Question

What is the equation of the line normal to `f(x)=(x^2-x)(x-2)` at `x=1`?

Answer 1

`y=x-1`

Explanation:

Equation of the line normal is `-y'_0(y-y_0)=x-x_0`

`y'=(2x-1)(x-2)+x^2-x=3x^2-6x+2`

`y'_0=-1; y_0=0; x_0=1`

`-(-1)*(y-0)=x-1`