SECTION 3.6 Rational Functions
283
/b) s(x\ = x2+x 20 = (^ 4) (x +5) = x_4, for x^ -5. Therefore, s(x) = x- 4, x^ -5. Since
w
v '
x + 5
x + 5
(_5) _ 4 = _g} the graph isthe line y = x - 4 with the point (-5, -9) removed.
t(x\ = 2*2 ~x~1 = (2g +*) (x ~*) = 2x +1, for a; ^ 1. Therefore, t(x) = 2x +1, x^ 1. Since
v '
x —1
x —1
2 (1) + 1 = 3,the graph isthe line y = 2x + 1 with the point (1,3) removed.
u(x)= x2X~2
= x X~2ns
= -,
for x^ 2. Therefore, u(x) = -,x
# 2. Whenx = 2, -x=l-, so the graph is
- 2x
(x —2)
x
x
the curve y = - with the point (2, |) removed.
y-
fMI.3)
h
l
X
84. (a)
Let / (x) = \.
Then r(x) = (x
, —2)
* x2 =/ (as - 2). From this form we see that
*
x^
the graph ofr is obtained from the graph of/ by shifting 2units to the right. Thus
r has vertical asymptote x = 2 and horizontal asymptote y = 0.
= 3^(x _|_ i) + 2. From this form we see that the graph ofs isobtained from
(b),(*) =23f+4*
+5=2 +
x2 + 2x+l " ' (x + 1)2
the graph of/ by shifting 1unit to the left, stretching vertically by afactor of 3, and shifting 2units vertically. Thus r
has vertical asymptote x = -1 and horizontal asymptote y = 2.
x2+2x + l | 2x2+4x+5
2x2+4x+2
I
I
-/l^T-S-n--
'
p— —i—I
I
l
l
l
l
l
I
I »
X
284
CHAPTER 3 Polynomial and Rational Functions
(c) Using long division, we see that p(x) = —
jtJC| A= -3 + ——^
; , which cannot be graphed by transforming
x2 —4x + 4
a:2 - 4x + 4
/ (x) = —. Using long divisionon q we have:
-3
-3
x2 - 4x +4 |-3x2+ 0x+ 2
So q (x) =
12a; - 3x2
x2-4x + 4
x2 - 4x + 4 | -3x2+12x+ 0
-3x2+12x-12
-3x2+12x-12
-12x+14
12
12
' (x-2f-2 = 12/ (x —2) - 3. From this form wesee that thegraph of q is obtained
= -3 +
from the graph of/ by shifting 2units to the right, stretching vertically by a factor of12, and then shifting 3 units
vertically down. Thus the vertical asymptote is x = 2and the horizontal asymptote is y = -3. We show y = p(x) just
to verify that we cannot obtain p (x) from y = —.
>•
1
y = q {x)
r
1
1 1>*
^
•>,
1
v
y=p (x)
Chapter 3 Review
1. P(x) = -x3 + 64
2. P (x) = 2x3 - 16 = 2(x - 2) (x2 + 2x + 4)
y
(0.64)
10
V-t.oi
1
3. P(x) = 2(x-l)"-32
\
4. P(x) = 64- (x- 3)'
(3-272.0)
CHAPTER 3
5. P (x) = 32+ (x - 1)£
Review
285
6. P (x) = -3 (x + 2)5 + 96
(0.0)
n
7. P (x) = x3 - 4x + 1. x-intercepts: -2.1,0.3, and 1.9.
8. P (x) = -2x3 + 6x2 - 2. x-intercepts: -0.5, 0.7, and
^-intercept: 1. Local maximum is (-1.2,4.1). Local
2.9. y-intercept: -2. Local maximum is (2,6). Local
minimum is (1.2, —2.1). y—•coasx—•oo;y
—• -co as
x —• —oo.
minimum is (0, —2). y —* oo as x —»• —oo and y —• —oo
as x —• oo.
-4
-6
9. P (x) = 3x4 - 4x3 - lOx - 1. x-intercepts: -0.1 and
10. P (x) = x5 + x4 - 7x3 - x2 + 6x+ 3. x-intercepts:
2.1. y-intercept: —1. Local maximum is (1.4, -14.5).
-3.0,1.3, and 1.9. y-intercept: -3. Local maxima are
There is no local maximum, y —* oo as x —• ±oo.
(-2.4,33.2) and (0.5,5.0). Local minima are (-0.6,0.6)
and (1.6, —1.6). y —* —oo as x —• —oo and y —* oo as x
20
—»• oo.
SO
° S.
