Answer
$\bf{Option \ (E)}$
Work Step by Step
1) The graph of the quadratic function $f(x) = ax^2+bx+c$ is a parabola that opens:
(a) Upward when $a \gt 0$
(b) Downward when $a \lt 0$
On comparing $f(x)=-x^2-1$ with $f(x) = ax^2+bx+c$, we get: $a=-1, b=0,c=-1; a \lt 0$. We can see that the given function shows a graph of a parabola that opens downward. This means that the vertex has a maximum value.
2) The minimum value of the graph is at its vertex, where $x=\dfrac{-b}{2a}$.
Thus, the given function has its vertex at $x=\dfrac{-0}{2(1)}=0$ and the minimum value is $f(0)=-(0)^2-1=-1$. So, the given function has its vertex at: $(0,−1)$
Thus, the parabola that opens downward and whose vertex is at $(0,−1)$ matches with $\bf{Option \ (E)}$.
