#### Answer

There are two $x$-intercepts: $-1$ and $1$.
The y-intercept is $1$.
Refer to the graph below.

#### Work Step by Step

In order to find the $x$-intercept, we have to plug $y=0$ into the equation, and the $x$-coordinate will give us the intercept itself.
$y=-x^2+1$
$0=-x^2+1$
$x^2=1$
$x=\pm1$
The same thing should be done for the $y$-intercept, however, here $x=0$ should be plugged into the equation.
$y=-x^2+1$
$y=-0^2+1$
$y=1$
There are two $x$-intercepts: $-1$ and $1$.
The y-intercept is $1$.
Few other points that will help us plot the graph:
If $x=2$
$y=-2^2+1=-3$
The point $(2,-3)$ is on the graph.
If $x=-2$
$y=-(-2)^2+1=-3$
The point $(-2,-3)$ is on the graph.
If $x=3$
$y=-3^2+1=-8$
The point $(3,-8)$ is on the graph.