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Math Help - derivative application

  1. #1
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    derivative application

    Show that the tangent to P: y= ax^2+bx+c with gradient m has y intercept c-(m-b)^2/4a. thank you.
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  2. #2
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    Quote Originally Posted by slaypullingcat View Post
    Show that the tangent to P: y= ax^2+bx+c with gradient m has y intercept c-(m-b)^2/4a. thank you.
    m = 2ax + b => x = ....

    Substitute that expression for x into y= ax^2+bx+c to get an expression for y. Now you know a point (x1, y1) on the tangent.

    Then the tangent is y - y1 = m(x - x1).

    Now substitute x = 0 and solve for y.
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  3. #3
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    Hello, slaypullingcat!

    The procedure is simple, but there's a LOT of algebra!
    I was lucky to get the answer on the first try . . .


    Show that the tangent to: . y \:=\:ax^2+bx+c with gradient m

    . . has y-intercept: . c-\frac{(m-b)^2}{4a}

    The derivative of: . y \:=\:ax^2+bx+c [1] .is: . y' \:=\:2ax + b

    If the gradient is m\!:\;\;2ax + b \:=\:m \quad\Rightarrow\quad x \:=\:\frac{m-b}{2a}

    Substitute into [1]: . y \:=\:a\left(\tfrac{m-b}{2a}\right)^2 + b\left(\tfrac{m-b}{2a}\right) + c \quad\Rightarrow\quad y \;=\;\frac{m^2-b^2 + 4ac}{4a}


    The tangent has point P\left(\tfrac{m-b}{2a},\;\tfrac{m^2-b^2+4ac}{4a}\right) . and slope m.


    Its equation is: . y - \tfrac{m^2-b^2+4ac}{4a} \;=\;m\left(x - \tfrac{m-b}{2a}\right)

    . . \text{which simplifies to: }\;y \;=\;mx + \underbrace{c - \tfrac{(m-b)^2}{4a}}_{y\text{-intercept}}

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  4. #4
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    That was a quality post, thank you very much. I understand it now.
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