## anonymous 3 years ago The graph of y=ax^(2)+bx+c is given as shown. Which of the following is / are true? I. a<0 II. b<0 III. c<0 A. I only B. I and II only C. I and III only D. II and III only E. I, II and III

1. anonymous

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2. Kira_Yamato

(C) c < 0 as graph cuts y-axis below the x-axis a < 0 because graph is frowning curve

3. anonymous

E) inverted U shape meaning a<0 intersecting y axis at negative means C is less than 0 c<0 and also there is shift along x axis in right direction means b<0

4. Kira_Yamato

Overlooked the detail for b<0... Sorry

5. anonymous

I am confusing why b>0 is correct

6. anonymous

the line of symmetry x = -b/2a, is a positive value, and we know a < 0 so b > 0

7. anonymous

I am sorry that I type it wrongly. It should be : "I am confusing why b<0 is correct"

8. anonymous

b > 0 is correct.

9. anonymous

why

10. anonymous

a<0 because parabola opens downward

11. anonymous

I know a<0 and c<0, I just don't know why b<0 is correct

12. anonymous

b < 0 is NOT correct.

13. anonymous

14. anonymous

yes. a < 0 and c < 0.

15. anonymous

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16. anonymous

@hashsam1 the line of symmetry is the line x = -b/2a. according to the graph it is to the right of the y-axis (so the sign of -b/2a is positive). we know that a < 0, so if b < 0 then the sign of -b/2a is -(-)/2(-) is negative.

17. phi

a simple example with 2 positive roots x=1 and x=2 the polynomial would be -(x-1)(x-2) = -x^2 +3x -2 this suggests a<0 b>0 c<0

18. anonymous

I understand now, thank you so much :)

19. anonymous

another question If f(x)=x-(1/x), then f(x)-f(1/x)= ?

20. anonymous

oops yes right i over looked the equation thinking its just a normal quadratic equation.

21. phi

If f(x)=x-(1/x), then f(x)-f(1/x)= ? replace f(x) with x- 1/x replace f(1/x) with 1/x - x

22. anonymous

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23. anonymous

yes

24. anonymous

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25. anonymous

yup, that's right.

26. anonymous

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27. anonymous

1/(1/x) = x. try again.

28. anonymous

oh -.-|dw:1352552602486:dw|

29. anonymous

you still need to combine x + x

30. anonymous

2x-(2/x)

31. anonymous

that's right!

32. anonymous

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33. anonymous

i'll accept either form.

34. anonymous

another question If f(x)=10^(2x), then f(4y)=?

35. anonymous

A. 10^(4y). B. 10^(2+4y) C. 10^(8y) D. 40^(y) E. 40^(2y)

36. anonymous

replace x by 4y, what do you get?

37. anonymous

B ?

38. anonymous

C it is.

39. anonymous

oh yes

40. anonymous

$\huge 10^{2(4y)}$

41. anonymous

f(x)=10^(2x) f(x)=10^(2(4y))

42. anonymous

If f(x)=x/(1-x), then f(1/x)f(-x)=?

43. anonymous

@gerryliyana

44. anonymous

45. anonymous

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46. anonymous

$\huge f(-x) = \frac{ (-x) }{ 1-(-x) }$

47. anonymous

errrr.....wait

48. anonymous

If f(x)=x/(1-x), then f(1/x)f(-x)= f(1/x) = (1/x)/(1-(1/x)) = 1/x((x-1)/x) f(-x) =(-x)/(1-(-x)) = -x/(1+x) f(1/x)f(-x) = 1/x((x-1)/x) times -x/(1+x) f(1/x)f(-x) = -x/(x((x-1)/x)(1+x))

49. anonymous

o.o errrr.... wait a moment

50. anonymous

f(1/x)f(-x) = -x/(x((x-1)/x)(1+x)) f(1/x)f(-x) = -x/((x-1)(x+1))

51. anonymous

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52. anonymous

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53. anonymous

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54. anonymous

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55. anonymous

is it necessary for you to give me a big cross....... I got it :)

56. anonymous

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57. anonymous

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58. anonymous

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59. anonymous

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60. anonymous

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61. anonymous

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62. anonymous

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63. anonymous

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64. anonymous

we're not discussing' we're exchanging notes. hahaha

65. anonymous

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66. anonymous

haha :))

67. anonymous

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68. anonymous

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69. anonymous

what's the answer to your problem $\huge f(\frac{ 1 }{ x })f(-x)$

70. anonymous

oh my god

71. anonymous

A. -1/2 B. -1 C. -(1-x)/(1+x) D. x/(1-x^(2)) E. x/(x^(2)-1) no answer since this is past paper, my teacher haven't give us the answers yet

72. anonymous

I still cannot get the answer

73. Callisto

Please start a new post for a new question. If possible, please also put them in the right categories (subtopics). Thanks!

74. anonymous

$\large f(1/x)f(-x) = \frac{ 1/x }{ 1-1/x }\frac{ (-x) }{ 1-(-x) }$ $\large = \frac{ -1 }{ 1+x -\frac{ 1 }{ x }-1 } \times \frac{ -x }{ -x }$

75. anonymous

so complicated....

76. anonymous

just carry out the multiplication and you'll get (D)

77. anonymous

|dw:1352556469608:dw| I get this only...

78. anonymous

f(1/x)f(-x) = -x/(x((x-1)/x)(1+x)) f(1/x)f(-x) = -x/((x-1)(x+1)) f(1/x)f(-x) = -x/(x^2 -1) f(1/x)f(-x) = -x/-(-x^2 +1) f(1/x)f(-x) = x/(1-x^2) the answer is D x/(1-x^2)

79. anonymous

you got it @kryton1212 ???? take it a simply way..,hehe

80. anonymous

confusing................. I think it later another question If a,b and c are all positive, which of the following may represent the graph of ax+by+c=0?|dw:1352556954406:dw|

81. anonymous

@hartnn would you kindly help me?

82. Callisto

@kryton1212 Please start a new post for a new question!

83. anonymous

the problem is I cannot draw picture in a new post

84. hartnn

ok, let this be last question discussed in this post, agreed @kryton1212 ? for this u know how to find intercepts ?

85. Callisto

You can post the question a draw it in the comment

86. anonymous

@hartnn that's ok. c=y-intecept

87. hartnn

nopes, c is y intercept in y=mx+c not here

88. hartnn

to get x intercept, put y=0 in ax+by+c=0 to get y intercept, put x=0 in ax+by+c=0

89. hartnn

now tell me what intercepts u get

90. anonymous

....oh yes. I don't know how to find the intercept in these gaphs

91. hartnn

to get x intercept, put y=0 to get y intercept, put x=0 in any graphs

92. anonymous

ax+by+c=0 put y=0??? ax+c=0?

93. hartnn

yes, thats right ax+c=0 x= ?

94. anonymous

x=-c/a?

95. hartnn

absolutely correct! now -c/a is positive or negative ?

96. hartnn

given a and c are positive....

97. anonymous

-ve

98. hartnn

so your x-intercept is negative, that is it will cut x axis in negative (left) side got this ? what about y-intercept ???

99. anonymous

so it may be B or E. y=-c/b is it E?

100. hartnn

yes, y intercept is also negative, so it will cut down. and yes its E clear as to how ?

101. anonymous

clear now, thank you

102. hartnn

make sure u ask next question in new post...

103. phi

@kryton1212 it would be better if you made each question a separate post.

104. anonymous

okok