 8.8] b [ 000, 000]

22 88 Chapter Polnomial, Power, and Rational Functions Section. Real Zeros of Polnomial Functions Quick Review = - = = 6 - = = = = = + - = = = = = + - = Section. Eercises. - -R f()=(-) +; - + +R f()=( -+)(+)-; f + = R f = ; f + = f - = R f = a ; b f + = > R f = ; f + - = R f f = ; + = = =

23 Section. Real Zeros of Polnomial Functions = = = = The remainder is f()=.. The remainder is f()=.. The remainder is f( )=. 6. The remainder is f( )=. 7. The remainder is f()=. 8. The remainder is f( )=. 9. Yes: is a zero of the second polnomial. 0. Yes: is a zero of the second polnomial.. No: when =, the second polnomial evaluates to 0.. Yes: is a zero of the second polnomial.. Yes: - is a zero of the second polnomial.. No: when = -, the second polnomial evaluates to.. From the graph it appears that (+) and (-) are factors f()=(+)(-)(-7) 6. From the graph it appears that (+) and (-) are factors f()=(+)(-)(-7) 7. (+)(-)(-)= (+)(-)(+)= (-) a - a - b b = = (+)(+)() a - b =(+)(+)(-) = Since f( )=f()=f()=0, it must be that (+), (-), and (-) are factors of f.so f()=k(+)(-)(-) for some constant k. Since f(0)=80, we must have k=. So f()=(+)(-)(-). Since f( )=f()=f()=0, it must be that (+), (-), and (-) are factors of f.so f()=k(+)(-)(-) for some constant k. Since f( )=, we must have k=, so f()=(+)(-)(-) ;. Possible rational zeros:, or,,, ;, ;, ;, ;6 ; is a zero. 6 ;, ;, ;7, ;. Possible rational zeros:, or,, 7, ;, ; 7, ;, ;, ; 7, ; ; is a zero. ;, ;, ;9. Possible rational zeros:, or,, 9,, ;, ; ;, ; 9 ; is a zero. ;, ;, ;, ;, ;6, ; 6. Possible rational zeros:, or, ;, ;, ;, ;6,,, 6,, ;, ;, ;, ;, ;, ; ; - and 6 are zeros Since all numbers in the last line are 0, is an upper bound for the zeros of f.

24 90 Chapter Polnomial, Power, and Rational Functions Since all values in the last line are 0, is an upper bound for the zeros of f() Since all values in the last line are 0, is an upper bound for the zeros of f() Since all values in the last line are 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().. 0 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() Since the values in the last line alternate signs, is a lower bound for the zeros of f().. B the Upper and Lower Bound Tests, is a lower bound and is an upper bound. No zeros outside window B the Upper and Lower Bound Tests, is a lower bound and is an upper bound. No zeros outside window Snthetic 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() , , Snthetic division shows that the lower/upper bounds tests were not met. There are zeros not shown (appro and 9.08), because and are not bounds for zeros of f() , , For #9 6, determine the rational zeros using a grapher (and the Rational Zeros Test as necessar). Use snthetic division to reduce the function to a quadratic polnomial, which can be solved with the quadratic formula (or otherwise). The first two are done in detail; for the rest, we show onl the snthetic division step(s). 9. Possible rational zeros: ;, ;, ;, ;6, or ;, ; ;, ;, ;, ;6, ; The onl rational zero is., ;. Snthetic division (below) leaves -, so the irrational zeros are ;. / Possible rational zeros: ;, ;, ;9. The onl rational zero is -. Snthetic division (below) leaves -, so the irrational zeros are ; Rational: -; irrational: ; Rational: ; irrational: ;

