**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