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Math Help - Show that the Tangent has a y intercept of...

  1. #1
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    Show that the Tangent has a y intercept of...

    Show that the tangent to P: ax^2 + bx + c with gradient m has a y intercept of c - ( (m - b)^2 / 4a )

    OK so far my working would be...

    find dydx to find gradient of tangent

    dydx = 2ax + b

    y intercept means x = 0, so when x =0, 2ax + b = b

    From here i am stuck.

    Any ideas would be appreciated
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  2. #2
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    I think that answer is wrong

    The tangent to the equation will have an equation in the form y = mx + yin

    (where yin= yitnercept, havent called it c as you already have c defined)

    therefore tangent equation is y = (2ax+b)x + yin

    you also have an equation y = ax^2 + bx + c

    you know these intercept so you can equate them, ie.

    ax^2 + bx + c = (2ax +b)x + yin

    then solve for yin

    Calypso
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  3. #3
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    Thanks Calypso

    so how does this look...

    The tangent to the equation will have an equation in the form y = mx + yin

    (where yin= yitnercept, havent called it c as you already have c defined)

    therefore tangent equation is y = (2ax+b)x + yin

    you also have an equation y = ax^2 + bx + c

    you know these intercept so you can equate them, ie.

    ax^2 + bx + c = (2ax +b)x + yin

    ax^2 + bx + c - (2ax +b)x = yin

    ax^2 + bx + c - 2ax^2 - bx = yin

    - ax^2 + c = yin

    ???
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  4. #4
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    Yes looks correct. However its strange that it is so different to what was stated as the correct answer. Where did you get that answer from of (c - ( (m - b)^2 / 4a ))
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