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anonymous

  • 5 years ago

At what point on y=(-3)/(x-1) does the tangent pass through the origin? Helpp!!!

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  1. anonymous
    • 5 years ago
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    First question: are you in calc 1?

  2. anonymous
    • 5 years ago
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    yeah i think lol high school calculus

  3. anonymous
    • 5 years ago
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    Same thing.

  4. anonymous
    • 5 years ago
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    Anyway, I am going to assume you guys are learnign about derivatives, correct?

  5. anonymous
    • 5 years ago
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    yep i found the derivative but idk what else to do afterwards

  6. anonymous
    • 5 years ago
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    So what technically (in case if your teacher did not make this clear), a derivative of a curve, is technically the slope, so to speak.

  7. anonymous
    • 5 years ago
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    So if you were to, i dont know, set it up in some sort of linear function using your (0,0) point I would imagine you could find an equation. :D

  8. anonymous
    • 5 years ago
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    0=m(0)+b, solve for b.

  9. anonymous
    • 5 years ago
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    hmm ok ill try it

  10. anonymous
    • 5 years ago
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    Make sense?

  11. anonymous
    • 5 years ago
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    but the thing is i have to find the point

  12. anonymous
    • 5 years ago
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    Hold on, comp acting funny

  13. anonymous
    • 5 years ago
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    Ah i see what you're asking.

  14. anonymous
    • 5 years ago
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    y =(-3)/(x-1) So y ' = 3/(x-1)^2 let the point of tangency (a, y(a)) ==> (a, (-3)/(x-1)))

  15. anonymous
    • 5 years ago
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    So the sloope at that point will be 3/(a-1)^2

  16. anonymous
    • 5 years ago
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    ok

  17. anonymous
    • 5 years ago
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    So the equation of tangent with slope 3/(a-1)^2 and passing through (a, (-3)/(x-1)) is y -(-3)/(x-1)= 3/(a-1)^2* (x - a)

  18. anonymous
    • 5 years ago
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    Since (0,0) is a point in this line you can substitute 0 for x and 0 for y.

  19. anonymous
    • 5 years ago
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    So we have 3/(-1)=3/(a-1)^2*(-a)

  20. anonymous
    • 5 years ago
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    -3=(-3a)/(a-1)^2 -3(a-1)^2=-3a -3(a^2-2a+1)=-3a a^2-2a+1=a a^2-3a+1

  21. anonymous
    • 5 years ago
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    =0*

  22. anonymous
    • 5 years ago
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    Hmmmm

  23. anonymous
    • 5 years ago
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    In theory that should have worked... where did i make an algebraic mistake?

  24. anonymous
    • 5 years ago
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    ok ill try do it the same way u did

  25. anonymous
    • 5 years ago
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    And hopefully you dont run into the error i am running into. This is strange.

  26. anonymous
    • 5 years ago
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    ok i got the answer thank you so much for pointing out what to do i have a test tomorrow and i was literally freaking out so ur a lifesaver

  27. anonymous
    • 5 years ago
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    Well, I am happy to hear my idea pushed you the right direction.

  28. anonymous
    • 5 years ago
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    Also, some pointers on calc in high school (thats when I took it):

  29. anonymous
    • 5 years ago
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    Breathe and practice similar problems =D

  30. anonymous
    • 5 years ago
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    Did the process make sense?

  31. anonymous
    • 5 years ago
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    yep it did and besides i got the correct answer :)

  32. anonymous
    • 5 years ago
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    Yay! Well I wish you luck on your exam, be sure to get some sleep (always a good thing).

  33. anonymous
    • 5 years ago
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    thanks a lot

  34. anonymous
    • 5 years ago
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    You're welcome.

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