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Tangents that pass through the origin

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data
    At how many points on the curve 4x[tex]^{}5[/tex] - 3x[tex]^{}4[/tex] + 15x[tex]^{}2[/tex] + 6 will the line tangent to the curve pass through the origin?

    2. Relevant equations

    3. The attempt at a solution

    I have no idea of how to even approach this... Erm, I fiddled around with the point-slope formula of a line

    y = mx - mx1 + y1

    where (x1, y1) is the point on the line/curve and m is the slope

    and in order for the line to pass through the origin, then mx1 + y1 have to cancel out... But I have no idea of how to find a place on the line where this occurs... Am I going about this all wrong/is there an easier way to do it? If not, where do I go from here?
  2. jcsd
  3. Nov 18, 2007 #2


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    I will assume that you mean "the curve y= [itex]4x^5- 3x^4+ 15x^2+ 6[/itex]" If x= [itex]x_1[/itex] then y= [itex]40x_1^5- 3x_1^4+ 15x_1^2+ 6[/itex] and m=y'= [itex]20x_1^4- 12x_1^3+ 20x_1[/itex]. Now use that "mx1+ y1= 0".
  4. Nov 18, 2007 #3


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    You are on the right track. So for the tangent line m=y1/x1. And the slope of the tangent line is dy/dx. So dy/dx=y/x. If you put the y polynomial into that you will have problems actually solving for x. The polynomial degree is 5 and it doesn't factor. On the other hand, the question asks you just to count the roots. I'm not sure I really know an elegant way to do that, except by messing around with it's derivatives to figure out where it's increasing and decreasing etc.
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