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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
    • 5 years ago
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    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
    • 5 years ago
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    (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
    • 5 years ago
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    what does y have to be in order for x to exist?

  4. anonymous
    • 5 years ago
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    I am not sure? this question is confusing

  5. myininaya
    • 5 years ago
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    can you have a negative under the square root?

  6. anonymous
    • 5 years ago
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    no

  7. myininaya
    • 5 years ago
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    you want 36+4y to be bigger than or equal to zero, correct?

  8. anonymous
    • 5 years ago
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    yes correct

  9. myininaya
    • 5 years ago
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    so solve this for y: 36+4y>=0 and you will have your range

  10. anonymous
    • 5 years ago
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    -9 >= y

  11. anonymous
    • 5 years ago
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    how do i write the range

  12. myininaya
    • 5 years ago
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    4y>=-36 so y>=-9

  13. anonymous
    • 5 years ago
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    what is the domain?

  14. myininaya
    • 5 years ago
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    solve for y

  15. myininaya
    • 5 years ago
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    to find range we solve for x to find domain we solve for y so solve the orignal equation for y

  16. anonymous
    • 5 years ago
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    so y >= -9 is the range or domain?

  17. anonymous
    • 5 years ago
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    are you still here?

  18. anonymous
    • 5 years ago
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    i think you are supposed to solve each of these for y and think of y as a function of x

  19. anonymous
    • 5 years ago
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    but there is something wrong here because you have these each set = 0

  20. anonymous
    • 5 years ago
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    that is what my project say

  21. myininaya
    • 5 years ago
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    y is range we got y>=-9

  22. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    now*

  25. myininaya
    • 5 years ago
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    not no lol

  26. anonymous
    • 5 years ago
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    LOL u guys are great help

  27. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    so the second one does not have a range

  30. anonymous
    • 5 years ago
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    range is also all real numbers except 0 because the numerator is \[4x^2+1\] which cannot be 0

  31. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    range is [-4,4] you find once you solve for x

  33. anonymous
    • 5 years ago
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    oops really? i will try it.

  34. myininaya
    • 5 years ago
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    oops o did somehting wrong its (-inf,-4]U[4,inf)

  35. anonymous
    • 5 years ago
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    is this for the second problem

  36. myininaya
    • 5 years ago
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    the problem that we are now we are finding range

  37. anonymous
    • 5 years ago
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    how do you like that. you get \[y^2-16\]under the radical so \[y^2-16\geq 0\] who knew?

  38. myininaya
    • 5 years ago
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    lol satellitle

  39. myininaya
    • 5 years ago
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    you remember all that inverse crap right?

  40. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    so the range is (-inf, -4)U(4,inf)

  42. anonymous
    • 5 years ago
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    is the last one \[5x^2y-x+2=0\]?

  43. anonymous
    • 5 years ago
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    yes you have the range correct.

  44. anonymous
    • 5 years ago
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    ok

  45. anonymous
    • 5 years ago
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    if i have the last one correct is is \[y=\frac{x-2}{5x^2}\] domain all numbers except 0

  46. anonymous
    • 5 years ago
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    and the range?

  47. anonymous
    • 5 years ago
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    honestly i am little confused by this. myininaya said solve for x. i can try it.

  48. anonymous
    • 5 years ago
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    oh ok!

  49. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    so i think i will stick to that answer, and go to bed.

  51. myininaya
    • 5 years ago
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    you want me to check your range? lol which one is this?

  52. myininaya
    • 5 years ago
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    ok the last one

  53. anonymous
    • 5 years ago
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    last one. i have confused myself.

  54. myininaya
    • 5 years ago
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    if i didnt make a mistake here is the range

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  55. anonymous
    • 5 years ago
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    i must be tired if i cannot multiply 4 by 5 by 2! i best retire.

  56. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    ok maybe but i am still skeptical. what do you get if x = 2?

  58. myininaya
    • 5 years ago
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    hmm you get (2,0)

  59. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    i should switch back to windows and run maple, graph the thing and then i would be more convinced.

  61. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    so the range is (-inf,1/40]

  63. anonymous
    • 5 years ago
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    i am afraid i do not believe that either.

  64. myininaya
    • 5 years ago
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    i believe in it lol

  65. anonymous
    • 5 years ago
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    as x goes to infinity \[\frac{x-2}{5x^2}\] goes to zero

  66. myininaya
    • 5 years ago
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    right there is a horizontal asymptote

  67. anonymous
    • 5 years ago
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    let x = 1000000

  68. anonymous
    • 5 years ago
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    bet you get something closer to 0 than 1/40

  69. anonymous
    • 5 years ago
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    oh wait. maybe it is always negative. oops. ok possible

  70. anonymous
    • 5 years ago
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    ok i am beginning to believe it. may be always less than 1/40. tomorrow i graph !

  71. myininaya
    • 5 years ago
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    hey satelittle you know what would really convince you? i do

  72. anonymous
    • 5 years ago
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    maple

  73. myininaya
    • 5 years ago
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    you know how to find the maximum right? that derivative crap?

  74. anonymous
    • 5 years ago
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    ok here we go.

  75. myininaya
    • 5 years ago
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    so you do :) yay!

  76. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    well 0 doesn't exist in the orignal function

  79. anonymous
    • 5 years ago
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    0 is out, 4 gives.... holy carp, what do you know.

  80. myininaya
    • 5 years ago
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    lol

  81. anonymous
    • 5 years ago
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    learn something new every day. thanks.

  82. myininaya
    • 5 years ago
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    who needs maple

  83. anonymous
    • 5 years ago
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    me. use it all the time when i am not running linux

  84. myininaya
    • 5 years ago
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    well it is a nice friend to have

  85. anonymous
    • 5 years ago
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    and so to bed, before i make more of a fool of myself. take care

  86. myininaya
    • 5 years ago
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    later

  87. myininaya
    • 5 years ago
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    i'm going too gn

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