Answer
Graph:
.
Work Step by Step
$f(x)=-x^{2}-x-1$
$f(x)=ax^{2}+bx+c,\qquad a=-1,b=-1,c=-1$
Coefficient $a$ determines which way the parabola opens.
Here $a=-1$, which is negative, so the parabola opens $down.$
Coefficient $c$ tells us the y-intercept.
Here, $(0,-1)$ lies on the graph.
The vertex lies on the axis of symmetry, $x=-\displaystyle \frac{b}{2a}$
$x=-\displaystyle \frac{-1}{-2}=-\frac{1}{2}$
At $x=-\displaystyle \frac{1}{2},\ f(-\displaystyle \frac{1}{2})=-(-\frac{1}{2})^{2}-(-\frac{1}{2})-1=-\frac{3}{4}$
$\mathrm{V}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{e}\mathrm{x}$:$\displaystyle \quad (-\frac{1}{2},-\frac{3}{4})$
For the x-intercepts:
$-x^{2}-x-1=0$
check the discriminant $D=b^{2}-4ac$
$D=1-4\lt 0$,
so there are no x-intercepts.
We find some points on the graph:
$f(1)=-1-1-1=-3$
$f(-2)=-3$, also, because of symmetry
(1 and -2 are $1\displaystyle \frac{1}{2}$ left and right of the axis of symmetry.)
With this information, we sketch the graph.
