The curve #C# has a equation #y=x^3-2x^2#. How do you show that #N#, the normal to #C# at point #(1,-1)#, has equation y=-2#?

1 Answer
Jim H
Nov 5, 2016

You can't show that. It is false. I suspect that there is a typographic error in the question.

Explanation:

For #y=x^3-2x^2#, we get derivative #y' = 3x^2-4x#

At the point #(1,-1)# the slope of the tangent line is #m = y']_(x=1) = 3-4=-1#.

Therefore the slope of the normal line is #-1/((-1)) = 1#.

The line through #(1,-1)# with slope #m=1# has equation

#y-(-1) = 1(x-1)# which is equivalent to

#y=x-2# #" "# (I suspect that #y=-2# is a typographic error in the question.)

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