Find points at which the tangent to the curve

The equation of the given curve is $y=x^{3}-3 x^{2}-9 x+7$.

$\therefore \frac{d y}{d x}=3 x^{2}-6 x-9$

Now, the tangent is parallel to the x-axis if the slope of the tangent is zero.

$\therefore 3 x^{2}-6 x-9=0 \Rightarrow x^{2}-2 x-3=0$

$\Rightarrow(x-3)(x+1)=0$

$\Rightarrow x=3$ or $x=-1$

When $x=3, y=(3)^{3}-3(3)^{2}-9(3)+7=27-27-27+7=-20$

When $x=-1, y=(-1)^{3}-3(-1)^{2}-9(-1)+7=-1-3+9+7=12$

Hence, the points at which the tangent is parallel to the x-axis are (3, −20) and

(−1, 12).