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Math Help - Equations of all tangents..

  1. #1
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    Red face Equations of all tangents..

    Find the equations of all the tangents to the graph of (fx) = x^2 - 4x + 25 that passes through the origin.

    Would I just use x=0?

    Thanks in advanced.
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  2. #2
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    No I don't think that is going to work.

    You need the tangent going through the origin which is y-0= m (x-0) (or for simplicity's sake y=mx) where m = f'(a) where a is the point in x where the tangent occurs.

    Find a the point where the line(s) and curve intersect
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  3. #3
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    Any point on that graph looks like (a,a^2-4a+25).
    So any tangent passing through the origin must have slope \dfrac{a^2-4a+25}{a}.

    Using derivatives we know that the slope at any point is 2a-4.

    Where are those equal?
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  4. #4
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    Quote Originally Posted by advancedfunctions2010 View Post
    Find the equations of all the tangents to the graph of (fx) = x^2 - 4x + 25 that passes through the origin.
    looks like there are two ...
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  5. #5
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    thanks for the help everyone..

    So I did this:

    2x - 4

    f'(0) = -4

    y=(0)^2 - 4(0) + 25
    y = 25

    y=mx+b
    25 = 14(0) +b
    b= 25

    And I'm not sure where I'm going..
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  6. #6
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    Quote Originally Posted by advancedfunctions2010 View Post
    thanks for the help everyone..

    So I did this:

    2x - 4

    f'(0) = -4

    y=(0)^2 - 4(0) + 25
    y = 25

    y=mx+b
    25 = 14(0) +b
    b= 25

    And I'm not sure where I'm going..
    Neither am I, why did you choose these calcs?
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  7. #7
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    I thought I was to take the deriviate and find the slope of the tangent line first? (which would be -4)
    and then use that to find the equation of the tangent which would be using b=25?
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  8. #8
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    Your two tangents occur if x=5~\&~x=-5.
    Now you reply with reasons why!
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  9. #9
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    I don't know.. I am very new and confused to calculus
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  10. #10
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    Quote Originally Posted by advancedfunctions2010 View Post
    I don't know.. I am very new and confused to calculus

    I understand it can be confusing

    Read this again

    Quote Originally Posted by Plato View Post
    Any point on that graph looks like (a,a^2-4a+25).
    So any tangent passing through the origin must have slope \dfrac{a^2-4a+25}{a}.

    Using derivatives we know that the slope at any point is 2a-4.

    Where are those equal?
    Now solve \dfrac{a^2-4a+25}{a}=2a-4

    Can you do that?
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  11. #11
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    note that a curve and a line tangent to the curve have the following two properties ...

    1) their slopes are equal

    2) the curve and tangent line are equal at the point of tangency


    using property (1), m = 2x-4

    a line through the origin has the form y = mx , so the line may be written y = (2x-4)x

    now, use property (2) to solve for the values of x where the curve = the tangent line.
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  12. #12
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    I am so sorry, I feel like I'm wasting everyone's time but I am really lost.
    I don't know how to respond, but I really appreciate your time and sincere ways of trying to help.. Im trying to do more research on the side to figure this out
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  13. #13
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    Quote Originally Posted by advancedfunctions2010 View Post
    I am so sorry, I feel like I'm wasting everyone's time but I am really lost.
    That is probably a vary true statement.
    Sometimes sites such as this are of absolutely no use to students such as yourself.
    Only a live sit-down with a real instructor/tutor will resolve the profound confusion.
    Please seek that for yourself.
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