•1
i
»
y
j
1
40
1
-10
20
10
-20
•4
p
-2
-1
»
/
/
. » 2
)
-10
-20
11. (a) Use the Pythagorean Theorem and solving for y2 we
have, x2 + y2 = 102 <& y2 = 100- x2.
12. (a) The area ofthe four sides is2x2 -F 2xy = 1200 &•
2xy = 1200 - 2x2
Substituting we get
(b) Domain is [0,10].
(d) The strongest beam has width 5.8 inches.
y=
600 - x2
Substituting we get
S = 13.8x (100 - x2) = 1380x - 13.8x3.
(c)
&
V
=^ =x2(60°^£!)=600a;_x3.
(b)
(c) V is maximized when x ss 14.14, y « 28.28
286
13.
CHAPTER3 Polynomialand Rational Functions
.. x2-Fx-12
x2 - 3x -F 5
14.
x-2
(x + 4)(x-3)
— = -
2
1
= x + 4. Factoring we
x — 3
have Q (x) = x + 4 and R (x) = 0. This can be
-3
1
, . _ t .
-^-
x — 3
2
-2
-1
3
confirmed with synthetic division:
3 I1
1
-12
12
Using synthetic division, we see that Q (x) = x —1 and
0
R(x)=3.
15.
x3 - x2 + llx + 2
16.
x —4
1
1
17.
x-F3
0
-10
3
-9
1-13
-19
-3
-1
11
4
12
92
3
23
94
-3
Using synthetic division, we see that
Using synthetic division, we see that Q(x) = x2 —x + 3
Q (x) = x2 + 3x -F 23andR (x) = 94.
and.R(x) = -19.
x4 - 8x2 + 2x + 7
18.
x + 5
-5 I 1
1
19.
x3 + 2x2 - 10
2x4 + 3x3 - 12
x + 4
-4
0
-8
-5
25
-85
415
-5
17
-83
422
0
0
-12
-8
20
-80
320
-5
20
-80
308
2
2
Using synthetic division, we see that
Using synthetic division, we see that
Q (x) = x3 - 5x2 -F 17x- 83and R (x) = 422.
Q(x) = 2x3 - 5x2 + 20x- 80and R (x) = 308.
2x3 + x2 - 8x + 15
20.
x2 + 2x - 1
2x -
x2 - x + 3
x2
3
|2x3 + x2
x2 -F 2x --1
x4 - 2x2 + 7x
2x3 + 4x2
- 8x + 15
x2 - x + 3
1*4
x4
- 2x
+
-
X
4
+ Ox3 - 2x2
+ 7x +
0
x3 + 3x3
-
-3x2 - 6x -F 15
x3 - 5x2 -F 7x
-3x2 - 6x +
x3 -
3
12
Therefore, Q (x) = 2x - 3, and R (x) = 12
x2 + 3x
-4x2 + 4x +
0
-4x2 + 4x -
12
12
Therefore, Q(x) = x2 + x - 4, and R (x) = 12.
21. P (x) = 2x3 - 9x2 - 7x + 13;find P (5).
5 I2
2
Therefore, P (5) = 3.
-9
-7
10
5
1
-2
13
22. Q(x) = x4 -F 4x3 + 7x2 + lOx + 15; find Q(-3)
-3 | 1
-10
1
4
7
10 15
-3
-3
-12
6
1
4
-2
21
Bythe Remainder Theorem, we have Q (—3) = 21.
CHAPTER 3
23. i isa zero of P (x) = 2x4 + x3 - 5x2 + lOx - 4 if
Review
287
24. x + 4 is a factor of
P (x) = x5 4- 4x4 - 7x3 - 23x2 + 23x+ 12if
P(§)=0.
\ | 2 1 -5
P(-4)=0.
10-4
-4 | 1
11-24
2
2-4
8
4-7 -23
-4
0
0
10-7
Since P(^)=0, |isa zero ofthe polynomial.
23
12
28
-20
-12
5
3
0
Since P (—4) = 0, x + 4 is a factor of the polynomial.
25. P (x) = x500 + 6x201 - x2 - 2x -F 4. The remainder from dividing P (x) by x - 1 is
P (1) = (l)500 + 6 (l)201 - (l)2 - 2 (1) + 4 = 8.
26. Let P (x) = x101 - x4 + 2. The remainder from dividing P (x) by x + 1 isP (-1) = (-1)101 - (-1)4 +2 = 0.
27. (a) P (x) = x5 - 6x3 - x2 + 2x + 18 has possible rational zeros ±1, ±2, ±3, ±6, ±9, ±18.
