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Normal Line and a Function

  1. Oct 1, 2005 #1
    Here is what I found, look at the end to see what my question is:

    At what point does the normal line to the graph of y= -4 - 3x + 4x^2 at (1, -3 ) intersect the parabola a second time?

    So far I have tried to find the derivative of the parabola which is:

    8x-3 (this corresponds to the slope of the tangent line). Therefore now I replaced x by 1 because of the point I am given. This gives:

    8(1)-3 = 5

    So that is the slope of the tangent line. Now to know the equation of this line I do:

    y=5x+b , replace x and y by the point I am given and solve for b.


    Thus, the equation is: y=5x-8

    The normal is perpendicular to the tangent at (1,-3) so I take the negative reciprocal of the slope to find the slope of the normal:

    5 becomes -1/5 (reciprocal). And now I get y=-1/5+b for the normal. Replace x and y by the point (1,-3):


    The equation for the normal is: y=-1/5x-14/5

    Now I want to find when this line and the parabola meet so:

    -1/5x-14/5 = -4 - 3x + 4x^2
    0 = -6/5 -14/5x + 4x^2
    Using (-b+/-sqrt(b^2-4ac))/2a
    x1 = 1
    x2 = -3/10

    x1 = 1 I already know because that is the point where the normal, the tangent, and the parabola intersect. And x2 = -3/10 is the other point where only the parabola and the normal intersect.

    To find the y-coordinate of -3/10 I just insert it into the equation and get:

    y= -137/50

    Thus the other point is: (-3/10,-137/50)

    I would just like to know if the above calculations are correct and that I am doing the proper steps.

    Thanks a lot,

    Cedrick O'Shaughnessy
  2. jcsd
  3. Oct 1, 2005 #2


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    Homework Helper

    That all looks good. I'm just not sure why you found the equation for the tangent line, since it doesn't seem like you needed it.
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