25 Section. Real Zeros of Polnomial Functions 9. Rational: and ; irrational: ; Rational: and ; irrational: ; Rational: - and ; irrational: none / Rational: ; irrational: about 0.68 / The suppl and demand graphs are shown on the window [0, 0] b [0, 00]. The intersect when p=$6.7, at which point the suppl and demand equal.7. [0, 0] b [0, 00] 9. Using the Remainder Theorem, the remainder is = Using the Remainder Theorem, the remainder is 6-7 = (a) Lower bound: Upper bound: The Upper and Lower Bound Tests are met, so all real zeros of f lie on the interval [, ]. (b) Potential rational zeros: Factors of 8 Factors of A graph shows that is most promising, so we verif with snthetic division: Use the Remainder Theorem: f - = 0 Z 0 f( 8)=,960,00 f - = 9 Z 0 f = 7 Z 0 f(8)=,78,0 f( 9)=,97 f(9)=9,89 Since all possible rational roots besides ield nonzero function values, there are no other rational roots. [, ] b [, 9] : ;, ;, ;9, ;8 ; (c) f = (d) From our graph, we find that one irrational zero of is L.0. (e) f L (a) D L t t +.79t The suppl and demand graphs, shown on the window [0, 0] b [0, 600], intersect when p=$ There S(p)=D(p)=00.. [, 8.] b [0, ] [0, 0] b [0, 600] (b) When t=0, D L 0.80 m. (c) The graph changes direction at t L.0 and at t L.8. Lewis is approimatel.7 m from the motion detector at t =.0 and.7 m from the motion detector at t =.8.

26 9 Chapter Polnomial, Power, and Rational Functions 6. False. -a is a factor if and onl if f(a)=0. So (+) is a factor if and onl if f( )=0. 6. True. B the Remainder Theorem, the remainder when f() is divided b - is f(), which equals. 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 b - is zero. So the answer is A. 66. B the Rational Zeros Theorem, ever rational root of f() must be among the numbers,, ;, ;. The answer is E. 67. f()=(+)( +-)- ields a remainder of when divided b either + or +-, from which it follows that + is not a factor of f() and that f() is not evenl divisible b +. The answer is B. 68. Answers A through D can be verified to be true. And because f() is a polnomial function of odd degree, its graph must cross the -ais somewhere. The answer is E. 69. (a) The volume of a sphere is V =. In this case, the pr radius of the buo is, so the buo s volume is. p weight (b) Total weight = volume # unit volume = volume # densit. In this case, the densit of the buo is so, the weight W b of the buo is d, W b = p #. d = dp (c) The weight of the displaced water is W H volume # O = densit. We know from geometr that the volume of a spherical cap is V = p so, 6 r + h h, WH O = p 6 r + # d = pd 6 r + (d) Setting the two weights equal, we have: W b = W H O pd = pd 6 r + = r + 0 = = = - + Solving graphicall, we find that L 0.67 m, the depth that the buo will sink. 70. The weight of the buo, W b, with densit is d, W b = p # So, d = pd. W b = W H O pd = pd 6 r + = r + 0 = = = - + Solving graphicall, we find that L 0.7 m, the depth the buo would sink. 7. (a) Shown is one possible view, on the window [0, 600] b [0, 00]. [0, 600] b [0, 00] (b) The maimum population, after 00 das, is 60 turkes. (c) P=0 when t L. about das after release. (d) Answers will var. One possibilit: After the population increases to a certain point, the begin to compete for food and eventuall die of starvation. 7. (a) d is the independent variable. (b) A good choice is [0, 7] b [0, ]. (c) s =. when d L ft. (found graphicall) 7. (a) sign changes in f(), sign change in f( )= + ++; 0 or positive zeros, negative zero. (b) No positive zeros, or negative zeros. (c) positive zero, no negative zeros. (d) positive zero, negative zero. 7. [, ] b [, 9] The functions are not eactl the same, when,we have f = - = + + = g The domain of f is -q,, q while the domain of g is -q, q. f is discontinuous at =. g is continuous.

