What is the process to solve these:Vertex, y-int., x-int, graph= -(x+1)^2+1

The intercepts are found by puttin $gx=0(y-\int )$ and $y=0(x-\int )$ into the parabola equation: ${\displaystyle y=-{(x+1)}^2+1.}$
When $x=0$, $y=-1+1=0$, when $y=0$, ${\displaystyle {(x+1)}^2=1}$, so $x+1=+-1$, and $x=-2$ and 0.
That gives us 0 as y-int and -2 and 0 as x-ints. The graph is an inverted U shape where the arms of the U intersect the x-axis at -2 and 0, so the U lies to the left of the y axis.
The x value for the vertex is midway between the x intercepts at $x=-1$ and $y=1(-1,1)$, and this can also be seen in the term $(x+1)$ as the displacement from the origin.