\(-x^2-5x=0\\
-x(x+5)=0\\
x=0 \vee x=-5\)
The x-intercepts are \((0,0)\) and \((-5,0)\).
\(\hbox{vertex}=(h,k)\\ h=\frac{x_1+x_2}{2},k=y(h)\\\\ h=\frac{0-5}{2}=\frac{-5}{2}=-2.5\\ k=y(-2.5)=-(-2.5)^2-5\cdot(-2.5)\\ k=-6.25+12.5\\ k=6.25\\\\ \hbox{vertex}=(2.5,6.25)\)
