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Math Help - Tangent -> Slop intercept form

  1. #1
    Junior Member
    Joined
    Mar 2010
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    Tangent -> Slop intercept form

    I'm struggling with the following. I have a question such as:

    Find an equation for the line tangent to y=3-5x^2 at (5,-122)

    I know how to calculate the slope which is:

    -50=(y-(-122))/(x-(5))

    And the above in slope intercept form is:

    -50x+128

    How was the slope intercept form calculated? I can't seem to figure this out.
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  2. #2
    Junior Member
    Joined
    Mar 2010
    Posts
    26
    I got it. The formula is y=mx+b

    In this case,

    y=-50x+b
    -122=-50(5)+b
    -122=-250+b
    -122+250=b
    128=b
    y=-50x+128
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