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anonymous

  • one year ago

Someone Please Help, Giving A Medal A.Find the slope of the tangent line to the graph of f at the given point. A.Find the slope-intercept equation of the tangent line to the graph of f at the given point.

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  1. anonymous
    • one year ago
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  2. welshfella
    • one year ago
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    find the derivative f(x) = 4x + 2 can you find f'(x)?

  3. anonymous
    • one year ago
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    -1/2+x/4

  4. anonymous
    • one year ago
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    no the derivative of that is 4

  5. welshfella
    • one year ago
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    yes thats right so the slope is - at every point on the graph as its a straight line

  6. welshfella
    • one year ago
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    * slope is 4

  7. welshfella
    • one year ago
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    the second function is:- f(x) = x^(1/2) can you find the derivative of this one?

  8. anonymous
    • one year ago
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    x^-1/2/2

  9. welshfella
    • one year ago
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    |dw:1435264790652:dw|

  10. welshfella
    • one year ago
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    so to find the slope at (16,4) plug x = 16 into the the above value for f'(x)

  11. welshfella
    • one year ago
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    then use the general form y-y1 = m(x-x1) and plug in slope m and the point (x1,y1) will be (16,4)

  12. anonymous
    • one year ago
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    for the first step is it 2

  13. welshfella
    • one year ago
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    1 / 2*x(1/2) = 1 / 2* (16)^1/2 = 1/ 2*4 = 1/8

  14. welshfella
    • one year ago
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    so m = 1/8

  15. anonymous
    • one year ago
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    im lost. I don't understand this at all.

  16. welshfella
    • one year ago
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    m stands for the slope in the formula the slope is given by the derivative of sqrt x - and is 1 / 2* x (1/2) o r 1 / 2 sqrtx here x = 16 ( point is (16/4) so m = 1 / 2 sqrt16 = 1 / 2*4 = 1/8

  17. anonymous
    • one year ago
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    ok but its two parts to the problem that's the answer to both?

  18. anonymous
    • one year ago
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    @Preetha

  19. anonymous
    • one year ago
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    @misssunshinexxoxo

  20. anonymous
    • one year ago
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    @peachpi

  21. anonymous
    • one year ago
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    @xapproachesinfinity

  22. anonymous
    • one year ago
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    @dan815

  23. anonymous
    • one year ago
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    @Compassionate

  24. anonymous
    • one year ago
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    @uri

  25. anonymous
    • one year ago
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    @surjithayer

  26. anonymous
    • one year ago
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    @LynFran

  27. LynFran
    • one year ago
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    ok what exactly u dont understand here?

  28. anonymous
    • one year ago
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    i dont understand the problem or the question this whole thing confuses me.

  29. LynFran
    • one year ago
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    ok lets go through this step by step from the beginning ok

  30. anonymous
    • one year ago
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    ok

  31. LynFran
    • one year ago
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    first of all the slope of a tangent line is the gradient of that line ok

  32. anonymous
    • one year ago
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    what is a gradient

  33. LynFran
    • one year ago
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    now to find the slope of the tangent line is y/x...|dw:1435279410553:dw|

  34. LynFran
    • one year ago
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    The Gradient (also called Slope) of a straight line shows how steep a straight line is.

  35. anonymous
    • one year ago
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    the question has two parts is 4 the answer for both

  36. LynFran
    • one year ago
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    |dw:1435279812417:dw|

  37. LynFran
    • one year ago
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    4 is just the slope of the tangent line we have to find the slope intercept equation at the given point ok to complete this question ok

  38. anonymous
    • one year ago
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    ok whats the next step

  39. LynFran
    • one year ago
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    |dw:1435280154077:dw|

  40. LynFran
    • one year ago
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    what equation u get??

  41. anonymous
    • one year ago
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    y=4x+2

  42. LynFran
    • one year ago
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    correct

  43. anonymous
    • one year ago
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    ok so thats the second part to the first question?

  44. LynFran
    • one year ago
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    we just did 2nd question and yes that was this part; .Find the slope-intercept equation of the tangent line to the graph of f at the given point.

  45. anonymous
    • one year ago
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    ok so 1.4 and 2. y=4x+2

  46. LynFran
    • one year ago
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    yes Find the slope of the tangent line to the graph of f at the given point. is 4

  47. LynFran
    • one year ago
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    for that question

  48. anonymous
    • one year ago
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    ok got it

  49. LynFran
    • one year ago
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    Did u do the other question?

  50. anonymous
    • one year ago
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    can you help me with that one too?

  51. LynFran
    • one year ago
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    ok

  52. LynFran
    • one year ago
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    |dw:1435281016248:dw|

  53. LynFran
    • one year ago
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    |dw:1435281095053:dw|

  54. LynFran
    • one year ago
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    now to take the derivative of it we get|dw:1435281258598:dw|

  55. LynFran
    • one year ago
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    |dw:1435281332523:dw|

  56. LynFran
    • one year ago
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    |dw:1435281411275:dw|

  57. anonymous
    • one year ago
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    ok im with you, so whats next

  58. LynFran
    • one year ago
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    well we have to plug in the value of x into the derivative we just found

  59. LynFran
    • one year ago
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    (16,4) and x=16 there ok so.. plug it into the derivative and let me know what u get

  60. LynFran
    • one year ago
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    |dw:1435282159975:dw|

  61. anonymous
    • one year ago
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    ok so when i plug in 16 and solve to find the derivative i get 0

  62. LynFran
    • one year ago
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    no u cant get 0 lets work it together

  63. anonymous
    • one year ago
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    ok

  64. LynFran
    • one year ago
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    |dw:1435282473866:dw|

  65. LynFran
    • one year ago
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    so 1/8 is this part Find the slope of the tangent line to the graph of f at the given point.

  66. anonymous
    • one year ago
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    ok

  67. LynFran
    • one year ago
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    so can u find this part now .Find the slope-intercept equation of the tangent line to the graph of f at the given point. |dw:1435282844490:dw|

  68. anonymous
    • one year ago
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    4-y1=1/8(16-x1)

  69. LynFran
    • one year ago
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    no we leave the y and x term alone and uses the y1 and x1

  70. LynFran
    • one year ago
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    |dw:1435283138396:dw|

  71. anonymous
    • one year ago
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    ok y-4=1/8(x-16)

  72. LynFran
    • one year ago
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    correct now the next step is to multiply out the bracket

  73. anonymous
    • one year ago
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    y-4=1/8x -2

  74. LynFran
    • one year ago
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    good next step is to bring the -4 over the equal sign ...remember when we bring a # over the equal sign , the sign of the # changes

  75. anonymous
    • one year ago
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    y=1/8x+2

  76. LynFran
    • one year ago
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    yes and u have officially completed the question ok now i would like to share this site with u it was very useful to me so https://www.symbolab.com/solver/calculus-calculator/%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft(%5Csqrt%7Bx%7D%5Cright)/?origin=button

  77. anonymous
    • one year ago
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    ok thank You so much for your help.

  78. LynFran
    • one year ago
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    ur welcome

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