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ISM: Precalculus Chapter 2

Chapter 2 Review Exercises 9. (-2+../-100)2 = (-2 +iMo)2

1. (8-3i)-(17-7i) =8-3i-17+7i = (-2 + 1Oi)2=-9 + 4i = 4 - 40i + (10i)2

2. 4i(3i- 2) = (4i)(3i) + (4i)( -2) = 4-40i-100mes

=12i2-8i = -96-40i

: of = -12-8i10. 4+~ 4+ i.J8 4+2i.fi

= 2+i.fi=---=3. (7- i)(2 + 3i) 2 2 2

= 7 .2 + 7(3i) + (-i)(2) + (-i)(3i) 11. 2 -2x+4 =0x= 14+ 21i - 2i + 3

x= -<-2)±j(-2)2_4(1)(4)= 17+ 19i 2(1)

4. (3-4;)2 =32 + 2.3(-4i) + (-4i)2 2±~4-16=9-24i-16 x=

2= -7 -24i 2±~-12

x=5. (7+8i)(7- 8i) = 72+ 82= 49+64 = 113 2

2± 2i/3ie 6 6 5-i x=6. 2-=_.--

x=1±rl3S+i 5+i 5-i30-6i The solution set is ~-i,l3, 1+ ~}=--

at 25+1L 30-6i 2-- 12. 2x -6x+5 =026f 15-3i -(-6) ± ~ (_6)2 - 4(2)(5)-- x=13 2(2)

15 3 . 6±"/36-40=---113 13 x=

4

3+4; 3+4i 4+2i6 ±.,J-4

7. X=--=--.-- 44-2i 4-2i 4+2i12+ 6i + 16i+ 8;2 6±2i

X=--16-4i2 4

12+22i-8 x =~± 2i

16+44 4

4+22i =~±.!.i-- 2 2

20Th luti . {3 1. 3 1-}1 11. e so ution set IS - - -1, 2'+2'1.=-+-1 2 2

5 10

8. ./-32 -../-18 = i.Jfi -i.J18 13• f(x)=-(X+l)2+4

= iJ}'6:i - im vertex: (-1,4)

=4i.fi - 3i.fix-intercepts:0=_(X+1)2 +4

= (4i - 3i).fi(x+ 1)2= 4

= i.fix+l=±2

x=-1±2

387

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Chapter 2 ISM: Precalculus

x = -3 or x = 1y-intercept:

1(0)=-(0+1)2+4=3The axis of symmetry is x = -1.

y

(~(. x

16./(x)=2x2-4x-6

1 (x) = 2 (X2- 2x +1) - 6 - 2

2(x-l)2 -8y

)x

)f(x) = -(x + 1)2 + 4 f(x) = 2x2 - 4x - 6

axis of symmetry: x = 1Domain: (-00,00) Range: [-8,(0)

17. l(x)=-x2+14x-106

Domain: (-00,00) Range: (-00,4]

14. f(x) = (x + 4l-2vertex: (--4, -2)x-intercepts:

0= (x+4)2 ~2

(x+4)2 = 2

x+4 =±.fi

x=-4±.fiy-intercept:

1(0)=(0+4)2 -2=14 =-1

a. Since a < 0 the parabola opens down withthe maximum value occurring at

b 14x=--=---=7.

2a 2(-1)The maximum value is 1(7).

1(7) = _(7)2 + 14(7) -106 = -57

h. Domain: (-00,00) Range: (-00, -57]

y

0x

18. l(x)=2x2+12x+703

a. Since a > 0 the parabola opens up with theminimum value occurring at

t:

(-4 )

(

uare yards 24. Y = (35 + x)(150 - 4x)length is 2y = 5250+ lOx -4x

-b -10 5x=- =-- =- =1.25 or 1 tree2a 2(-4) 4

The maximum number of trees should be 35 + 1= 36 trees.

maximum number of trees should be 35 + 1= 36trees.

y = 36(150 - 4x) = 36(150 - 4'1) = 5256The maximum yield will be 5256 pounds.

25. I (x) = _x3 + 12x2- x

The graph rises to the left and falls to the right

)= 7. and goes through the origin, so graph (c) is the

:luct are 7best match.