(b) Since P (x) has2 variations insign, there areeither 0 or2 positive real zeros. SinceP(—x) = -x5+6x3-x2-2x+18
has 3 variations in sign, there are 1 or 3 negative real zeros.
28. (a) P (x) = 6x4 + 3x3 + x2 + 3x + 4 has possible rational zeros ±1, ±2, ±4, ±± ±± ±|,±± ±±
(b) Since P (x) has novariations insign, there arenopositive real zeros. Since P (-x) = 6x4 - 3x3 + x2 - 3x + 4 has 4
variations in sign, there are 0, 2, or 4 negative real zeros.
29. (a) P (x) = x3 - 16x = x (x2 - 16)
= x (x —4) (x + 4)
30. (a) P (x) = 2x3 - 16 = 2 (x3 - 8)
= 2(x - 2) (x2 + 2x+ 4)
has zeros -4,0,4 (all of multiplicity 1).
(b)
We use the quadratic formula on the remaining
y*
quadratic: x = ~2±^4~16, which is not real. So the
only real zero is 2 (multiplicity 1).
(b)
,*
288
CHAPTER 3 Polynomial and RationalFunctions
31. (a)P(x) =x4+x3-2x2=x2(x2+x-2)
32. (a) P(x) = x4 - 5x2 +4 = (x2 -4) (x2 - l)
= x2 (x + 2) (x - 1)
= (x - 2) (x + 2) (x - 1) (x + 1)
The zeros are 0 (multiplicity 2), —2 (multiplicity 1),
Thus, the zeros are —1,1, -2, 2 (all of multiplicity 1).
and1 (multiplicity 1)
(b)
(b)
y,
y*
33. (a) P (x) = x4 - 2x3 - 7x2 + 8x + 12. The possible rational zeros are±1, ±2, ±3, ±4, ±6, ±12. P has 2 variations in
sign, so it has either 2 or 0 positive real zeros.
1 I 1 -2
-7
1
-1
-8
0
-1
-8
0
12
1
8 12
2 I 1 -2
2
-7
8
12
0
-14
-12
0-7
-6
P (x) = x4 - 2x3 - 7x2 + 8x+ 12 = (x- 2) (x3 - 7x - 6).Continuing:
211 0 -6 -6
2
3
1
0
-7 -6
1
3
0
(b)
4-4
1 2 - 2 -10
so x = 3 is a root and
P (x) = (x- 2) (x - 3)(x2 + 3x+ 2)
= (x - 2) (x - 3) (x + 1) (x + 2)
Therefore the real roots are -2, —1,2, and 3 (all
of multiplicity 1).
34. (a) P (x) = x4 - 2x3 - 2x2 + 8x - 8
= x2 (x2 - 2x+ 2) - 4 (x2 - 2x+ 2)
= (x2-4)(x2-2x + 2)
= (x - 2)(x+ 2) (x2 - 2x+ 2)
The quadratic is irreducible, so the real roots are
±2 (each of multiplicity 1).
(b)
x = 2 is a root.
CHAPTER 3
Review
289
35. (a) P (x) = 2x4 + x3 + 2x2 —3x —2. The possible rational roots are ±1, ±2, ±|. P has one variation in sign, and
hence 1 positive real root. P (—x) has 3 variations in sign and hence either 3 or 1 negative real roots.
1 I 2 1 2 -3
2
2
3
3
5
-2
2
0
=» x = 1 is a zero.
P (x) = 2x4 + x3 + 2x2 - 3x- 2 = (x - 1) (2x3 + 3x2 + 5x+ 2). Continuing:
1|2
3
5
2
-2
-1
-4
1
4
-2
2
2 | *•
2
3
5
2
-1
-1
-2
2
4
-2 2
3
(b)
5
-4 2
-14
2 - 1 7 -12
0 =>x = -±
P (x) = (x - 1)(x + \) (2x2 + 2x+ 4). The
quadratic is irreducible, so the real zeros are 1
and —\ (each ofmultiplicity 1).
36. (a) P (x) = 9x5 - 21x4 + 10x3 + 6x2 - 3x - 1. The possible rational zeros are ±1, ±\, ± j. P has 3variations in sign,
hence3 or 1 positive real roots. P (—x) has 2 variations in sign, hence 2 or 0 real negative roots.
11 9 —21
10
9
-12
9
-12
6 -3 -1
-2
- 2 4 1 0 =• x = 1 is a zero.
P (x) = (x - 1)(9x4 - 12x3 - 2x2 + 4x+ 1). Continuing:
11 9 —12 —2
9
9
P(x) = (x-l)2(9x3
3x2
4
1
-3 -5 -1
-3 -5 -1
5x-l).
