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anonymous

  • 5 years ago

Determine the equations of both lines that are tangent to the graph of f(x) = x^2 and pass through the point (1,-3).

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  1. myininaya
    • 5 years ago
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    f'(x)=2x f'(a)=2a so the equation of the line is y=2ax+b we know a point on this line (1,-3) so we -3=2a(1)+b -3=2a+b ....

  2. myininaya
    • 5 years ago
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    thinking...

  3. anonymous
    • 5 years ago
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    One would have a negative slope, the other would be positive

  4. myininaya
    • 5 years ago
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    right

  5. anonymous
    • 5 years ago
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    Would those two lines be orthogonal to each other?

  6. myininaya
    • 5 years ago
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    that looks possible it looks like it could form a 90 degree angles

  7. anonymous
    • 5 years ago
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    Ok so then they must have reciprocal slopes

  8. myininaya
    • 5 years ago
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    opposite reciprocal slopes

  9. myininaya
    • 5 years ago
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    wait i htink i'm fixing to figure something out....

  10. myininaya
    • 5 years ago
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  11. myininaya
    • 5 years ago
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    now we need to figure the y intercept for each line which is easy since we know a point on both lines. both lines have the point (1,-3)

  12. myininaya
    • 5 years ago
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    y=6x+b (1,-3) -3=6(1)+b -3=6+b -3-6=b -9=b so y=6x-9 is the line that is tangent to the point (3,9) on the curver y=x^2 that passes through (1,-3) y=-2x+b (1,-3) -3=-2(1)+b -3+2=b -1=b so y=-2x-1 is the line tangent to the point (-1,1) on the curve y=x^2 that passes through (1,-3)

  13. myininaya
    • 5 years ago
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    we can check if you want Let's find the tangent line at (3,9) and see if passes through (1,-3) so f'(x)=2x f'(3)=6 y=6x+b 9=6(3)+b 9-18=b b=-9 so y=6x-9 is (1,-3) on that line -3=6(1)-9=-3 so yes! now let's check the tangent line at (-1,1) and see if passes through (1,-3) so f'(x)=2x f'(-1)=-2 y=-2x+b 1=2+b b=-1 y=-2x+-1 is (1,-3) on that line -3=-2(1)+-1=-3 so yes! :) we are finished!

  14. myininaya
    • 5 years ago
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    so the lines arent orthogonal

  15. myininaya
    • 5 years ago
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    do you have any questions?

  16. anonymous
    • 5 years ago
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    One question

  17. anonymous
    • 5 years ago
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    Why did you choose the general points (a,a^2) to be X1,Y1 respectively? Does it make a difference if I were to use them as X2,Y2 in the slope formula?

  18. myininaya
    • 5 years ago
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    (a^2-(-3))/(a-1)=(-3-a^2)/(1-a) we can make both sides look the same (-1)/(-1)=1 so if we multiply our second fraction by (-1)/(-1) we would get the first fraction or vice versa

  19. myininaya
    • 5 years ago
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    (y1-y2)/(x1-x2)=(y2-y1)/(x2-x1) (-1)/(-1)*(y1-y2)/(x1-x2)=(y2-y1)/(x2-x1)

  20. myininaya
    • 5 years ago
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    a-b is the same as -(b-a)

  21. myininaya
    • 5 years ago
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    (6-2)/(3-1)=4/2=2 (2-6)/(1-3)=-4/(-2)=2

  22. myininaya
    • 5 years ago
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    cool?

  23. anonymous
    • 5 years ago
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    Yes thank you very much. I have just a few more questions if you don't mind helping me out with them.

  24. myininaya
    • 5 years ago
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    go for it

  25. anonymous
    • 5 years ago
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    If f(a) = 0 and f'(a) = 6, find the limit h --> 0 of: f(a + h) ------- 2h

  26. myininaya
    • 5 years ago
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    i got it scanning it now

  27. myininaya
    • 5 years ago
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  28. myininaya
    • 5 years ago
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    i just used the definition of derivative

  29. myininaya
    • 5 years ago
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    and plugged and what they gave me and solved for limh->0 (f(a+h)/(2a))

  30. myininaya
    • 5 years ago
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    do you understand both problems we did?

