x 16 d( x) 16 n( x) 36 d( x) zeros: x 2 36 = 0 x 2 = 36 x = ±6 Section Yes. Since 1 is a polynomial (of degree 0), P(x) =

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1 9 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS Section -. Yes. Since is a polynomial (of degree 0), P() P( ) is a rational function if P() is a polynomial.. A vertical asymptote is a vertical line a that the graph of a function approaches ever more closely, but does not cross, as approaches a from left and/or right, while function values increase or decrease without bound. 5. An oblique asymptote is a line of non-zero slope that the graph of a function approaches ever more closely as increases and/or decreases without bound. 7. This graph has a vertical asymptote, and a horizontal asymptote y. This corresponds to g(). 9. This graph has a vertical asymptote, and a horizontal asymptote y. This corresponds to h().. (A) (B) (C) (D). (A) (B) (C) (D) 5. f() 9 n ( ) Domain: d() d( ) zero: 0 domain: all real numbers ecept 0, (, 0) (0, ) intercepts: n() 9 zero: intercept: n( ) 9. r() Domain: d() d( ) zeros: 0 ±i No real zeros domain: all real numbers intercepts: n() zeros: 0 ( )( ) 0, intercepts:,. f() 5 n ( ) d( ) vertical asymptotes: d() zero: vertical asymptote horizontal asymptotes: Since n() and d() have the same degree, the line y 5 is a horizontal 6 n( ) 7. h() Domain: d() d( ) zeros: 0 ± domain: all real numbers ecept ± (, ) (, ) (, ) intercepts: n() 6 zero: 6 intercept: 6. F() 6 n( ) Domain: d() 6 6 d( ) zeros: ±6 domain: all real numbers ecept ±6 (, 6) ( 6, 6) (6, ) intercepts: n() 6 No real zeros no intercepts n( ) 5. s() 6 d( ) vertical asymptotes: d() 6 zeros: ± vertical asymptotes:, horizontal asymptote: Since the degree of n() is less than the degree of d(), the ais is a horizontal horizontal asymptote: y 0

2 SECTION - 9 n( ) 7. p() d( ) vertical asymptotes: d() zero: 0 vertical asymptote: 0 horizontal asymptote: Since the degree of n() is greater than the degree of d(), there are no horizontal asymptotes. 8 n( ) 9. h() 6 d( ) vertical asymptotes: d() 6 zeros: 6 0 ( ) 0 0, vertical asymptotes:, 0 horizontal asymptote: Since n() and d() have the same degree, the line y is a horizontal. The graph has more than one horizontal. The graph has a sharp corner at (0, 0). 5. ( ) if 0. The graph of f is the same as the graph of g ecept that f has a hole at (0, ). n( ) 9. f() d( ) Intercepts. There are no real zeros of n(). No intercept f(0) y y intercept Vertical asymptotes. d() zeros: Horizontal asymptotes. Since the degree of n() is less that the degree of d(), the ais is a horizontal. f() n( ) d( ) Intercepts. Real zeros of n() 0 intercept f(0) 0 y 0 y intercept The graph crosses the coordinate aes only at the origin. Vertical asymptotes. d() zeros: Horizontal asymptotes. Since n() and d() have the same degree, the line y is a horizontal ( )( 8) 8 if. The graph of f is the same as the graph of g ecept that f has a hole at, 6. Complete the sketch. Plot a few points. f ( ) Common Error: Entering / in the calculator. Parentheses are needed: /( ).

3 9 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS n( ). g() d( ) Intercepts. Real zeros of n() 0 ± intercepts g(0) is not defined no y intercepts Vertical asymptotes. d() zeros: 0 0 Horizontal asymptotes. Since n() and d() have the same degree, the line y is a horizontal 9 5. f() 9 n( ) d( ) Intercepts. There are no real zeros of n() 9. No intercept f(0) y intercept Vertical asymptotes. d() 9 zeros: ± Horizontal asymptotes. Since the degree of n() is less than the degree of d(), the ais is a horizontal 7. f() n( ) d( ) Intercepts. Real zeros of n() 0 intercept f(0) 0 y 0 y intercept The graph crosses the coordinate aes only at the origin. Vertical asymptotes. d() zeros: 0 ± Horizontal asymptotes. Since the degree of n() is less than the degree of d(), the ais is a horizontal 9. g() n( ) d( ) Intercepts. There are no real zeros of n(). No intercept g(0) y intercept Vertical asymptotes. There are no real zeros of d() No vertical asymptotes Horizontal asymptotes. Since the degree of n() is less than the degree of d(), the ais is a horizontal Complete the sketch. Plot a few additional points