0
x = 1 is a zero again
(b)
Continuing:
1I9 -3 -5 -1
9
6
1
0 =£• x = 1 is a zero yet again.
P (x) = (x- l)3 (9x2 + 6x + 1)
= (x - l)3 (3x+ l)2
So the real zeros of P are 1 (multiplicity 3) and
-\ (multiplicity 2).
37. (2 - 3i) + (1 + U) = (2 + 1) + (-3 + 4) i = 3 + i
38. (3 - 6i) - (6 - 4i) = 3 - 6i - 6 + 4z = (3 - 6) + (-6 + 4) i = -3 - 2%
39. (2 + i) (3 - 2i) = 6 - 4f + 3i - 2z2 = 6 - i + 2 = 8 - i
40. 4t (2 - ±i)= 8i - 2i2 = 8i + 2 = 2 + 8z
290
CHAPTER 3 Polynomial and Rational Functions
41 4+ 2i _ 4+ 2* 2+ i _ 8+ 8i + 2i2 _ 8+ 8i - 2 _ 6+8i _ 6 • 8•
' 2-i ~ 2-i ' 2+ i~
4-i2
~
4+1
~
5 ~s + 5z
42 8+ 3i _ 8+ 3i 4-3i _ 32 - 12i - 9i2 _ 32 - 12i +9 _ 41 - 12i _ 41 _ 12 •
' 4 + 3i ~ 4 + 3i *4-3z ~
16 - 9i2
~
16 + 9
~
25
~ 25 ~ 25*
43. i25 = i24i=(^)6i = (lfi = i
44. (1 + i)3 = 1 + 3i + 3i2 + i3 = 1 + 3i - 3 - i = -2 + 2i
45. (1 - v/=T) (1 + V^T) = (1 - t) (1 +1) = 1+ i - i - i2 = 1+ 1 = 2
46. V-iO • \T^ = VlOi •2y/Wi = 20i2 = -20
47. Since the zeros are —|, 2,and 3,a factorization is
P(x) = C(x + i)(x-2)(x-3) = iC(2x+l)(x2-5x + 6)
= \C (2x3 - 10x2 + 12x + x2 - 5x + 6) = \C (2x3 - 9x2 + 7x + 6)
Since the constant coefficient is 12, \ C(6) = 12
& C = 4, and so the polynomial is P (x) = 4x3 - 18x2 + 14x +12.
48. Since the zeros are ±3i and 4 (which is a double zero), a factorization is
P (x) = C(x - 4)2 (x- 3i) (x+ 3i) = C(x2 - 8x + 16) (x2 + 9) = C(x4 - 8x3 + 25x2 - 72x + 144). Since all
ofthecoefficients are integers, we can choose C = 1,soP (x) = x4 —8x3 + 25x2 —72x+ 144.
49. No, there is no polynomial of degree 4 with integercoefficients that has zeros i, 2i, 3i and Ai. Since the imaginary zeros
of polynomialequations with real coefficients come in complexconjugate pairs, there would have to be 8 zeros, which is
impossible for a polynomial of degree 4.
50. P (x) = 3x4 + 5x2 + 2 = (3x2 + 2) (x2 + l). Since 3x2 + 2 = 0and x2 + 1 = 0 have no real zeros, itfollows that
3x4 + 5x2 + 2 has no real zeros.
51. P (x) = x3 - 3x2 - 13x+ 15has possible rational zeros ±1, ±3, ±5, ±15.
1 I 1 -3 -13
1
15
1
-2
-15
-2
—15
0
=»• x = 1 is a zero.
So P (x) = x3 - 3x2 - 13x + 15 = (x - 1) (x2 - 2x- 15) = (x - 1)(x - 5)(x + 3).Therefore, the zeros are -3,1,
and 5.
52. P (x) = 2x3 + 5x2 - 6x - 9 has possible rational zeros ±1, ±3, ±9, ±5, ±f, ±f. Since there isone variation in sign,
there is a positive real zero.
112 5 —6 —9
1
3|2 5-6-9
6
33
81
§12 5 -6 -9
3
12
9
2 7 1-8
2 11 27 72 =^ x = 3 isan upper bound
2 8 6 0 => x = § isa zero.
So P (x) = 2x3 + 5x2 - 6x- 9 = (2x - 3)(x2 + 4x+ 3) = (2x - 3)(x + 3)(x+ 1). Therefore, the zeros are -3,
-landf.
CHAPTER 3
Review
291
53. P (x) = x4 + 6x3 + 17x2 + 28x+ 20has possible rational zeros ±1, ±2, ±4, ±5, ±10, ±20. Since all ofthe coefficients
are positive, there are no positive real zeros.
