jmc

The $x$-coordinate of the vertex of the parabola is $\frac{-b}{2a}=\frac{0}{2(1)}=0$. The vertex is then $(0,-1)$. The intersections of the line $y=r$ with $y=x^2-1$ are found by setting the $y$ values equal to each other, so $$\begin{array}{rl}r& =x^2-1\\ \Rightarrow r+1& =x^2\\ \Rightarrow \pm \sqrt{r+1}& =x.\end{array}$$So the vertices of our triangle are $(0,-1)$, $(-\sqrt{r+1},r)$, and $(\sqrt{r+1},r)$. If we take the horizontal segment along the line $y=r$ to be the base of the triangle, we can find its length as the difference between the $x$-coordinates, which is $\sqrt{r+1}-(-\sqrt{r+1})=2\sqrt{r+1}$. The height of the triangle is the distance from $(0,-1)$ to the line $y=r$, or $r+1$. So the area of the triangle is $$A=\frac{1}{2}bh=\frac{1}{2}(2\sqrt{r+1})(r+1)=(r+1)\sqrt{r+1}.$$This can be expressed as $(r+1{)}^{\frac{3}{2}}$.

We have $8\le A\le 64$, so $8\le (r+1{)}^{\frac{3}{2}}\le 64$. Taking the cube root of all three sides gives $2\le (r+1{)}^{\frac{1}{2}}\le 4$, and squaring gives $4\le r+1\le 16$. Finally, subtract $1$ to find $3\le r\le 15$. In interval notation, this is $\boxed{[3,15]}$.