27 Section. Comple Zeros and the Fundamental Theorem of Algebra Divide numerator and = denominator b. - Write coefficients of zero of divisor - dividend. line for products line for sums - Quotient, remainder 8 Cop into the first quotient position. Multipl # = and add this to -. Multipl - # and add this to = -. Multipl # and add this to. The last line tells = 8 us a - ba - + b = Graph both functions in the same viewing window to see if the differ significantl. If the graphs lie on top of each other, then the are approimatel equal in that viewng window. 77. (a) g()=f(), so the zeros of f and the zeros of g are identical. If the coefficients of a polnomial are rational, we ma multipl that polnomial b the least common multiple (LCM) of the denominators of the coefficients to obtain a polnomial, 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 ;, ;, ;, ;, ;6, ; ;, ;, ;, ;, ;6, ;0, ;, ;0, ;, ;, ;, ;, ;, ;, ;, ;, ;, ;, ;, ;, ; 0 6, ; 6, ;, ;. There are no rational zeros. 78. Let f = - = + -. Notice that is a zero of f. B the Rational Zeros Theorem, the onl possible rational zeros of f are ; and ;. Because is none of these, it must be irrational. 79. (a) Approimate zeros:.6,.07, 0.90,.9 (b) f L g() = (c) Graphicall: Graph the original function and the approimate factorization on a variet of windows and observe their similarit. Numericall: Compute fc and gc for several values of c. Section. Comple Zeros and the Fundamental Theorem of Algebra Eploration. fi = i - ii + = = 0; f -i = -i - i -i + = = 0; no.. gi = i - i + + i = - - i + + i = 0; g - i = - i - - i + + i = -i + i = 0; no.. The Comple Conjugate Zeros Theorem does not necessaril hold true for a polnomial function with comple coefficients. Quick Review.. - i i.... (-)(+) 6. (+)(-) ; - 7. = = ; 9 i ; = = = - ; 7 i i - - i = i = - i + i - i = - i + i - i = - i + 6i - i = 7 + i + i - i = + i # + i - i + i + 0i + i + i = i = 6 = - + i ;, ; ;, ;, or ;, ;, ;, ; = i = + i = ; 9 ; 7 ;, ; ;, ;, ;, or ;, ;, ;, ;, ;, ;

28 9 Chapter Polnomial, Power, and Rational Functions 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)=(+)( +) = The factored form shows the zeros to be = and = ; i. The real zero = is the -intercept of the graph.. (-)(-)(+i)(-i) =( -+)( +) = The factored form shows the zeros to be = (multiplicit ) and = ;i. The real zero = is the -intercept of the graph.. (-)(--i)(-+i) =( -)[-(+i)][-(-i)] =( -)[ -(-i++i) +(+)] =( -)( -+)= The factored form shows the zeros to be =0, =, and =_i. The real zeros =0 and = are the -intercepts of the graph. In # 6, an constant multiple of the given polnomial is also an answer.. (-i)(+i)= + 6. (-+i)(--i)= (-)(-i)(+i)=(-)( +9) = (+)(-+i)(--i) =(+)( -+)= (-)(-)(-i)(+i) =(-)(-)( +) = (+)(-)(-+i)(--i) =(+)(-)( -+) = (-)(--i)(-+i) =(-)( -6+)= (+)(--i)(-+i) =(+)( -+)= ++0. (-) (+) = (+) (-)= (-) (--i)(-+i) =(-) ( -6+0) =( -+)( -6+0) = (+) (++i)(+-i) =(+) ( ++) =( ++)( ++) = In #7 0, note that the graph crosses the -ais at oddmultiplicit zeros, and kisses (touches but does not cross) the -ais where the multiplicit is even. 7. (b) 8. (c) 9. (d) 0. (a) In # 6, the number of comple zeros is the same as the degree of the polnomial; the number of real zeros can be determined from a graph. The latter alwas differs from the former b an even number (when the coefficients of the polnomial are real).. comple zeros; none real.. comple zeros; all real.. comple zeros; real.. comple zeros; real.. comple zeros; real. 6. comple zeros; real. In #7, look for real zeros using a graph (and perhaps the Rational Zeros Test). Use snthetic division to factor the polnomial 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 onl real zero. Dividing f() b - leaves ++ (below). The quadratic formula gives the remaining zeros of f(). 0 0 Zeros: =, = ; 9 i f() = - c - a - c - a + 9 i bd - 9 i bd = i i 7 8. Zeros: = (graphicall) and = (appling ; i the quadratic formula to -7+) f() = - c - a 7 c - a 7 + i bd - i bd = i i 9. Zeros: =_(graphicall) and = - ; i (appling the quadratic formula to ++6) f()=(-)(+) c - a - + i bd c - a - - i bd = i + - i