-ill
6
1
5
17
28
20
-5
-12
-16
12
16
4
-2 I 1
6
17
28
20
-2
-8
-18
-20
4
9
10
0
1
=> x = -2 is a zero.
P (x) = x4 + 6x3 + 17x2 + 28x + 20 = (x + 2) (x3 + 4x2 + 9x+ 10). Continuing with the quotient, we have
-2 I 1
1
4
9
10
-2
-4
-10
2
5
0
^> x = -2 is a zero.
Thus P(x) = x4 + 6x3 + 17x2 + 28x + 20 = (x+ 2)2 (x2 + 2x + 5). Now x2 + 2x + 5 = 0 when
x= "2±^42"4(5)(1)- = ^^ = -1 ±2%. Thus, the zeros are -2(multiplicity 2) and -1 ±2%.
54. P (x) = x4 + 7x3 + 9x2 - 17x - 20has possible rational zeros ±1, ±2, ±4, ±5, ±10, ±20.
ill 7 9 -17 -20 2ll 7 9 -17 -20
1
8
17
0
1 8
17
0
-20
1
-111
2
18
54
74
9
27
37
54 =$>x = 2isan
1
7 9 -17 -20
-1
-6
-3
20
6
3
-20
0
=> x = -1 is a zero.
upper bound.
So P (x) = x4 + 7x3 + 9x2 - 17x - 20 = (x+ 1) (x3 + 6x2 + 3x- 20). Continuing with the quotient, we have
—111 6 3-20
-1
1
-5
—211 6 3-20
2
-2
5 - 2 -18
1
-8
-411 6 3-20
10
-4
4 - 5 -10
-8
20
12-5
0
=> x = -4 is a zero.
So P (x) = x4 + 7x3 + 9x2 - 17x - 20 = (x+ 1) (x+ 4) (x2 + 2x- 5). Now using the quadratic formula on
x2 + 2x —5 we have: x =
——
=
= —1± y/6. Thus, thezerosare —4, —1, and —1 ± y/6.
55. P (x) = x5 - 3x4 - x3 + llx2 - 12x+ 4 has possible rational zeros ±1, ±2, ±4.
ill -3 -1 11 -12
1-2-3
1-2-3
8
4
8
-4
-4
0
=• x = 1 is a zero.
P (x) = x5 - 3x4 - x3 + 1lx2 - 12x + 4 = (x- 1) (x4 - 2x3 - 3x2 + 8x- 4). Continuing with the quotient, we have
ill -2 -3
8-4
1-1-4
1—1—4
4
4
0
^
x = 1 is a zero.
x5 - 3x4 - x3 + llx2 - 12x + 4 = (x- l)2 (x3 - x2 - 4x + 4) = (x- l)3 (x2 - 4)
= (x - l)3 (x - 2)(x + 2)
Therefore, the zeros are 1 (multiplicity 3), —2, and 2.
56. P(x) = x4 - 81 = (x2 - 9) (x2 + 9) = (x - 3)(x+ 3)(x2 + 9) = (x - 3)(x+ 3)(x - 3«) (x+ 3*) . Thus, the
zeros are ±3, ±3i.
292
CHAPTER3 Polynomial and Rational Functions
57. P (x) = x6 - 64 = (x3 - 8) (x3 + 8) = (x - 2) (x2 + 2x + 4) (x + 2) (x2 - 2x + 4). Now using the quadratic
formula to find the zeros of x2 + 2x + 4, we have
x=
*-£•
= ~2±^% = -1 ± y/3i, and using the quadratic formula to find the zeros ofx2 - 2x + 4, we have
x= 2±v/4-4(4)(i) _ 2±^gi = 1±^ Therefore, the zeros are 2, -2,1 ± V5i, and -1 ± y/3i.
58. P (x) = 18x3 + 3x2 - 4x - 1has possible rational zeros ±1, ±±,±J, ±±,±|, ±i
1118 3-4-1
| 18 3-4-1
18
21
17
18 21 17 16 ^> x = 1 is an upper bound.
9
6
1
18
__ 12
__
2_
. x_ = 52 is a zero.
0_ =»
So P (x) = 18x3 + 3x2 - 4x - 1 = (2x - 1) (9x2 + 6x + l) = (2x - 1) (3x + l)2. Thus the zeros are \ and -\
(multiplicity 2).
59. P (x) = 6x4 - 18x3 + 6x2 - 30x + 36 = 6(x4 - 3x3 + x2 - 5x + 6) has possible rational zeros ±1,±2,±3,±6.
116 —18
6 -30
6-12
6
-12
36
-6 -36
-6 -36
0
=» x = 1 is a zero.
