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Math Help - Finding points where a tangent lines is parallel to another line

  1. #1
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    Finding points where a tangent lines is parallel to another line

    #8. find each point at which the tangent line to the curve y1=2x +(4/x) +1 is parallel to the line y2+2x=6.

    How would I do this? Take y1' and get y1'(4) and set it equal to the slope of y2 and then plug in y1 for x and y values? The answer is (1, 7) and (-1, -5).
    Last edited by hazecraze; October 15th 2009 at 07:58 AM.
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  2. #2
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    Solve the second equation for "y=". Then read the slope value off the equation. (It'll be the value multiplied on the "x".)

    Differentiate the first equation with respect to x. Set equal to the given slope value, and solve for the corresponding x-value(s).

    Plug these x-values into the first equation, and solve for the corresponding y-value(s), and thus the point(s) in question.

    (I get the same answer as you've listed.)
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    At first I confused my self by thinking it was y1=2x +(4/x +1) and got (1,3), but I see that it was y1=2x +(4/x) +1. Also, after differentiating and setting equal to -2 and then simplifying, I got -(x^3+2x^2+x+1). With synthetic division and a (x-1) root, I got -(x-1)=0, which was the x value of (1,7), but for the other part: (x^2+x), I factored out an x to get x(x+1), so wouldn't there also be a point 0, in addition to the -1, thus giving (0,1) and then the (1,7) and (-1,-5)?
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  4. #4
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    Quote Originally Posted by hazecraze View Post
    #8. find each point at which the tangent line to the curve y1=2x +(4/x) +1 is parallel to the line y2+2x=6.

    How would I do this? Take y1' and get y1'(4) and set it equal to the slope of y2 and then plug in y1 for x and y values? The answer is (1, 7) and (-1, -5).
     slop\ of\ line \ y_{2}+2x=6 is m_{2}=-2\quad (\because y=mx+c)
     slop\ of \ curve \ y_{1}= 2x +\frac{4}{x} +1 \ is
     m_{1}=\frac{dy}{dx}=\quad2- \frac{4}{x^2}
     since\ the\ tangent\ line\ y_{1}\ to\ the\ curve\ is\  parallel\ to\ the\ line\ y_{2}
     m_{1}=m_{2} \quad \ so \ that\ \quad \quad2- \frac{4}{x^2} = -2
     or \ x= \pm 1
    on substituting in equation of curve we get
    for \ x=1 \quad y=7 \ and\ for \ x=-1 \quad y=-5
    at (1, 7) and (-1, -5) ,the tangent line to the curve \ y_{1}= 2x +\frac{4}{x} +1 is parallel to the line  y_{2}+2x=6
    Last edited by ramiee2010; October 15th 2009 at 09:25 AM. Reason: attach graph
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  5. #5
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    Oh, well that was much more simplistic than what I did.
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