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Math Help - find tangent lines

  1. #1
    Super Member bigwave's Avatar
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    Cool find tangent lines

    given

    y=x^2 and y=-x^2+6x-5

    I know the derivatives are 2x and -2x+6

    how are the equations of 2 tangent lines found to both graphs

    the answers are: y=2x-1 and y=4x-4
    Last edited by bigwave; February 7th 2010 at 08:40 PM. Reason: added "to both graphs"
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  2. #2
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    Quote Originally Posted by bigwave View Post
    given

    y=x^2 and y=-x^2+6x-5

    I know the derivatives are 2x and -2x+6

    how are the equations of 2 tangent lines found

    the answers are: y=2x-1 and y=4x-4
    Hi

    The equation of the tangent line at a point whose abscissa is x_0 is
    y = f'(x_0)(x-x_0) + f(x_0)

    f(x)=x^2
    f'(x)=2x
    The equation of the tangent line at the point whose abscissa is x_0 = 1 is
    y = f'(1)(x-1) + f(1)
    y = 2(x-1) + 1
    y = 2x-1
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  3. #3
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    earboth's Avatar
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    Quote Originally Posted by bigwave View Post
    given

    y=x^2 and y=-x^2+6x-5

    I know the derivatives are 2x and -2x+6

    how are the equations of 2 tangent lines found

    the answers are: y=2x-1 and y=4x-4
    1. You have the parabolas:

    p: y = x^2
    and
    q: y = -x^2+6x-5

    2. Let P(p, p) denote the tangent point on the parabola p and Q(q, -q+6q-5) the tangent point on the parabola q.

    3. The tangent to p at P has the equation:

    t_p: y = 2px - p^2
    and the tangent to q at Q has the equation

    t_q: y = (-2q+6)x +q^2-5

    4. Both equations describe the same line. Thus

    \left|\begin{array}{rcl}2p&=&-2q+6 \\p^2+q^2&=&5 \end{array}\right.

    5. Solve this system of simultaneous equations for p and q. Resubstitute the result into the equations of t_p and t_q.

    6. For confirmation: Draw the 2 parabolas and the tangents.
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  4. #4
    Super Member bigwave's Avatar
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    Quote Originally Posted by running-gag View Post
    Hi

    The equation of the tangent line at a point whose abscissa is x_0 is
    y = f'(x_0)(x-x_0) + f(x_0)

    f(x)=x^2
    f'(x)=2x
    The equation of the tangent line at the point whose abscissa is x_0 = 1 is
    y = f'(1)(x-1) + f(1)
    y = 2(x-1) + 1
    y = 2x-1
    where did you get abscissa is x_0 = 1
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  5. #5
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    Quote Originally Posted by bigwave View Post
    given

    y=x^2 and y=-x^2+6x-5

    I know the derivatives are 2x and -2x+6

    how are the equations of 2 tangent lines found

    the answers are: y=2x-1 and y=4x-4
    Here's a tip- in the future tell us what the problem really says! Here, you have never said exactly which tangent lines you want to find. Every graph has an infinite number of tangent lines.

    I suspect that the problem is asking for lines that at tangent to both of these graphs. Suppose y= mx+ b is the equation of such a tangent line. At the point, (x_0, y_0) where that line is tangent to y= x^2 we must have y_0= x_0^2= mx_0+ b and y'= 2x_0= m. At the point, (x_1, y_1) where it is tangent to x^2+ 6x- 5, we must have y_1= x_1^2+ 6x_1- 5= mx_1+ b and y'= 2x_1+ 6= m so we have 4 equations to solve for m, b, x_1, and x_2
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  6. #6
    Super Member bigwave's Avatar
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    yes the 2 tangent lines would be tangent to both graphs.

    it was baffeling to determine the slope of the 2 lines just based on the derivatives and not knowing the points of contact..

    appreciate the replys
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