So P(x) = 6x4 - 18x3 + 6x2 - 30x + 36 = (x- 1)(6x3 - 12x2 - 6x- 36) = 6(x - 1) (x3 - 2x2 - x - 6).
Continuing with the quotient we have
ill -2 -1 -6
1
-1
-2
1-1-2-8
2I1 —2 —1 —6
0
3I1 —2 —1 —6
-2
3
10-1
-1 --8
8
3
11
12
2
6
0
=>• x = 3 is a zero.
So P(x) = 6x4 - 18x3 + 6x2 - 30x + 36 = 6(x - 1) (x- 3)(x2 + x + 2). Now x2 + x + 2 = 0 when
x = ~ld:VV(1)(2) = =*%&, and so the zeros are 1,3, and =**&.
60. P (x) = x4 + 15x2 + 54 = (x2 + 9) (x2 + 6). Ifx2 = -9, then x = ±3i. Ifx2 = -6, then x = ±V6~t. Therefore, the
zeros are ±3z and ±\/6i.
61. 2x2 = 5x + 3
<&
2x2 - 5x - 3 = 0. Thesolutions
are x = -0.5, 3.
62. LetP (x) = x3 + x2 - 14x - 24. Thesolutions to
p (x) = o are x = -3, -2, and 4.
10
2
/
v"
•4
/
'
•"
•»
1
CHAPTER 3
63. x4 - 3x3 - 3x2 - 9x - 2 = 0 has solutions x
-0.24, 64. x5 = x + 3
&
x5 - x - 3 = 0. We graph
P (x) = x5 —x —3. Theonlyreal solution is 1.34.
4.24.
10
•»
\
/
•2
-1
/
/
/
'
°
1 /
2
-4
-4
"*
-18
-12
3x-12
65. r (x) = ——jp.
When x = 0, we have r (0) = ——^ = —12, so the y-intercept
is —12. Since y = 0, when 3x - 12 = 0
•*»
x = 4, the x-intercept is 4. The
vertical asymptote is x = -1. Because the degree of the denominator and
numerator are the same, the horizontal asymptote isy = | = 3.
66.
r (x) =
-J. When x = 0,we have r (0) = -z = j, sothe y-intercept is \.
(x + 2)
2
Since the numerator is 1, y never equals zero and there is no x-intercept. There is a
vertical asymptote at x = —2. The horizontal asymptote is y = 0 because the
degree ofthe denominator is greater than the degree of the numerator.
67.
r (x) =
Review
x-2
x-2
x2 - 2x - 8
(x + 2) (x - 4)'
When x = 0, we have
r (0) = Ef = \> sothe y-intercept is \. When y = 0,we have x - 2 = 0
-o-
x = 2, so the x-intercept is 2. There are vertical asymptotes at x = —2 and x = 4.
>*
293
294
68.
CHAPTER 3 Polynomial and Rational Functions
r(x) =
2x2 - 6x - 7
_
. When x = 0, we have r (0) = ^| = |, so the y-intercept
isy = |. We use the quadratic formula tofind the x-intercepts:
x= -(-,)±V(.ga.4(2)(_2 = 6^52 = 3^ ^ me x4Btmxfila ^
x « 3.9 and x « —0.9. The verticalasymptoteis x = 4. Becausethe degree of the
numerator is greater than the degree of the denominator, there is no horizontal
asymptote. By long division,we have r (x) = 2x + 2 H
-, so the slant
x —4
asymptote is r (x) = 2x + 2.
69' r(x) =2^2+T =^"tog +l 3)" whena: =0»wehaver(0) =^.sothe
y-intercept is -9. When y = 0, we have x2 - 9 = 0
-^
x = ±3 so the
x-intercepts are —3 and3. Since 2x2 + 1 > 0, thedenominator is never zeroso
there are no vertical asymptotes. The horizontal asymptote is aty = | because the
degree of the denominator and numerator are the same.
70.
r (x\ —X + 27 xiTU*.n
n ..,=
u„» „r (0)
fr\\ =_ 2Z
27 so
„^ »u^
.. :«»«^^«* is
:<. ..y =_ 27
When x = 0,
wehave
they-intercept
X+ 4
When y = 0, wehave x3 + 27 = 0
«*•
x3 = -27 =• x = -3. Thus the
x-intercept is x = —3. The vertical asymptote is x = -4. Because the degree of
the numerator is greater than the degree of the denominator, there is no horizontal
asymptote. Bylong division, we have r (x) = x2 —4x + 16 —
37
x + 4
. So the end
behavior ofy islike theend behavior ofg (x) = x2 - 4x - 16.
