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Posted by on Friday, October 23, 2009 at 10:36pm.

Find the equation of the normal to the curve y = 3x^2 - 5x that is perpendicular to the line y = x + 4.

  • Maths - , Friday, October 23, 2009 at 10:46pm

    Find the slope of the tangent first.

    the perpendicular to the line x+4 has a slope of -1.

    so the normal to the curve has the same slope.
    y'=6x-5 and then the perpendicular to this must have a slope of -1/6x

    so -1/6x=-1 or x= 1/6

    y=mx+b slope is -1
    y=-x+b but the point of tangencey is x=1/6
    y=-1/6+b
    Now for y, in the curve y=3x^2-5x, when x is 1/6, then y= 3/36-30/36=-27/36
    so solve for b.

    All this depends on my intrepretation of normal to the curve that is perpendicular to the line...

    That is most unusual wording.

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