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Math Help - Verification requested

  1. #1
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    Verification requested

    Let D(n) denote change in some integer n; let A(x) and A(y) denote either the corresponding x or y co-ordinate at some point A on the euclidean plane;

    Question: Considering two points P and Q upon the curve y= x^2-2x-3 of the euclidean plane what is the slope of the secant line PQ at the fixed point P as Q approaches P?

    slope m of secant line PQ = D(y)/D(x)

    so, if D(y)= Q(y) - P(y) then Q(y)= P(y) + D(y)

    and if D(x)=Q(x) - P(x) then Q(x)=P(x) + D(x)

    i.e. D(y) = [x^2 + 2xD(x) + D(x)^2 - 2x - 2D(x) - 3] - [x^2 - 2x - 3]

    and furthermore D(y)/D(x)= 2x + D(x) - 2

    am uncertain of final solution and would like to know if this answer implies the limit of the slope of the secant line as D(x) approaches 0 is 2x - 2
    Last edited by Foxlion; January 15th 2011 at 07:22 AM.
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  2. #2
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    Quote Originally Posted by Foxlion View Post
    Let D(n) denote change in some integer n; let A(x) and A(y) denote either the corresponding x or y co-ordinate at some point A on the euclidean plane;

    Question: Considering two points P and Q upon the curve y= x^2-2x-3 of the euclidean plane what is the slope of the secant line PQ at the fixed point P as Q approaches P?

    slope m of secant line PQ = D(y)/D(x)

    so, if D(y)= Q(y) - P(y) then Q(y)= P(y) + D(y)

    and if D(x)=Q(x) - P(x) then Q(x)=P(x) + D(x)

    i.e. D(y) = [x^2 + 2xD(x) + D(x)^2 - 2x - 2D(x) - 3] - [x^2 - 2x - 3]

    and furthermore D(y)/D(x)= 2x + D(x) - 2

    am uncertain of final solution and would like to know if this answer implies the limit of the slope of the secant line as D(x) approaches 0 is 2x - 2
    The problem asked what happens "as Q approaches P", which means, yes, D(x) goes to 0 and the the slope of the secant line goes to 2x- 2. That, by the way, implies that the equation of the tangent line, at (x_0, y_0) is y= (2x_0- 2)(x- x_0)+ y_0.

    I presume that this is the very beginning of a Calculus or Analysis course and you will so be introduced to the "derivative" as "slope of the tangent line".
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    I presume that this is the very beginning of a Calculus or Analysis course and you will so be introduced to the "derivative" as "slope of the tangent line".
    you presume correct, thank you.
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