4 SECTION - 95 n( ) 5. f() ( 5) d( ) Intercepts. Real zeros of n() 0 intercept f(0) 0 y 0 y intercept The graph crosses the coordinate aes only at the origin. Vertical asymptotes. d() ( 5) zeros: ( 5) Horizontal asymptotes. Since n() and d() have the same degree, the line y is a horizontal 5. To have zeros,,, and, the numerator must have factors ( ), ( ), ( ), and ( ). To have a horizontal asymptote at y, the degree of the denominator should be the same as the numerator, and the leading coefficient of the numerator should be times as large. To have no vertical asymptote, the denominator should have no real zeros. ( )( )( )( ) ( )( ) f() or f() will work. r( ) 55. To get y 5 as an oblique asymptote, our function should look like f() 5 where the q( ) degree of q() is greater than the degree of r(). If q() 0, we'll have 0 as vertical In that case r() will have to be constant so that its power is less than q(); r() 00 will do. 00 ( 5)( 0) 00 ( 5)( 0) 00 f() Let f() p( ) q( ) Common Error: It is not correct to multiply both sides by. The zero of p() is 0. The zero of q() is. These two zeros partition the ais into the three intervals shown in the table. A test number is chosen from each interval to determine the sign of f(). Interval Test number f() Sign of f (, 0) (0, ) (, ) The equality is satisfied at 0, but not at. We conclude that the solution set is [0, ) > 0 Let f() 5 p( ) q( ) 6 5 The zeros of p() 6 ( )( ) are and. The zero of q() 5 is 5. These three zeros partition the ais into the four intervals shown in the table. A test number is chosen from each interval to determine the sign of f().

5 96 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS Interval Test number f() Sign of f (, ) 5, 5 0 8, 5 5 (, ) 5 9 We conclude that the solution set is, (, ). 5 0 p( ) 6. Let f() q( ) The zeros of p() 8 0 ( )( 0) are and 0. The zero of 0 q() is 0. 0 These three zeros partition the ais into the four intervals shown in the table. 0 A test number is chosen from each interval to determine the sign of f() Interval Test number f() Sign of f (, ) 9 (, 0) (0, 0) 9 (0, ) The equality holds at and 0, but not at 0. We conclude that the solution set is [, 0) [0, ) 5 6. < < 0 5 9( ) < 0 ( ) 9 ( ) < 0 p( ) Let f() q( ) 9 ( ). The zeros of p() 9 9 ( )( ) are and. The zeros of q() ( ) ( )( ) are 0,, and. These five zeros partition the ais into the si intervals shown in the table. A test number is chosen from each interval to determine the sign of f().

6 SECTION - 97 Interval Test number f() Sign of f 7, 6, (, 0) 6 (0, ) , 65., 7 6 We conclude that the solution set is, (0, ),. 7 > 0 zeros of the numerator 7 0 a, b 7, c 7 7 ()() () 6.5, 0.59 zeros of the denominator: The three zeros partition the ais into the four intervals shown in the table below. A test number is chosen from each interval to determine the sign of the rational epression. Interval (, 6.5) ( 6.5, ) (, 0.59) ( 0.59, ) Test number f() 7 8 Sign of f The epressions is positive on ( 6.5, ) ( 0.59, ). We need to eclude all endpoints, so this is the solution. Graphical check: The graph is above the ais on ( 6.5, ) ( 0.59, ).

7 98 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS (Simplify) zeros of the numerator: a, b 9, c ( )( 5) , 8.05 ( ) zeros of the denominator: 0 These three zeros partition the ais into the four intervals shown in the table below. A test number is chosen from each interval to determine the sign of the rational epression. Interval (, 0) (0, 0.595) (0.595, 8.05) (8.05, ) Test number f() 5 0 Sign of f The epression is negative on (, 0), (0, 0.595), and (8.05, ). It's equal to zero when and 8.05, so we include those endpoints, but we eclude 0 since the epression is undefined there. The solution is (, 0) (0, 0.595] [8.05, ). Graphical check: The graph of y 9 5 is below the graph of y on (, 0), (0, 0.595) and (8.05, ) and they intersect at and > 0 (Simplify) zeros of the numerator: > zeros of the denominator: 5 0( 5) These two zeros partition the ais into the intervals shown in the table > 0 below. A test number is chosen from each interval to determine the sign of the 5 5 rational epression > > 0 5 Interval (, 5) (5, 7.9) (7.9, ) Test number f() Sign of f The epression is positive on (5, 7.9). We need to eclude the endpoints, so this is the solution.

8 SECTION - 99 Graphical check: 7. The graph of y is above the graph of y 0 on (5, 7.9). 5 7 (Simplify) zeros of the numerator: zeros of the denominator: ( ) 0 7( ) 0, 0 These three zeros partition the ais into the four intervals shown in the table ( ) ( ) below. A test number is chosen from each interval to determine the sign of the 7 7 rational epression. 0 ( ) 7 0 ( ) Interval (,.) (., ) (, 0) (0, ) Test number 0.5 f() 5 Sign of f The epression is positive on (,.) (, 0). We include. because it makes the epression zero, but eclude 0 and as they make the epression undefined. The solution is (,.] (, 0). Graphical check: The graph of y is above the graph of y 7 on (,.] (, 0). 7. f() n( ) 75. p() d( ) n( ) d( ) Vertical asymptotes. Real zeroes of d() Vertical asymptotes. There are no real zeros of d(). No vertical asymptotes. Horizontal Since the degree of n() is Horizontal asymptotes. Since the degree of n() is greater than the degree of d(), there is no greater than the degree of d(), there is no horizontal horizontal

