Precalculus (10th Edition)

$E$
If $f(x)=ax^2+bx+c$, then if $a\gt0$, the graph opens up, hence it has a maximum, if $a\lt0$, the graph opens down, hence it has a minimum. The minimum value of the graph is at $x=-\frac{b}{2a}$. Let's compare $f(x)=-x^2-1$ to $f(x)=ax^2+bx+c$. We can see that $a=-1$, $b=0$, $c=-1$. $a\lt0$, hence the graph opens down, hence it's vertex is a maximum. The maximum value is at $x=-\frac{b}{2a}=-\frac{0}{2\cdot(1)}=0.$ Hence the maximum value is $f(0)=-(0)^2-1=-1.$ Thus, the vertex is at $(0,-1)$. The vertex is at $(0,-1)$ and the graph opens down. Hence it is graph $E$.