71. r(x) =
x-3
From the graph we see that the
2x + 6
2x — 7
72. r (x) = —2—-. From thegraph weseethat the
x
x-intercept is 3, the y-intercept is -0.5, there is a vertical
+ y
x-intercept is 3.5, the y-intercept is —0.78, there is a
asymptote at x = —3 and a horizontal asymptote at
horizontal asymptote at y = 0 and no vertical asymptote,
y = 0.5, and there is no local extremum.
the local minimum is (—1.11, —0.90),and the local
maximum is (8.11,0.12).
_J
•10
4
* , -4
1
/ ""
•»
'l S*
•1
< -u
p
•4
-4
•1
CHAPTER 3
73.
r(x) =
x3 + 8
x2-x-2'
Review
295
From the graph we see that the x-intercept is —2, the
y-intercept is —4, there are vertical asymptotes at x = —1 and x = 2 and a
horizontal asymptote at y = 0.5. r has a local maximum of (0.425, —3.599) and a
local minimumof (4.216,7.175). By using long division,we see that
/ (x) = x + 1 +
74.
r (x) =
10-x
x2 - x - 2
, so / has a slant asymptote of y = x + 1.
2x3 - x2
—. Fromthe graph we see that the x-intercepts: are 0 and 5, the
SO
\l J
y-intercept is 0, there is a vertical asymptote at x = —1, the local minimum is
40
(0.425,-3.599) and the local maxima are (-1.57,17.90) and (0.32, -0.03).
JO
Using long division, weseethat r (x) = 2x2 —3x + 3 —
So the end
x + 1
\
.* -4
2£
2
,<
«
behavior of r isthesame as theendbehavior oft?(x) = 2x2 —3x + 3.
75. The graphs of y = x4 + x2 + 24x and y = 6x3 + 20 intersect when x4 + x2 + 24x = 6x3 +20
x4 - 6x3 + x2 + 24x - 20 = 0. Thepossible rational zeros are ±1, ±2, ±4, ±5, ±10, ±20.
1 I 1 -6
1
24
-20
1
-5
-4
20
-5
-4
20
0
1
<&
=> x = 1 is a zero.
So x4 - 6x3 + x2 + 24x - 20 = (x - 1)(x3 - 5x2 - 4x+ 20) = 0. Continuing with the quotient:
1 I 1 -5
1
-4
-4
-8
20
2 I 1 -5
-4
20
-8
2
-6
-20
-3
-10
0
12
1
=*• x = 2 is a zero.
So x4 - 6x3 + x2 + 24x - 20 = (x - 1)(x - 2) (x2 - 3x- 10) = (x - 1) (x - 2) (x - 5)(x + 2) = 0. Hence, the
pointsof intersection are (1,26), (2,68), (5,770), and (-2, -28).
296
CHAPTER 3 Polynomial and Rational Functions
Chapter 3 Test
1. f(x) = - (x + 2)3 + 27has y-intercept y = -23 + 27 = 19 and x-intercept
where - (x + 2)3 =-27 <^> x = 1.
2 I 1 0 -4
2. (a)
2 5
x3 + 2x2
(b)
+
2
4
0
4
2x5 + 4x4 -- x3
12
0
2
9
2x5
-
l
2
x2
+
Ox +
7
+
7
- x3
Therefore, the quotient is
4x4
x2
Q(x) = x3 + 2x2 + 2, and theremainder is
4x4
2x2
-
x2
R (x) = 9.
x2
i
—
2
15
2
Therefore, the quotient is Q(x) = x3 + 2x2 + \ and the
remainder is R(x) = -y.
3. (a) Possible rational zeros are: ±1, ±3, ±\, ±|.
(b)
-12
-5
-4
3
-2
7
-3
2
-7
x = —1 is a zero.
P (x) = (x + 1) (2x2 - 7x+ 3) = (x + 1) (2x - 1) (x - 3)
= 2(x + l)(x- §)(x-3)
(c) The zeros ofP are x = —1, 3, |.
