#### Answer

$(-\infty,\ -1] \cup\ [6,\ \infty)$

#### Work Step by Step

We see from the graph that the equation $y=x^{2}-4x$ lies above (i.e. is greater than) the equation $y=x+6$ for $-\infty\lt x\lt 1$ and $6\lt x\lt \infty$, i.e. $(-\infty,\ -1] \cup\ [6,\ \infty)$.