**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:
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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