Answer
$\{x|x\lt -8\text{ or }x\gt 0\}=(-\infty,-8)\cup(0,\infty)$
Work Step by Step
We are given the inequality:
$x^2+8x>0$
$f(x)=x^2+8x$
Determine the intercepts:
$y$-intercept: $f(0)=0^2+8(0)=0$
$x$-intercepts: $x^2+8x=0$
$x(x+8)=0$
$x=0$ or $x+8=0$
$x=0$ or $x=-8$
The $y$-intercept is 0; the $x$-intercepts are 0 and -8.
Determine the vertex:
$x_V=-\dfrac{b}{2a}=-\dfrac{8}{2(1)}=-4$
$y_V=f(x_V)=(-4)^2+8(-4)=-16$
The vertex is $V(-4,-16)$
Graph the function.
The graph is above the $x$-axis for:
$x>0$
The solution set is:
$\{x|x\lt -8\text{ or }x\gt 0\}=(-\infty,-8)\cup(0,\infty)$
