4U review package solutions - MHFWA Review Questions Polynomial Functions(Units 1&2\nswer Section MULTIPLECHOICE 1 13 2 A 3 c 4 B 5 A 5 c 7 D 8 c 9 B 10

4U review package solutions - MHFWA Review Questions...

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Unformatted text preview: MHFWA Review Questions - Polynomial Functions (Units 1&2) \nswer Section MULTIPLECHOICE 1. 13 2. A 3. c 4. B 5. A 5. c 7. D 8. c 9. B 10. A 11. D 12. A 13. c 14:; 0k 15. D 16. B 17. A 18; D 19. D 20. c 21. B ,HORTANSWER 22. f(x) increases for xeR; g(x) increases for x < 3; and h(x) increases for x < —1 and x > 2.5. 23- XX) - a); g(x) - d); h(x) - 6) 24. A local maximum is a point at which a curve changes from increasing to decreasing, whereas the local maximum value is the y-coordinate of that point. 25. When a function changes fiom decreasing to increasing, a local minimum occurs. 26. for) = —2x3 — 22:2 +1le— 6 27. Zeros are 3, —2, and —5. k> 0 :> asx ——> oo,j(x) —> oo 28. r = 9 29. a) x + 1 is the divisor f; k=~l 30. 3]. 32. 33. 34. 35. 36. 37. 38. 39. 40. The given zeros: fix) = k(x —- 1)(x + 1)(x + 2) 1x4+lx3+2x2+0x+1 J, l J, l 1 0 1 - Z 3 i (remove literal coefficients) 1 J, .1 Q :2 1 2 l (multiply number at lower left by k) 0 (add) ' l l 12 i (insert literal coefficients) lx3 + 0x2 + 2x— 2 remainder 3 3 'L _ 3 all-la +2>x +l _. «5+2/k «2 2,, b)(x+1)(1x3+2x—2)+3=x‘+x3+2x2+1 ' adj yes The remainder is 13. 1+] 2 x +2x—3 j 3 2x+1 2x +5x2—4x—5 remainder—2 1 2x +1 is the divisor 2 k= -5 2x3 + 5x2 — 4x — 5 J, ~L J, J, (remove literal coefficient) 2 5 —4 ~—5 J, —_1 i _3~ (multiply number at lower left by k) 2 4 6 —2 (add) Jr J, l i (+ 2, coefficient of x in divisor) 1 2 ‘ —3 t J, ‘I, J, J, (insert literal coefficients) 1x2+2x—3 remainder—2 xs—sz—x+5=(x—1)(x—5)(x+ l) 4x3+4x’—x—1=(2x—1)(2x+1)(x+ 1) x3 — 1000 = (x — 10)(x2 +10x +100) x4—7x2— 6x =x(x+1)(x—3)(x+ 2) x=—l, 1,3 x(2x+ l)(x—4)>0 1 The zeros of the corresponding equation are —— 2,,O and 4. x(2x+ l)(x—4)> Owhen 3 <1: <0 andx> 4. y—intercept: flO) = ——6 flO) = k(0 —1)(O + 1)(0 + 2) —6 = k(—2) 3 = k fix) = 3(x — 1)(x + 1)(x 47 2)isthe required function. l 41. a) When j(x) is divided by 2:: — 1, the remainder = 45] flx)=4x4—x3+2xZ—1 [I] [I] [I] [I] 1 1 1 3 f— =4 — - — +2 — —1=——-+——1=—— 2 15 8 4 4 8 2 3 3 Remainde = ”8: 1 b) Ifo—l is afactor of4x4—x3+2xZ—~1,thenf[§] = D. 1 3 - 4 3 2 However,f -— =-E,s<>2x—1 IS notafactor of4x —x +2x —1. 2 42. fix)=x3—3x2+5x—3 fl1)=1—3 +5—3=0$x—- 1 isafactor x2—2x+3 x— 1 if — 3::2 + 5x— 3 remainder o (x—1)(x2—2x+3)=0 x—1=Oorx2—2x+3=0 x-— —bt\}bz—4ac x=lor —————-—- 2a _2_23_m . f a?) 2(1) ”J gig/firm. \. ‘ 0 W7“ _2i'\f:§ f/ _ 2 ,} ,‘ / ) ZiZINf—i e C y = 2 . =1:I:2‘ 2 :vr=l,1+:'«J(-2:T a’ /i / LL“- Review Questions Rational Functions knswer Section MULTIPLE CHOICE P°>‘9‘EJ‘PP’5\":" w ”2’ m UWUW>OUWO>OWUJU SHORT ANSWER 15. ANS: The degree of the numerator must be exactly one more than the degree of the denominator. 16. ANS: The zeros are x= 9 and x = —2. 17. ANS: The vertical asymptotes are x = l] and x = 5. 18. ANS: The domain is {XIXaé 1,): e R}. 19. ANS: The x-intercepts are x = U, 3,-— 5. 20. ANS: 1 The y-intercept is .31 . 21. 22. 23. 24. 25. 26. 27. ANS: There is a vertical asymptote at x = 3 and a horizontal asymptote at y = 1. ANS: The domain is {xix #3 1,— 2, x e R} . There is only one vertical asymptote at x =-- —2. ANS: In order to have vertical asymptotes of x = 5 and X = —3, the denominator must have the factors (x— 5) and (PH- 3). In order to have a horizontal asymptote at y = I), the degree of the numerator must be less than the degree of the denominator. To ensure an x-intercept of 2, the numerator must have (2: — 2) as a factor. These various factors give us the following rational function: f(x) = ———:———. This function simplifies to (x — 5X2: + 3) x — 2 fix) = T—. x — 2x— 15 ANS: The function has a vertical asymptote at x = —2. There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator. In order to find oblique asymptotes, long division is required. X + l a x2 + 2x x - 2 x + 2 — 4 x2 + 3x— 2 4 ..f(x) = --—--— is equal to x+ l — -—-——-. x + 2 x + 2 Hence, the equation of the oblique asymptote is y = x + 1. ANS: . Since there are vertical asymptotes at x = —-3 and x = 2, the denominator of the function must include 2 5; OH 3X2: - 2}. This simplifies to x + x— 6. So far, the function is fix) = —§———--—. There appears to be a y— x + x— 6 intercept at )2 = --1. Set x equal to zero while letting f(x) = -1. This gives a value of 6 for k. The equation is 6 an = ~2———. x + x — 6 ANS: There is no oblique asymptote. There is a horizontal asymptote at y = l, which can be found by creating x—Z x+2 tables. There is only one vertical asymptote at x = 1. When the function is factored into f(x) = E—j—i—E—zg x— x- and then simplified, the only remaining factor in the denominator is x.—..i, giving only. one vertical asymptote. 28. 29. ANS: In order to have restrictions on x, we must have the factors (x + 2)(x~ 7) in the denominator, but since there is no asymptote when x = —2, we should have x+ 2 in the numerator as well. The function would then be x+ 2 1 {(3’) = ——£——)——-. This function would look very similar to f(x) = — but there would be a hole in the (x — 7)(x + 2) x - T graph where x = -2. ANS: x = ————-—— fl ) (x+ 5)(x— 3) The domain is {xix :6 -5, 3, x e R} . The x-intercept ofx = 5 is found by setting fix) equal to zero. The y- I intercept of}! = 3- is found by setting x equal to zero. There is a vertical asymptote at x = 3 and x = —5. There is a horizontal asymptote at y --_ 0‘, which can be found by creating tables. ...
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  • Spring '15
  • Rational function, horizontal asymptote, Jr J

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