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

y=$- \frac{1}{4} \left(x - 2\right)$
First find f'(x) = $3 {x}^{2} - 4 x$. 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 $- \frac{1}{4}$.
At x=2, f(x)=0. Hence the normal at this point (2,0) would be y-0= $- \frac{1}{4}$ (x-2)
On simplification it is y=$- \frac{1}{4} \left(x - 2\right)$ in the point slope form.