The point at which the tangent to the curve $y=x^2-4 x$ is parallel to x-axis, is

(0, 4) (-2, 4) (2, 4) (2, -4)

(2, -4)

Let $\left(x_1, y_1\right)$ be the required point. Then, $\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=0 \Rightarrow 2 x_1-4=0 \Rightarrow x_1=2$ Since $\left(x_1, y_1\right)$ lies on $y=x^2-4 x$. ∴  $y_1=x_1{ }^2-4 x_1 \Rightarrow y_1=4-8=-4$ Hence, the coordinates of the point are (2, -4)