Question

What is the equation of the line normal to `f(x)=x^2(x-2)` at `x=2`?

Answer 1

y=`-1/4 (x-2)`

Explanation:

First find f'(x) = `3x^2 -4x`. This represents the slope of any tangent line to f(x). At x=2, the slope of the tangent line would be 12-8=4. The slope of the normal at the same point would be `-1/4`.

At x=2, f(x)=0. Hence the normal at this point (2,0) would be y-0= `-1/4` (x-2)

On simplification it is y=`-1/4 (x-2)` in the point slope form.