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Christos

  • 2 years ago

Limits, Can you help me find out this specific thing I dont know how to find out http://screencast.com/t/PUvesX4N

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  1. agent0smith
    • 2 years ago
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    Once you find f'(x), plug in a to get the slope of the line, m. Then you want to get it in the form y=mx+b, by using the fact it passes through the point (a, f(a))

  2. BTaylor
    • 2 years ago
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    Do you know how to take the derivative? If so, then first take the derivative of the function, then evaluate it at the given a -- this is "m", the slope of the tangent line. Now, evaluate the function at the given a. This we'll call "y1" The tangent line is: y-y1 = m (x-a)

  3. Christos
    • 2 years ago
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    Can please someone tell me what is a slope just so I get the bigger picture of whats going on?

  4. agent0smith
    • 2 years ago
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    Slope is the steepness of the line, or rate of change. Remember back in algebra, equations y=mx+b have a slope of m.

  5. Christos
    • 2 years ago
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    @BTaylor what do you mean by "Now, evaluate the function at the given a. " ?

  6. BTaylor
    • 2 years ago
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    so, for number 9, f(x) = 2x^2. The given "a" is 1. So evaluating the function at a means you plug in a for x, and get 2(1)^2, which is 2.

  7. agent0smith
    • 2 years ago
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    That means plug in x=1 (since a=1) into f = 2x^2

  8. BTaylor
    • 2 years ago
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    @agent0smith is right. To visualize this tangent line, click this link: http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=tangent%20line%20to%202x%5E2%20at%20x%3D1

  9. Christos
    • 2 years ago
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    I got y1 = 0

  10. Christos
    • 2 years ago
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    y1 = -2

  11. Christos
    • 2 years ago
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    y1 = 2 :D **

  12. BTaylor
    • 2 years ago
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    the last one is right. y1 = 2.

  13. Christos
    • 2 years ago
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    so that's the answer?? y1 = 2 and I am done? Nothing else needed?

  14. agent0smith
    • 2 years ago
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    No... did you find the slope?

  15. Christos
    • 2 years ago
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    its 4

  16. BTaylor
    • 2 years ago
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    yes. because the derivative is 4x, evaluated at x=1 gives you 4(1) =4

  17. agent0smith
    • 2 years ago
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    So then put it in point-slope form: y-y1 = m (x-x1) where x1 = 1, y1= 2

  18. Christos
    • 2 years ago
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    yea

  19. Christos
    • 2 years ago
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    I did all that and I got y1 = 2 in the end, I am done ?

  20. agent0smith
    • 2 years ago
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    Wait, no, where is your x going? y-y1 = m (x-a) y - 2 = 4(x-1)

  21. Christos
    • 2 years ago
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    x-1 = 0

  22. Christos
    • 2 years ago
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    x = 1

  23. agent0smith
    • 2 years ago
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    x is x.

  24. agent0smith
    • 2 years ago
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    a is 1.

  25. Christos
    • 2 years ago
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    then why y is not y too and had to be replaced

  26. agent0smith
    • 2 years ago
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    y is y. y-2 = 4 (x-1)

  27. agent0smith
    • 2 years ago
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    You're using point slope form to write the equation of a line: http://www.purplemath.com/modules/strtlneq2.htm

  28. Christos
    • 2 years ago
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    Nice! I see now! Thank you both! I wish I could give more medals!

  29. agent0smith
    • 2 years ago
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    Your line has a slope of m=4, and passes through the point (1, 2) then you just plug it into the point slope formula.

  30. Christos
    • 2 years ago
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    oh i see

  31. agent0smith
    • 2 years ago
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    And again, this is what you have found: http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=tangent%20line%20to%202x%5E2%20at%20x%3D1

  32. Christos
    • 2 years ago
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    yes

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