## Precalculus (10th Edition)

$\{x|x\lt -8\text{ or }x\gt 0\}=(-\infty,-8)\cup(0,\infty)$
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: $\{x|x\lt -8\text{ or }x\gt 0\}=(-\infty,-8)\cup(0,\infty)$