·7=-49. 26. g(x) = x6-6X4 +9x2

The graph rises to the left and rises to the right,so graph (b) is the best match.

27. h(x) = x5 -5x3 +4xThe graph falls to the left and rises to the rightand crosses the y-axis at zero, so graph (a) is thebest match.

28. I(x) = _X4 +1fly) f'o:lIll(" +,... +\... ...•.1~.c.._.- 1 •

30. In the polynomial, f{x) = _X4 +21x2 +100,the leading coefficient is -1 and the degree is 4.Applying the Leading Coefficient Test, weknow that even-degree polynomials withnegative leading coefficient will fall to the leftand to the right. Since the graph falls to theright, we know that the elk population will dieout over time.

31. f(x) = -2(x-l)(x+2)2(x+5)lX = 1, multiplicity 1, the graph crosses the x-axisx = -2, multiplicity 2, the graph touches thex-axisx = -5, multiplicity 5, the graph crosses thex-axis

32. f(x) = x3- 5x2 - 25x + 125

= X2(X -5) -25(x- 5)

= (x2 -25)(x-5)

= (x+5)(x-5)2x = -5, multiplicity 1, the graph crosses the x-axisx = 5, multiplicity 2, the graph touches the x-axis

33. f (x) = Xl - 2x -1

f(1) = (1)3- 2(1) -1 = -2

f(2) = (2)3 - 2(2) -1 = 3The sign change shows there is a zero betweenthe given values.

•I(x) = x3 - xl

35. f(x) = 4x-x3

a. Since n is odd ato the left and fa

b. f(-x)=-4x+Jf(-x)=-f(x)origin symmetry

c. f(x) = x(x2 -4)zeros: x = 0, 2, -:

y

I(x) = 4x - x3

36. f(x) = 2Xl +3x2 -8x-

a. Since h is odd andto the left and rises

b. fe-x) = -2x3 +3x:

ts ISM: Precalculus Chapter 2

37. g(x)=-x4+2SX2 39. f (x) = 3x4 -ISx3

a. The graph falls to the left and to the right. a. The graph rises to the left and to the right.

b. fe-x) = _(_X)4 +2S(_X)2

= _X4 +2Sx2 = f(x)

y-axis symmetry

b. f( -x) = 3(-xt -IS( _X)2 = 3x4 + ISx3

fe-x) "'-f(x), fe-x) "'-- f(x)no symmetry

C. _X4 +2Sx2 = 0

_X2 (X2 -2S) = 0

_X2 (x-S)(x+s) = 0

zeros: x = -S, 0, S

C 3x4 -ISx3 = 0

3x3 (x-S) = 0zeros: x = 0, S

y

y

x

_xf(x) = 3xA - ISi3

40. f ( x ) = 2X2 ( x_I)3 (x + 2)Since an > 0 and n is even.jix) rises to the leftand the right.x = 0, x = 1, x =-2The zeros at 1 and -2 have odd multiplicity sofix) crosses the x-axis at those points. The root at°has even multiplicity so fix) touches the axis at(0,0)

f(O) = 2(0)2 (0 _1)3 (0 + 2) = 0

The y-intercept is 0.

f(x) = -xA + 2sx2

a. The graph falls to the left and to the right.

b. fe-x) = -(-xt +6(-X)3 -9(-x)

= _X4 - 6x3 - 9x2 f( -x) "'-f(x)

f(-x)"'--f(x)no symmetry

c. =_X2(X2-6x+9)=0

_X2 (x-3}(x-3) = 0

zeros: x = 0, 3

y

xy

x f(x) = 2r (x - 1)3(x + 2)

41. f(X) = _x3 (X+4)2 (x-I)

Since an < 0 and n is even, fix) falls to the leftand the right.x = 0, x = -4, x = 1The roots at ° and 1 have odd multiplicity so fix)crosses the x-axis at those points. The root at -4has even multiplicity so fix) touches the axis at(-4,0)

f(O) = _(0)3 (0+4)2 (0-1) = 0

I(x) = - x4 + (jx3 - 9x2

391

02007Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.