review 2 - rational functions - Review Questions Rational Functions Multiple Choice x2 5X 4 1 What is the x-intercept of the function x = 1 x a

review 2 - rational functions - Review Questions Rational...

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Unformatted text preview: Review Questions Rational Functions Multiple Choice . _ x2 — 5X+ 4 1. What is the x-intercept of the function fix) = —1? x’ a. x=landx=4 c. x=~4 b x=l d x=4 _ . x + 3 2. What is the domain of the function fix) = 3 ? x — 1 a. {xlxe—lfifiefi} C. {Xix:‘=-—1,XER} 1)- {xix¢ 1,}: e R} d. {xlme —3,x e R} x2 — 3x+ 2 3. What is the oblique asymptote for fix) = ——--—2-? x + a. y= 12 c. 12 y = x - 5 + —— x + 2 b. y = x— 5 d. 12 y _ x+ 2 4. Which of the following functions does not have a vertical asymptote? a. x+ 3 c. 2 3; = X _ 4 2 y = x — 5 . x+ 2 b. x d. 2 )3 = 2 Y = ‘— (X “ 3) x 5. Which of the following functions does not have a horizontal asymptote? a. 3 2 c. x+ 1 x — 3x + 3x— 1 y: ____. r = -—-—;--— x— 1 x —- 5 b x2 — 1 d 32:2 — 5x+ 2 Y = 3’ = x3 + 8 2.3:2 — 5 (x — 3)(x + 2) 6. What are the zeros of f(x} = -—--——? x(x + 5:311}: — 1) a. x: ,~2,~5,1,D c. x=3,—2 b. x=—5,l d. x=U,—5,l . . . . six} . 7. Which of the followmg statements 18 false for the function f(x) = m where both g and h are polynomial . 2: functions? a. It is possible to have both a vertical asymptote and a horizontal asymptote. b. It is possible to have both an oblique asymptote and a horizontal asymptote. c. It is possible to have both a vertical asymptote and an oblique asymptote. d. It is impossible to have an oblique asymptote, a horizontal asymptote, and a vertical asymptote. x _ 2 __ 8. What is the horizontal asymptote of f (x) = 1‘.7 x + a.y2 eyfl b.y—1 d.y1 x 9. What is the vertical asymptote of f (I) = ———-—? x — 8x+ 15 a. x=3andx=5 c. x=5 b. I = 3 d- X = U x2 — 10. What are the vertical asymptotes of fix) = 2——--—-———? x ~7X+ 10 a. x=2andx=5 c. x=3andx=—3 b. x=—2andx=—5 d. x=1 11. Which of the following functions has an oblique asymptote? a. 2 c. x+ 3 1’ 2 2 x — 5x-— 14 x —4 b. x2 _ 4 d. (x— 3): y— x+3 3L (x-3)(x+4) 12. Which of the following functions has a vertical asymptote at x = 2 and a horizontal asymptote of y = 1? a. 2 c. x+ 3 x — 6x+ 9 y = 2 Y = — - 2:2 + 3x+ 2 x 4 b. y= 3 d. X2 _ 9 x— 2 J; = 7“” x — 4x + 4 13. Which of the following functions has {xix e R} as its domain? a. x2+5x—6 0' x2+3x+2 3km” W“?— x—l :1: ~4x+3 b' x2+x d' x2+6x+8 y=T_—_ y=__2__..___._. x +X+3 x +5x+4 14. What is the domain of the following function? a. {xlXCUUPfiDLXER} 0. {xIxeR} 13- {xlpne ~3,2,x e R} d- {xlx¢ ~2,3,x e R} Short Answer 15. Under what condition does a rational function have an oblique asymptote? (x— 9)(x+ 2) 7 16. What are the zeros for the function fix] = 2:0: - 1) (x— 2)(x+ 5) 9 l7. What are the vertical asymptotes of the function fix) = ( 5) x x — . . XiIi-l) 18. What is the domain of the function f(x) = ——-9 (x—1)(x2 + 4). . . IU‘[email protected]+$ 19. What are the x—intercepts of the function f(x) = -————? x — 4 _ x2 —- 2x+ 1 20. What is the y—intercept of the function fix) = —3? x + 21. Draw the graph of a rational function that has a vertical asymptote at x = 3 and a horizontal asymptote at y = 2. x + 2 22. What are all the asymptotes of the function f(x) = —3? x _. 2 _. 23. What is the domain of the function fix) = —-—~——? How many vertical asymptotes does this function (x— l)(x + 2) have? ' 24. Create a rational function with vertical asymptotes at x = 5 and x = —3, a horizontal asymptote at y = U, and an x-intercept of 2. Describe your process in creating the function. 2 x + 3x— 2 25. Find the equations of all asymptotes for the rational function fix) = ——2 x + . . . k 26. Flnd the equation for the followmg graph in the form f(x) = 2 x + bx + C 2 27. For the rational function f(x) = ‘, determine the equations of all asymptotes. x2 ~ 3x+ 2 28. Create a rational function that has a domain of {xlx ¢ —2,T,x e R} and a vertical asymptote at X = 7". Describe the graph of this function. x—S x2 +2x— 15 29. Using domain, intercepts, and asymptotes, draw the graph of f (2:) = Review Questions Rational Functions Answer Section MULTIPLE CHOICE ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: 10. ANS: 11. ANS: 12. ANS: 13. ANS: 14. ANS: 9°.\‘.m§:"r°~9’!"’f‘ 5° UwUUfi>OUWO>OUfiwU 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 = U and x = 5. 18. ANS: The domain is {XIX :5 1,x e R}. 19. ANS: The x—intercepts are x = 0, 3,— 5. 20. ANS: 1 The y—intercept is .37. 21. ANS 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 #5 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 (X + 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 (x - 2) as a factor. These 2: — 2 various factors give us the following rational function: f(x) = —-——. This function simplifies to (x — 5)(x + 3) x—2 rtx)=,—. 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 x+2 X2+3X—2 2 2: +22: :1 m >1 N + —4 xg+3x—2 4 .'.f(x) = ---—- is equal to x+ 1 ~ ~———-. x+2 x+2 Hence, the equation of the oblique asymptote is y = PH 1. ANS: Since there are vertical asymptotes at x = —3 and x = 2, the denominator of the function must include 2 k (x+ 3X2: — 2). This simplifies to .1: + x— 6. So far, the function is f(x) = a—w' There appears to be a y— ): + x - 6 intercept at y = —1. Set x equal to zero while letting f(x) = ~ 1. This gives a value of 6 for k. The equation is 6 fix} = —,— x + x — t5 ANS: There is no oblique asymptote. There is a horizontal asymptote at y = l, which can be found by creating x— 2 x+ 2] tables. There is only one vertical asymptote at x = 1. When the function is factored into f(x) = E————5fl x — x — and then simplified, the only remaining factor in the denominator is x —-. 1, giving only one vertical asymptote. 28. 29. ANS: In order to have restrictions on x, we must have the factors (I + 2X7: ~ 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 {(1') = LL. This function would look very similar to for) = but there would be a hole in the (x - "fjlfix + 2) x ~ 7 graph where x = —2. ANS: ft ) X— 5 x = _—_.. (x + 5) (x — 3) The domain is {XIX aé -5, 3, x E R} . The x-intercept of x = 5 is found by setting fix) equal to zero. The y- l intercept of y = —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‘ f U? which can be found by creating tables. 2t ...
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  • Spring '15
  • Rational function, horizontal asymptote

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