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anonymous

  • one year ago

Find the equation of the tangent line to f(x) = 9 − x^2 through the point (5, 0) that touches f (x) at a point below the x-axis.

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  1. dan815
    • one year ago
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    first you need to find the slope at this point

  2. anonymous
    • one year ago
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    -2x ?

  3. dan815
    • one year ago
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    that is the function of all the slopes wrt to x

  4. dan815
    • one year ago
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    f(x) is telling you all the heights at every x f'(x) is teelling you all the slopes at every x now at what x do we need to know our f'(x)

  5. anonymous
    • one year ago
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    the point where it touches f(x) which is below the x axis?

  6. dan815
    • one year ago
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    take a look at your point

  7. dan815
    • one year ago
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    (5, 0)

  8. dan815
    • one year ago
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    whats the x value of the point

  9. anonymous
    • one year ago
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    5

  10. dan815
    • one year ago
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    right so we need to know f'(5) = ?

  11. anonymous
    • one year ago
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    But that point doesn't even lie on the curve...

  12. dan815
    • one year ago
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    we just want the slope there at x=5

  13. anonymous
    • one year ago
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    so that would be -2(5) = -10, right?

  14. dan815
    • one year ago
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    oh wait 5,0 isnt on this curve really?

  15. dan815
    • one year ago
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    u are right lol xD

  16. anonymous
    • one year ago
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    Yeah cause try plugging it in the function

  17. anonymous
    • one year ago
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    XD

  18. anonymous
    • one year ago
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    I think we should first find a point that apparently lies somewhere under the x-axis

  19. dan815
    • one year ago
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    k lets graph and see what exactly is going on here

  20. dan815
    • one year ago
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    |dw:1435217659239:dw|

  21. dan815
    • one year ago
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    |dw:1435217751709:dw|

  22. dan815
    • one year ago
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    okay so eventually this has to happen, they want us to find this x,y

  23. anonymous
    • one year ago
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    Yep!

  24. anonymous
    • one year ago
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    So are we solving for the secant line?

  25. anonymous
    • one year ago
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    between (5,0) and (a, 9-a^2)

  26. dan815
    • one year ago
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    you already got that the slope was F'(x)=-2x

  27. dan815
    • one year ago
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    right

  28. anonymous
    • one year ago
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    Right

  29. dan815
    • one year ago
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    now write the slope with those 2 points

  30. ganeshie8
    • one year ago
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    Any line through \((5,0)\) is of form \[y=m(x-5)\tag{1}\] we want that line to "just touch" below parabola \[y=9-x^2\tag{2}\] simply set both equations equal to each other and use the discriminant to find the slope, \(m\)

  31. dan815
    • one year ago
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    and equate it to the F'(x)

  32. dan815
    • one year ago
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    |dw:1435217973229:dw|

  33. anonymous
    • one year ago
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    Yeah lol i was typing the same xD

  34. anonymous
    • one year ago
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    We get a quadratic equation when we solve for a... that means we'll get 2 values for a...

  35. anonymous
    • one year ago
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    a^2 - 10a + 9 = 0

  36. anonymous
    • one year ago
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    thats a 9 :P

  37. dan815
    • one year ago
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    yes

  38. ganeshie8
    • one year ago
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    Yes, two tangent lines can be drawn to a parabola form an external point

  39. anonymous
    • one year ago
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    lol xD ok so a=1 and x=9 ? Is that right?

  40. anonymous
    • one year ago
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    i mean a* = 9

  41. ganeshie8
    • one year ago
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    |dw:1435218241639:dw|

  42. ganeshie8
    • one year ago
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    a=9 is right, but how did u figure ?

  43. anonymous
    • one year ago
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    but neither of the 2 points i found for 'a' lie under the x-axis

  44. anonymous
    • one year ago
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    I solved the quadratic a^2 -10a +9

  45. ganeshie8
    • one year ago
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    okay i see you simply discarded a=1 without any reason

  46. anonymous
    • one year ago
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    nope i wrote that as well lol ^^ check again :P

  47. anonymous
    • one year ago
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    What happens next?

