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iop360

  • 3 years ago

find the equations of three tangent lines to the graph of y = 4x^3 -40x^2 + 125x - 126 which pass through point (1, -1) (but not on the curve)

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  1. iop360
    • 3 years ago
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    \[y = 4x^3 -40x^2 + 125x -126\]

  2. allank
    • 3 years ago
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    Cool question. I'd start by calculating the first derivative of y which will give me the slope of the lines we are looking for.

  3. allank
    • 3 years ago
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    Can you take it from here?

  4. iop360
    • 3 years ago
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    yea

  5. allank
    • 3 years ago
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    Great. Good luck. :)

  6. iop360
    • 3 years ago
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    thanks

  7. allank
    • 3 years ago
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    You're welcome :)

  8. iop360
    • 3 years ago
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    i get \[y' = 12x^2 - 80x +125\] what now

  9. iop360
    • 3 years ago
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    im supposed to get 3 slopes

  10. iop360
    • 3 years ago
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    factor right?

  11. iop360
    • 3 years ago
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    \[y = (2x-5)(6x-25)\]

  12. iop360
    • 3 years ago
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    y prime i mean

  13. iop360
    • 3 years ago
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    slopes so far are 5/2, 25/6, and...?

  14. allank
    • 3 years ago
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    Just a sec iop360...

  15. iop360
    • 3 years ago
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    actually is that even right?

  16. iop360
    • 3 years ago
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    ok

  17. allank
    • 3 years ago
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    Now, what we want to do is first get the positions where the tangents touch the curve...

  18. allank
    • 3 years ago
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    Think about it...how would we do that?

  19. allank
    • 3 years ago
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    Hint: Simultaneous equations.

  20. iop360
    • 3 years ago
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    so use m = m with one m representing the first derivative i just calculated and the other m representing rise/run

  21. iop360
    • 3 years ago
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    equate the two and solve for x?

  22. allank
    • 3 years ago
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    Sorry for taking forever to reply...

  23. iop360
    • 3 years ago
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    \[\frac{ (4x^3 - 40x^2 +125x -126) - (-1) }{ x-1 }= 12x^2 -80x + 125\]

  24. iop360
    • 3 years ago
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    its alright

  25. allank
    • 3 years ago
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    Okay, we have the gradient: m = 12x^2-80x+125 and we have a point (1,-1). I'm thinking we can use the general form of an equation of a line: y-y1 = m(x-x1) where y1 and x1 are -1 and 1 respectively, and m is the gradient. I'm not sure whether that will work. Lemme try....

  26. iop360
    • 3 years ago
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    we would have to find m first

  27. allank
    • 3 years ago
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    True. Trying something different...

  28. iop360
    • 3 years ago
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    i think i found the slopes, tell me what you have too slope = 0, 5/2, 4

  29. allank
    • 3 years ago
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    Well, I have used a different method, where I first determine the positions where the tangents touch the curve. Want to finish off and compare answers?

  30. iop360
    • 3 years ago
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    ok

  31. iop360
    • 3 years ago
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    you mean now find the line equations?

  32. allank
    • 3 years ago
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    Yep.

  33. iop360
    • 3 years ago
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    k

  34. iop360
    • 3 years ago
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    y = -1 y = (5/2)x - (7/2) y = 4x - 5

  35. iop360
    • 3 years ago
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    doesnt seem right when i graph it...

  36. iop360
    • 3 years ago
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    hopefully u have a better one

  37. allank
    • 3 years ago
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    Well, I've graphed yours and It seems okay. :) My equation is giving me imaginary solutions -_-

  38. iop360
    • 3 years ago
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    u graphed mine?

  39. allank
    • 3 years ago
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    Yep.

  40. iop360
    • 3 years ago
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    it doesnt look tangent to the curve

  41. allank
    • 3 years ago
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    Plotting on a calculator?

  42. iop360
    • 3 years ago
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    yes

  43. iop360
    • 3 years ago
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    they just all intersect together and dont touch the curve

  44. allank
    • 3 years ago
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    Darn. Okay, maybe I'm just sleepy to notice that on a computer grapher...so wanna see my method? Though it may not lead to an answer...

  45. iop360
    • 3 years ago
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    sure

  46. iop360
    • 3 years ago
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    perhaps i should bump this

  47. allank
    • 3 years ago
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    Okay, so we have the slope m = 12x^2-80x+125. Remember that the value of m depends on the x value on the curve.... Knowing the point (1,-1) , the equation of the tangent(s) will be y+1 = m(x-1) Plugging in m: y = m(x-1)-1 y = (12x^2-80x+125)(x-1) -1 y = 12x^3 - 80x^2 +125x - 12x^2 + 80x -125 - 1 y = 12x^3 - 92x^2 + 205x - 126 (equation 1) So, we have an equation that depends on the very same x value on the curve as the original equation: y = 4x^3 -40x^2 + 125x - 126 (equation 2) Two nice simultaneous equations that will give the x values on the curve where the tangents touch the curve: y = 12x^3 - 92x^2 + 205x - 126 (equation 1) y = 4x^3 -40x^2 + 125x - 126 (equation 2)

  48. allank
    • 3 years ago
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    I hope I've done the math right. Subtracting the second from the first eliminates y and the 126. y = 12x^3 - 92x^2 + 205x - 126 (equation 1) y = 4x^3 -40x^2 + 125x - 126 (equation 2) 0 = 8x^3 +52x^2 + 330x And solving for x gives x=0 and two other imaginary solutions. *sigh*

  49. iop360
    • 3 years ago
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    i see

  50. allank
    • 3 years ago
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    Do bump @iop360. I'll sleep on it and in case you haven't gotten a valid response by the time I'm awake, I will post the solution.

  51. iop360
    • 3 years ago
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    alright

  52. iop360
    • 3 years ago
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    ill leave this question open

  53. iop360
    • 3 years ago
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    thanks for helping

  54. iop360
    • 3 years ago
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    hey, i got the question right! if you want to know how, the values for x i found (0, (5/2), and 4) were actually the x coordinates of the tangents so i plugged these values in the original equation to get their respective y values after that i used each (x,y) coordinates of the three tangents with the given point in the question (1,-1) in the rise/run formula to get the slope for each line

  55. iop360
    • 3 years ago
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    plug in the found slopes in point-slope formula and it should be right

  56. allank
    • 3 years ago
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    Nice. Good job! :)

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