Answer
$\{x|x\lt 0\text{ or }x\gt 4\}=(-\infty,0)\cup(4,\infty)$
Work Step by Step
We are given the inequality:
$x^2-4x>0$
$f(x)=x^2-4x$
Determine the intercepts:
$y$-intercept: $f(0)=0^2-4(0)=0$
$x$-intercepts: $x^2-4x=0$
$x(x-4)=0$
$x=0$ or $x-4=0$
$x=0$ or $x=4$
The $y$-intercept is 0; the $x$-intercepts are 0 and 4.
Determine the vertex:
$x_V=-\dfrac{b}{2a}=-\dfrac{-4}{2(1)}=2$
$y_V=f(x_V)=2^2+4(2)=-4$
The vertex is $V(2,-4)$
Graph the function.
The graph is above the $x$-axis for:
$x<0$ or $x>4$
The solution set is:
$\{x|x\lt 0\text{ or }x\gt 4\}=(-\infty,0)\cup(4,\infty)$
