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

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)