  31. myininaya
    • 5 years ago
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    oops solved for limh->0(f(a+h)/(2h))

  32. anonymous
    • 5 years ago
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    I don't understand how you did the second question.

  33. myininaya
    • 5 years ago
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    what is definition of the derivative at a of f

  34. anonymous
    • 5 years ago
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    Wait, ok wouldn't it be 12 for the answer because: 6 = f(a+h)/h * (1/2) 6(2) = f(a+h)/h

  35. anonymous
    • 5 years ago
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    Sorry I don't know what happened

  36. myininaya
    • 5 years ago
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    doesn't the question above ask for lim h->0 (f(a+h)/(2a))

  37. myininaya
    • 5 years ago
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    oops the bottom is suppose to read 2h

  38. myininaya
    • 5 years ago
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    so we have 6=limh->0(f(a+h)/(2h))

  39. myininaya
    • 5 years ago
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    oops 6=limh->0(f(a+h)/h) do you understand this far?

  40. anonymous
    • 5 years ago
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    yes

  41. myininaya
    • 5 years ago
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    ok but the want 1/2* limh->0(f(a+h)/h) so if we multiply the above equation on one side by 1/2 don't we have to do it to the other side?

  42. anonymous
    • 5 years ago
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    yes

  43. myininaya
    • 5 years ago
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    ok 6*(1/2)=3

  44. anonymous
    • 5 years ago
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    Got it. Thank you.

  45. anonymous
    • 5 years ago
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    Ok last question for now is this: Determine the value of "a', given that the line "ax - 4y + 21 = 0" is tangent to the graph y = a/x^2 at x = -2

  46. myininaya
    • 5 years ago
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    if you write that tangent line in y=mx+b form we can find the slope of the line

  47. myininaya
    • 5 years ago
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    so the equation of that line can be written y=ax/4-21/4 where a/4 is the slope of this tangent line

  48. anonymous
    • 5 years ago
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    The slope of the line is a right?

  49. anonymous
    • 5 years ago
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    ok so then we equate the two equations together

  50. anonymous
    • 5 years ago
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    right?

  51. myininaya
    • 5 years ago
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    so you are talking about finding the derivative of that function y=a/x^2 at x=-2 and then setting that equal to the slope of tangent line and solving for a? that's what I'm working on right now

  52. anonymous
    • 5 years ago
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    yes and I got 7 when equating the two equations together. But I don't understand how we can equate a y with a y' (derivative)

  53. myininaya
    • 5 years ago
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    so we get a=a so this equation holds for any a is what I get

  54. anonymous
    • 5 years ago
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    Oh so you are making them equal to each other by solving for a?

  55. anonymous
    • 5 years ago
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    a = , a =, and then equating them?

  56. myininaya
    • 5 years ago
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    derivative means slope this slope is suppose to be the same as the slope tangent to whatever point we are talkng about so we can find the slope of the tangent line and find the derivative of the curve at the given point and set them equal to find what a should be but the resulting equation is a=a this equation holds for every number 3=3 4=4 it is never not true

  57. myininaya
    • 5 years ago
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    a/4=-2a/(x^3) at x=-2 is what I did

  58. anonymous
    • 5 years ago
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    how did you get a/4?

  59. myininaya
    • 5 years ago
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  60. myininaya
    • 5 years ago
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    a/4 is the slope of the tangent line

  61. anonymous
    • 5 years ago
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    OOHHHH I understand So we can take a/4 to be the slope, which must be the same as the graph's slope at that point?

  62. myininaya
    • 5 years ago
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    right just like the very very question we did sort of

  63. anonymous
    • 5 years ago
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    Got it. Thank you so much for your help. I have to go now. I really appreciate you taking your time to do this for me.

  64. myininaya
    • 5 years ago
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    no problem become my fan :)

  65. anonymous
    • 5 years ago
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    Sure, see you.

  66. myininaya
    • 5 years ago
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    later

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