4. (a) (3 - 2i) + (4 + 2,i) = (3 + 4) + (-2 + 3) i = 7 + i
(b) (3 - 20 - (4 + 3i) = (3 - 4) + (-2 - 3) z = -1 - 5/
(c) (3 - 20 (4 + 30 = 12 + 9i - 8i - 6r = 12 + i - 6 (-1) = 18 + i
(d)
3 - 2i _ (3 - 2Q (4 - 3Q _ 12- 17/ - (ir _ ± _ n
4 + 3i
(4 + 30 (4 - 30
16 - 9/2
**
(e) /'8 = (i2)24 = (-D24 = 1
(f) \/2-
5) (V§ + 7=2) = (x/2 - i\/2) (V§ + i\/2) = v^-s/§ + z(\/2) - i%/2\/8 - r (v/2)"
= \/T6 + 2/ - \/T6i + 2 = 6-2i
CHAPTER 3
Test
297
5. P (x) = x3 —x2 - 4x - 6. Possible rational zeros are: ±1, ±2, ±3, ±6.
ill -1 -4 -6
0
211 —1 —4 —6
2
-4
1 0 - 4 -10
1
311 —1 —4 —6
-4
1 - 2 -10
12
6
6
2
0 =* x = 3 is a zero.
So P(x) = (x —3) (x2 + 2x+ 2). Using the quadratic formula on the second factor, we have
-2 ± s/22 - 4 (1)(2)
-2±y/=4 _ -2±2y^T
2(1)
= -1 ± i. So zeros of P (x) are 3, —1 —i, and —1 + i.
6. P (x) = x4 —2x3 + 5x2 —8x + 4. The possible rational zeros of P are: ±1, ±2, and ±4. Since there arefour changes in
sign, P has 4, 2, or 0 positive real zeros.
1 I 1 -2
5-8
1
-1
1-1
4
4
-4
4-4
0
So P(x) = (x —1)(x3 —x2 + 4x—4). Factoring the second factor by grouping, we have
P (x) = (x - 1) [x2 (x- 1) + 4(x- 1)] = (x- 1) (x2 + 4) (x- 1) = (x- l)2 (x- 2t) (x+ 2i).
7. Since 3i is a zero of P(x), —3i is also a zero of P(x). And since —1 is a zero of multiplicity 2,
P (x) = (x+ l)2 (x- 3i) (x+ 3i) = (x2 + 2x + l) (x2 + 9) = x4 + 2x3 + 10x2 + 18x + 9.
8. P (x) = 2x4 - 7x3 + x2 - 18x+ 3.
(a) Since P (x) has 4 variations in sign, P (x) can have 4, 2, or 0 positive real zeros. Since
P (—x) = 2x4 + 7x3 + x2 + 18x+ 3 has novariations insign, there are no negative real zeros.
(b)
2
2
-7
18
8
20
8
2
11
15
Since the lastrowcontains no negative entry, 4 is an upper bound for the realzerosof P (x).
-1 I 2 -7
2
1 -18
3
-1
9
-10
28
-9
10
-28
31
Sincethe last row alternates in sign, -1 is a lowerbound for the realzeros of P (x).
(c) Usingthe upperand lowerlimit from part (b), we
(d) Local minimum (2.8, -70.3).
graphP (x) in the viewing rectangle [—1,4] by
\
[—1,1]. The two real zeros are 0.17 and 3.93.
*"
\
\
M
*°
X?
OS
•2
"
•20
-W
02
•60
*
•42
•Ob
2
-80 J
^V
*
&
CHAPTER 3 Polynomial and Rational Functions
298
9. r (x) =
2x
1
, .
X2 —X - 2'" ^
x3+27 ., .
x
X2+4
xxTi
+ 2
and u (x) =
x2 + x - 6
x—*u
c2 - 25
(a) r (x) hasthe horizontal asymptote y = 0 because thedegree of the denominator is greater thanthe degree of the
numerator, u (x) has the horizontal asymptote y = \ = 1because the degree ofthe numerator and the denominator are
the same.
(b) Thedegree of thenumerator of s (x) is onemore than thedegree of thedenominator, so s hasa slantasymptote.
(c) The denominator of s (x) is never0, so s has no vertical asymptote.
m\ i \
z2+x-6
(x+ 3)(x-2) •
(d) «(*) = x2_25 = (x_5)(g +5)- W1™* =<>>behave
u(x) = fjjg = 2g, so the y-intercept is y = ^. When y = 0, we have x = -3 or
x = 2, so the x-intercepts are —3 and 2. The vertical asymptotes are x = —5 and
x = 5. The horizontal asymptote occurs aty = \ = 1 because the degree ofthe
denominator and numerator are the same.
xl -
(e)
x + 2
2x -
5
x3 + Ox2 - 9x +
0
x3 + 2x2
-2x2 - 9x
-2x2 - 4x
-5x +
0
-5x -
10
-10
Thus P (x) = x2 - 2x - 5 and t (x) =
x6 -9x
x + 2
have the same end behavior.