Answer
$E$
Work Step by Step
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 $E$.
