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Zekarias

  • 3 years ago

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

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  1. Zekarias
    • 3 years ago
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    I will post the answer exactly after one hour.

  2. mathslover
    • 3 years ago
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    Hint: Never challenge OpenStudy

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

  4. TheViper
    • 3 years ago
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    Right @mathslover :)

  5. hartnn
    • 3 years ago
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    so its not D

  6. sriramkumar
    • 3 years ago
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    @hartnn that means???

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

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

  9. KKJ
    • 3 years ago
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    I will the slope of the line through the two points

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

  11. hartnn
    • 3 years ago
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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.

  12. sauravshakya
    • 3 years ago
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    PLZ DONT POST THE SOLUTION

  13. Zekarias
    • 3 years ago
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    I am here to post the answer.

  14. Zekarias
    • 3 years ago
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    ok 5min

  15. sauravshakya
    • 3 years ago
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    ok

  16. hartnn
    • 3 years ago
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    whats wrong with my answer? and method ??

  17. Zekarias
    • 3 years ago
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    w/c is your ans?

  18. hartnn
    • 3 years ago
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    B)

  19. hartnn
    • 3 years ago
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    and p=-7

  20. Zekarias
    • 3 years ago
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    correct but the approach not.

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

  22. Zekarias
    • 3 years ago
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    yape

  23. hartnn
    • 3 years ago
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    then ?

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

  25. Zekarias
    • 3 years ago
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    Time is up @sauravshakya

  26. sauravshakya
    • 3 years ago
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    OK.......

  27. Zekarias
    • 3 years ago
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    Here is the answer

  28. Zekarias
    • 3 years ago
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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.

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

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

  31. sauravshakya
    • 3 years ago
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    Because the line l and the required line are parallel

  32. sauravshakya
    • 3 years ago
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    So, both of their slope is -1

  33. hartnn
    • 3 years ago
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    okk.

  34. hartnn
    • 3 years ago
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    it wasn't that difficult.

  35. Zekarias
    • 3 years ago
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    yes but need smart approach

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

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

  38. sauravshakya
    • 3 years ago
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    agreed with @hartnn

  39. sauravshakya
    • 3 years ago
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    that saves time in exam

  40. Zekarias
    • 3 years ago
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    For exam only

  41. hartnn
    • 3 years ago
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    atleast i got the answer...

  42. Zekarias
    • 3 years ago
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    yape

  43. Zekarias
    • 3 years ago
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    ok thanks for your help....... closing

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