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Math Help - equation of tangent lines

  1. #1
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    equation of tangent lines

    Find the equation of 2 tangent lines to the graph of f(x)=2x-x^2 that passes through the point (1,5)
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  2. #2
    MHF Contributor Calculus26's Avatar
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    the tangent line at a point (a,f(a)) is

    y= f ' (a) (x-a) +f(a)

    for your problem

    y=(2-2a)(x-a) + 2a -a^2

    For the tangent line to pass through (1,5)

    5 = (2-2a)(1-a) + 2a -a^2

    solve this for a

    See attachment
    Attached Thumbnails Attached Thumbnails equation of tangent lines-tanlines.jpg  
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  3. #3
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    Here's Fermat's method of finding tangents that pre-dates the calculus.

    Any non-vertical line through (1, 5) can be written as y= m(x-1)+ 5. Of course, to be a tangent line, it must touch the parabola at some point: we must have m(x- 1)+ 5= 2x- x^2 or x^2+ (m-2)x+ 5-m=0. A root of that equation would be the x value of the point of tangency. In fact, to be tangent that equation must have a double root. That means we must have x^2+ (m-2)x+ 5-m= (x- a)^2= x^2- 2ax+ a^2 for some a. Comparing coefficients, m- 2= -2a and 5- m= a^2. From the first equation, m= 2- 2a. Putting that into the second equation, 5- (2- 2a)= 3+ 2a= a^2. a^2- 2a- 3= (a- 3)(a+ 1)= 0. Use those values of a to find m, the slopes of the two tangent lines, and so the tangent lines.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    I would have been a serf in those days
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