29 Section. Comple Zeros and the Fundamental Theorem of Algebra 9 0. Zeros: = and = (graphicall) and = (appling the quadratic formula to ; i ++=( ++)) / f()= c - a - i bd c - a + i bd = i + - i. Zeros: = - 7 and = (graphicall) and = ; i (appling the quadratic formula to =6( -+)). 7/ / f()=(+7)(-)[-(-i)] [-(+i)] =(+7)(-)(-+i)(--i). Zeros: = - and = (graphicall) and = ; i =(+)(-)(-+i)(--i) (appling the quadratic formula to =( -+)) / f()=(+)(-)[-(-i)] [-(+i)] =(+)(-)(-+i)(--i) In # 6, since the polnomials coefficients are real, for the given zero z=a+bi, the comple conjugate z =a-bi must also be a zero. Divide f() b -z and -z to reduce to a quadratic.. First divide f() b -(+i) (sntheticall). Then divide the result, +( +i) -+(-i),b -(-i). This leaves the polnominal -. 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() b -i. Then divide the result, +i --i,b+i. This leaves the polnomial -. Zeros: = ;, =;i i i 6 i 8 i i 0 i i i i 0 i 0 0 f()= i + i. First divide f() b -(-i).then divide the result, +( -i) -+6+i, b -(+i). This leaves -. Zeros: = ;, = ; i - i i i i i 0 + i - - i i + i i 0-0 f()= - + [-(-i)] [-(+i)] = - + (-+i)(--i) 6. First divide f() b -(+i). Then divide the result, +( +i) -+-i, b -(-i). This leaves -. Zeros: = ;, = ; i + i i i i - - i 0 - i - + i - - i - i i 0-0 f()= - + [-(-i)] [-(+i)] = - + (-+i)(--i)

30 96 Chapter Polnomial, Power, and Rational Functions For #7, find real zeros graphicall, then use snthetic division to find the quadratic factors. Onl the snthetic divison step is shown. 7. f()=(-)( ++) f()=(-)( ++) f()=(-)( ++) f()=(-)( ++) f()=(-)(+)( +) f()=(-)(+)( +) p. Solve for h: (h -h )(6.)= ()(0), so that p h -h =60. Of the three solutions (found graphicall), onl h.776 ft makes sense in this setting. p. Solve for h: (h -h )(6.)= ()(), so that p h -h =60. Of the three solutions (found graphicall), onl h 6. ft makes sense in this setting.. Yes: (+)( +)= + ++ is one such polnomial. Other eamples can be obtained b multipling an other quadratic with no real zeros b (+). 6. No; b the Comple Conjugate Zeros Theorem, for such a polnomial, if i is a zero, so is i. 7. No: if all coefficients are real, -i and +i must also be zeros, giving zeros for a degree polnomial. 8. Yes: f()=(--i)(-+i)(--i) (-+i)= is one such polnomial; all other eamples would be multiples of this polnomial. 9. 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()= f() must have the form a(--i)(-+i) (--i)(-+i); since f(0)=a( -i)( +i)( -i)( +i) =a()() =0a=0, we know that a=. Multiplied out, this gives f()= (a) The model is D 0.080t +0.96t -.6t [, 9] b [0, ] (b) Sall walks toward the detector, turns and walks awa (or walks backward), then walks toward the detector again. (c) The model changes direction at t.8 sec (D. m) and t.6 sec (when D.6 m).. (a) D 0.t -.79t+. [, 9] b [0, 6] (b) Jacob walks toward the detector, then turns and walks awa (or walks backward). (c) The model changes direction at t. (when D.0m).. False. Comple, nonreal solutions alwas come in conjugate pairs, so that if -i is a zero, then +i must also be a zero.. False. All three zeros could be real. For instance, the polnomial f()=(-)(-)= - + has degree, real coefficients, and no non-real zeros. (The zeros are 0,, and.). Both the sum and the product of two comple conjugates are real numbers, and the absolute value of a comple number is alwas real. The square of a comple number, on the other hand, need not be real. The answer is E. 6. Allowing for multiplicities other than, then, the polnomial can have anwhere from to distinct real zeros. But it cannot have no real zeros at all. The answer is A. 7. Because the comple, non-real zeros of a real-coefficient polnomial alwas come in conjugate pairs, a polnomial of degree can have either 0,, or non-real zeros. The answer is C. 8. A polnomial with real coefficients can never have an odd number of non-real comple zeros. The answer is E.