#### Answer

The other point of intersection is $(\frac{3}{2}, \frac{5}{4})$
We can see the sketch below.

#### Work Step by Step

$y = x^2-1$
$\frac{dy}{dx} = 2x$
The slope of the graph at $x = -1$ is: $~~\frac{dy}{dx} =2(-1) = -2$
The slope of the normal line is $~~-(\frac{1}{-2}) = \frac{1}{2}$
We can find the equation of the normal line:
$(y-0) = \frac{1}{2}[x-(-1)]$
$y = \frac{1}{2}x+\frac{1}{2}$
We can find the points of intersection:
$x^2-1 = \frac{1}{2}x+\frac{1}{2}$
$2x^2-2 = x+1$
$2x^2-x-3 = 0$
$(x+1)(2x-3) = 0$
$x = -1, \frac{3}{2}$
When $x = \frac{3}{2},~~$ then $~~y = (\frac{3}{2})^2-1 = \frac{5}{4}$
The other point of intersection is $(\frac{3}{2}, \frac{5}{4})$
We can see the sketch below.