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## anonymous 5 years ago use algebraic methods to determine the domain and range for the function from to defined by the given equation. a. x^2 – 6x –y =0 b. 4x^2 – xy +1 = 0 c. 5x^2(y) – x + 2 = 0

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1. anonymous

use algebraic methods to determine the domain and range for the function from x to y defined by the given equation. a. x^2 – 6x –y =0 b. 4x^2 – xy +1 = 0 c. 5x^2(y) – x + 2 = 0

2. myininaya

(a) x^2-6x-y=0 we can solve this for x by using the quadratic formula $x=\frac{6 \pm \sqrt{36-4(-y)}}{2}=\frac{6 \pm \sqrt{36+4y}}{2}$

3. myininaya

what does y have to be in order for x to exist?

4. anonymous

I am not sure? this question is confusing

5. myininaya

can you have a negative under the square root?

6. anonymous

no

7. myininaya

you want 36+4y to be bigger than or equal to zero, correct?

8. anonymous

yes correct

9. myininaya

so solve this for y: 36+4y>=0 and you will have your range

10. anonymous

-9 >= y

11. anonymous

how do i write the range

12. myininaya

4y>=-36 so y>=-9

13. anonymous

what is the domain?

14. myininaya

solve for y

15. myininaya

to find range we solve for x to find domain we solve for y so solve the orignal equation for y

16. anonymous

so y >= -9 is the range or domain?

17. anonymous

are you still here?

18. anonymous

i think you are supposed to solve each of these for y and think of y as a function of x

19. anonymous

but there is something wrong here because you have these each set = 0

20. anonymous

that is what my project say

21. myininaya

y is range we got y>=-9

22. anonymous

ok fine. so first one is $x^2-6x-y=0$ so $y=x^2-6x$ as a function $f(x)=x^2-6x$ domain all real numbers since it is a polynomial range is -9 to infinity

23. myininaya

no we find domain solve x^2-6x-y=0 for x and you will see the domain is all real numbers because x^2-6x is a polynomial

24. myininaya

now*

25. myininaya

not no lol

26. anonymous

LOL u guys are great help

27. anonymous

second one. $4x^2-xy+1=0$ $xy=4x^2+1$ $y=\frac{4x^2+1}{x}$ domain all real numbers except 0

28. myininaya

all you have to do is solve both for x and y when you solve for x you can find the range when you solve for y you can find the domain

29. anonymous

so the second one does not have a range

30. anonymous

range is also all real numbers except 0 because the numerator is $4x^2+1$ which cannot be 0

31. anonymous

no, range is just the possible y values. y can be anything except 0 since the numerator of your fraction is never 0

32. myininaya

range is [-4,4] you find once you solve for x

33. anonymous

oops really? i will try it.

34. myininaya

oops o did somehting wrong its (-inf,-4]U[4,inf)

35. anonymous

is this for the second problem

36. myininaya

the problem that we are now we are finding range

37. anonymous

how do you like that. you get $y^2-16$under the radical so $y^2-16\geq 0$ who knew?

38. myininaya

lol satellitle

39. myininaya

you remember all that inverse crap right?

40. anonymous

oh of course. this thing is a rational function with a slant asymptote. so it does skip values, never between -4 and 4. must be late!

41. anonymous

so the range is (-inf, -4)U(4,inf)

42. anonymous

is the last one $5x^2y-x+2=0$?

43. anonymous

yes you have the range correct.

44. anonymous

ok

45. anonymous

if i have the last one correct is is $y=\frac{x-2}{5x^2}$ domain all numbers except 0

46. anonymous

and the range?

47. anonymous

honestly i am little confused by this. myininaya said solve for x. i can try it.

48. anonymous

oh ok!

49. anonymous

if i use the quadratic formula i see under the radical $1-20y$ meaning $1-20y\geq 0$ $y\leq \frac{1}{20}$ would held if i had the picture of this thing to see if that makes sense.

50. anonymous

so i think i will stick to that answer, and go to bed.

51. myininaya

you want me to check your range? lol which one is this?

52. myininaya

ok the last one

53. anonymous

last one. i have confused myself.

54. myininaya

if i didnt make a mistake here is the range

55. anonymous

i must be tired if i cannot multiply 4 by 5 by 2! i best retire.

56. myininaya

yeah you were close then we also have y in the denominator and we know we cannot divide by 0 and when y is 0 the denominator is 0 so our range does not include y=0

57. anonymous

ok maybe but i am still skeptical. what do you get if x = 2?

58. myininaya

hmm you get (2,0)

59. anonymous

looks like if x = 2, then y = 0 original equation is $5x^2y-x+2=0$ replace x by 2 get $20y-2+2=0$ $20y=0$ $y=0$

60. anonymous

i should switch back to windows and run maple, graph the thing and then i would be more convinced.

61. myininaya

well and also the quadratic formula just failes when a=0 in this case when 5yx^2=0 since y=0 a=0 so the quadratic formula won't work for that part of the answer

62. myininaya

so the range is (-inf,1/40]

63. anonymous

i am afraid i do not believe that either.

64. myininaya

i believe in it lol

65. anonymous

as x goes to infinity $\frac{x-2}{5x^2}$ goes to zero

66. myininaya

right there is a horizontal asymptote

67. anonymous

let x = 1000000

68. anonymous

bet you get something closer to 0 than 1/40

69. anonymous

oh wait. maybe it is always negative. oops. ok possible

70. anonymous

ok i am beginning to believe it. may be always less than 1/40. tomorrow i graph !

71. myininaya

hey satelittle you know what would really convince you? i do

72. anonymous

maple

73. myininaya

you know how to find the maximum right? that derivative crap?

74. anonymous

ok here we go.

75. myininaya

so you do :) yay!

76. myininaya

f'(x)=[5x^2-10x(x-2)]/[(5x^2)]^2=[-5x^2+20x]/(25x^4) a critical numbers isfour f(4)=1/40

77. anonymous

well it is nice and late and i have had a few beers. but it looks like extreme value at 0 and 4

78. myininaya

well 0 doesn't exist in the orignal function

79. anonymous

0 is out, 4 gives.... holy carp, what do you know.

80. myininaya

lol

81. anonymous

learn something new every day. thanks.

82. myininaya

who needs maple

83. anonymous

me. use it all the time when i am not running linux

84. myininaya

well it is a nice friend to have

85. anonymous

and so to bed, before i make more of a fool of myself. take care

86. myininaya

later

87. myininaya

i'm going too gn

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