find the gradient to the curve y = x^3 - 3x^2 at the point x = 3
then find the equations of the tangent and the normal to the curve at the point where x = 3

thanks

2. $f(x) = x^3 - 3x^2$
$f'(x) = 3x^2 - 6x$
$f'(3) = 3(3)^2 - 6(3) = 9$

So the slope of your tangent is $9$, and the slope of your normal is $-\frac{1}{9}$.

Also, since
$f(3) = (3)^3 - 3(3)^2 = 0$,
both equations will pass through the point $(3, 0)$.

Let us know if you need more help from there.