9 00 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS Oblique Thus, f(). Hence, the line y is an oblique Oblique asymptote: Thus, p(). Hence, the line y is an oblique 5 n( ) 77. r() d( ) Vertical asymptotes. Real zeros of d() 0 Horizontal Since the degree of n() is greater than the degree of d(), there is no horizontal Oblique Thus r() 5. Hence the line y is an oblique 79. n( ) Hence, the line y is an oblique f() Complete the sketch. Plot a few points. d( ) Intercepts. There are no real zeros of n(). f ( ) No intercept. f(0) is not defined. No y intercept. 5 Vertical asymptotes. Real zeros of d(). 0 Horizontal Since the degree of n() is greater than the degree of d(), there is no 5 horizontal Oblique f() n( ) Oblique asymptote: k() d( ) Intercepts. Real zeros of n() 0 ( )( ) 0, intercepts k(0) y intercept Vertical asymptotes. Real zeros of d() 0 Horizontal Since the degree of n() is greater than the degree of d(), there is no horizontal Thus, k().

10 SECTION - 0 Hence, the line y is an oblique 8 n( ) 8. F() Hence, the line y is an oblique d( ) Intercepts. Real zeros of n() ( )( ) No real zeros intercept F(0) is not defined. No y intercept. Vertical asymptotes. Real zeros of d(). 0 Horizontal Since the degree of n() is greater than the degree of d(), there is no horizontal 8 8 Oblique F(). 85. f(). f() is not defined if 0, that is, Domain: (, ) (, ) ( )( ) f() ( ) f() The graph is a straight line with slope and y intercept, ecept that the point (, ) is not on the graph. There are no asymptotes. 87. r(). f() is not defined if 0, that is, ± Domain: (, ) (, ) (, ) r() ( )( ) r() The graph is the same as the graph of the function, ecept that the point, is not on the graph. Intercepts: y. No intercept. Vertical asymptote: Horizontal asymptote: y 0 Complete the sketch. Plot a few additional points

11 0 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS 89. N(t) 50 t t 0 t Intercepts: Real zeros of 50t: t 0 N(0) 0 Vertical asymptotes: None, since, the only zero of t, is not in the domain of N. Horizontal asymptote: N 50. As t, N N(t) 5t 0 t t Intercepts: Real zeros of 5t 0, t. None, since 6, the only zero of 5t 0, is not in the domain of N. Vertical asymptotes: None, since 0, the only zero of t, is not in the domain of N. Horizontal asymptote: N 5. As t, N 5 Complete the sketch. Plot a few points. t N( t) (A) C (n) C( n), n 5n 5n 75,500 n n n (B) The minimum value of the function C (n) is C c, where a 5 and c,500 a min C (n) C,500 5 (C) Intercepts: Real zeros of,500 75n 5n. None. No n intercepts. 0 is not in the domain of n, so there are no C intercepts. Vertical asymptotes: Real zeros of n. The line n 0 is a vertical Sign behavior: C is always positive since n 0. Horizontal asymptote: None, since the degree of C(n) is greater than the degree of n. C 00 C (0) This minimum occurs when n 0, after 0 years. Oblique asymptote: The line C 5n 75 is an oblique 95. (A) Since Area length width, length Area width 5. Then total length of fence width length L() (B) can be any positive number, thus, domain (0, )

12 SECTION -5 0 c (C) The minimum value of the function L() is L where a and c 50. a 50 min L() L L( 5 ) L(5) This minimum occurs when 5. Width 5 feet. Length 5 5 feet. 5 (D) Intercepts: Real zeros of 50. None, hence, no intercepts. 0 is not in the domain of L, so there are no L intercepts. Vertical asymptotes: Real zeros of. The line 0 is a vertical Sign behavior: L() is always positive since > 0. Horizontal asymptote: None, since the degree of 50 is greater than the degree of. Oblique asymptote: The line L is an oblique Section -5. y increases, for if >, k > k so y > y.. y decreases, for if >,, k k, so y < y. 5. If y varies directly with, then y k. If 0, y k F k 9. R kst. L km k. A kc d 5. P k 7. h s. Write u k v. Substitute u and v and solve for k. k k k The equation of variation is u When v 0, u 0. v. m 9. R k d. D k y z 5. Write L k. Substitute L 9 and M 9 and solve M for k. 9 k 9 k 9 79 The equation of variation is L 79. M When M 6, L Write Q k mn P P and solve for k. (6) k 9k. Substitute Q, m, n 6, and 9. t k T. L k wh. N F d k 9 The equation of variation is Q 9 When m, n 8, and P, Q 9 mn P. (8)