  48. ganeshie8
    • one year ago
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    Okay just making sure :) plugin x=a=9 into the equation, you do get a point under the x axis

  49. anonymous
    • one year ago
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    I do? Ohh thats right! xD

  50. anonymous
    • one year ago
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    So that means I get 2 tangent lines?

  51. ganeshie8
    • one year ago
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    plugin x=a=1, you get y = 9-x^2 = 9-1 = 8 that means the tangent line touches the parabola at (1, 8) which is above x axis so discard this

  52. anonymous
    • one year ago
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    Ohh nice!!

  53. anonymous
    • one year ago
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    I love discarding stuff in math xD haha

  54. ganeshie8
    • one year ago
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    lol we're very much alike :P

  55. dan815
    • one year ago
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    i am not following exactly what u are doing here ganeshie y=m(x-5) y=9-x^2

  56. dan815
    • one year ago
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    are u then setting m=-2x to solve it

  57. dan815
    • one year ago
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    arent there multiple solutions to inf solutions to m that intersect with this poiint on the parabola, how are you limiting to the case where its just touching

  58. ganeshie8
    • one year ago
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    nope set both equations equal to each other and use discriminant : m(x-5) = 9-x^2 x^2 + mx - (5m+9) = 0 For this quadratic to have a single solution, the discriminant must be 0 : m^2 + 4(5m+9) = 0

  59. ganeshie8
    • one year ago
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    solving that gives the slope of two tangent lines passing thru (5, 0)

  60. dan815
    • one year ago
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    oh i see! what you mean i didnt understand what discriminant meant lol

  61. dan815
    • one year ago
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    nice trick

  62. anonymous
    • one year ago
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    neither did I o.o XD lol

  63. Astrophysics
    • one year ago
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    Lol that had me confused to but I was too scared to ask, nice one.

  64. ganeshie8
    • one year ago
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    thats okay, stick to dan's method as it is more calculussy

  65. Astrophysics
    • one year ago
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    I feel like an idiot now xD

  66. anonymous
    • one year ago
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    No I totally understood everything we did!! I just didn't get what you were explaining to dan at the end xP

  67. anonymous
    • one year ago
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    @Astrophysics pshhhh in that case i'm more of an idiot that everyone! XD lol

  68. dan815
    • one year ago
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    basically what he did was, well normally to a line passing through that point (5,0) to your parabola there will be 2 intersections everywhere except for 1 place

  69. ganeshie8
    • one year ago
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    reminds me of 3 idiots movie

  70. anonymous
    • one year ago
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    Ahhh i see

  71. dan815
    • one year ago
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    |dw:1435219094311:dw|

  72. anonymous
    • one year ago
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    Lmao XD truuueee :D so much humour the night before my exam :D lol

  73. anonymous
    • one year ago
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    hmmm i see it now

  74. ganeshie8
    • one year ago
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    just to conclude : x=a = 9 gives slope=-2x=-2*9 = -18 so the equation of desired tangent line is y = -18(x-5)

  75. dan815
    • one year ago
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    right, and u know how from the quadratic equation the 2 differnet solutions come from that +/- part in squareroot

  76. dan815
    • one year ago
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    |dw:1435219226919:dw|

  77. anonymous
    • one year ago
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    Yep! Got it all :)

  78. dan815
    • one year ago
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    k good

  79. anonymous
    • one year ago
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    Thanks so much!!! :)

  80. dan815
    • one year ago
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    yeah sure =]

  81. Astrophysics
    • one year ago
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    Thank you teachers!

  82. anonymous
    • one year ago
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    Yeshh :D

  83. Astrophysics
    • one year ago
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    Are all these questions part of a practice exam?

  84. anonymous
    • one year ago
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    Yeah, they're some sample/practice questions... or whatever they call it XD lol

  85. anonymous
    • one year ago
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    @Astrophysics :D sorry about the late response, wasn't paying attention lol

  86. Astrophysics
    • one year ago
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    Lol no worries was helping someone anyways, haha and good luck, calc exams have always been long >.<

  87. anonymous
    • one year ago
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    thank you! :) Yeah it's gonna be 2.5 hours 8:30 in the morning >.<

  88. Astrophysics
    • one year ago
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    Eek, haha get some sleep as well! Later :P

  89. anonymous
    • one year ago
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    I'd better xD lol otherwise its a big fat 0 xD haha

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