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Zekarias Group Title

I AM CERTAIN THAT NO ONE WILL ANSWER THIS QUESTION... Suppose that l be a line through the points (0,3)and (5,-2) and tangent to the graph of the function \[F(x)=\frac{ x^2-p }{ 3x+1 }\] where ‘p’ is real number. What is the equation of a line parallel to l and tangent to the graph of F. A) 4x+4y+7 = 0 B) 9x+9y+37 = 0 C) x+y-3 = 0 D) 4x+6y-37 = 0

  • 2 years ago
  • 2 years ago

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  1. Zekarias Group Title
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    I will post the answer exactly after one hour.

    • 2 years ago
  2. mathslover Group Title
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    Hint: Never challenge OpenStudy

    • 2 years ago
  3. sriramkumar Group Title
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    m= -2-3/5-0 = -1 the equation will be y-3= -1(x-0) x+y-3 =0

    • 2 years ago
  4. TheViper Group Title
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    Right @mathslover :)

    • 2 years ago
  5. hartnn Group Title
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    so its not D

    • 2 years ago
  6. sriramkumar Group Title
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    @hartnn that means???

    • 2 years ago
  7. hartnn Group Title
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    that means answer can be a,b or c

    • 2 years ago
  8. sriramkumar Group Title
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    yep.... but the answer hasn't yet finished... @hartnn

    • 2 years ago
  9. KKJ Group Title
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    I will the slope of the line through the two points

    • 2 years ago
  10. hartnn Group Title
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    p=-7 B)9x+9y+37=0

    • 2 years ago
  11. hartnn Group Title
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    solve y=3-x=F(x)=.... u get two roots of x, equate them because the line is tangent and can have only one intersection point. from this i got p=-7. now answer can only be either A or B so solving both of them with F(x), gave only B) option with one intersection point, A) doesn't even intersect.

    • 2 years ago
  12. sauravshakya Group Title
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    PLZ DONT POST THE SOLUTION

    • 2 years ago
  13. Zekarias Group Title
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    I am here to post the answer.

    • 2 years ago
  14. Zekarias Group Title
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    ok 5min

    • 2 years ago
  15. sauravshakya Group Title
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    ok

    • 2 years ago
  16. hartnn Group Title
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    whats wrong with my answer? and method ??

    • 2 years ago
  17. Zekarias Group Title
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    w/c is your ans?

    • 2 years ago
  18. hartnn Group Title
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    B)

    • 2 years ago
  19. hartnn Group Title
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    and p=-7

    • 2 years ago
  20. Zekarias Group Title
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    correct but the approach not.

    • 2 years ago
  21. hartnn Group Title
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    didn't u get p=-7, the same way as i got ?

    • 2 years ago
  22. Zekarias Group Title
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    yape

    • 2 years ago
  23. hartnn Group Title
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    then ?

    • 2 years ago
  24. Zekarias Group Title
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    you use the options to answer the question, isn't it?

    • 2 years ago
  25. Zekarias Group Title
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    Time is up @sauravshakya

    • 2 years ago
  26. sauravshakya Group Title
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    OK.......

    • 2 years ago
  27. Zekarias Group Title
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    Here is the answer

    • 2 years ago
  28. Zekarias Group Title
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    First of all what you have to do is finding l so as to get p \[l:\frac{ y-3 }{ x-0 }=\frac{ -2-3 }{ 5-0 }\] \[l:y=-x+3\]thus \[F(x)=\frac{ x^2-p }{ 3x+1} =-x+3=y\]simplifying the above expression \[4x^{2}-8x-p-3=0\]Now since l is tangent to F(x), the quadratic equation above has only one solution. That means \[b^{2}-4ac=0\] \[(-8)^{2}-4(4)(-p-3)=0\]thus \[(-8)^{2}-4(4)(-p-3)=0\]\[p=-7\]Now we get F(x) to be \[F(x)=\frac{ x^2+7 }{ 3x+1 }\]Therefore lets solve F’(x)=-1 so as to get the coordinate of the unknown line. \[F'(x)=\frac{ 2x(3x+1)-3(x^{2}+7) }{ (3x+1)^2 }=-1\] \[3x^{2}+2x-21=-9x^{2}-6x-1\] \[12x^{2}+8x-20=0\]thus \[x=1 or x=\frac{ -5 }{ 3 }\]and \[F(1)=2 and F(\frac{ -5 }{ 3 })=\frac{ -22 }{ 9 }\]Thus we got this two points (1,2) and (-5/3,-22/9). The two possible lines therefore are \[\frac{ y-2 }{ x-1 }=-1{\rightarrow}y=-x+3\]w/c is the given line l. And \[\frac{ y-(-\frac{ 22 }{9 }) }{ x-(-\frac{ 5 }{ 3 })}=-1{\rightarrow}9y+9x+37=0\]w/c is the answer.

    • 2 years ago
  29. hartnn Group Title
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    why f'(x) = -1 ?

    • 2 years ago
  30. Zekarias Group Title
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    b/c the slope at that point is -1

    • 2 years ago
  31. sauravshakya Group Title
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    Because the line l and the required line are parallel

    • 2 years ago
  32. sauravshakya Group Title
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    So, both of their slope is -1

    • 2 years ago
  33. hartnn Group Title
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    okk.

    • 2 years ago
  34. hartnn Group Title
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    it wasn't that difficult.

    • 2 years ago
  35. Zekarias Group Title
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    yes but need smart approach

    • 2 years ago
  36. sauravshakya Group Title
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    YEP....... Actually I did a algebra mistake, as usual

    • 2 years ago
  37. hartnn Group Title
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    lol! checking options(when given) is the smartest approach!! :P

    • 2 years ago
  38. sauravshakya Group Title
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    agreed with @hartnn

    • 2 years ago
  39. sauravshakya Group Title
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    that saves time in exam

    • 2 years ago
  40. Zekarias Group Title
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    For exam only

    • 2 years ago
  41. hartnn Group Title
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    atleast i got the answer...

    • 2 years ago
  42. Zekarias Group Title
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    yape

    • 2 years ago
  43. Zekarias Group Title
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    ok thanks for your help....... closing

    • 2 years ago
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