Question 747913
This parabola is in standard form - which is nice:

{{{
f(x)=(x-1)^2-2
}}}

{{{
f(x)=a(x-h)^2+k
}}}

{{{
V(h,k)
}}}



so our vertex is (1,-2)


to get the y-intercept set x=0, then



{{{f(0)=(0-1)^2-2=1-2=-1}}}


to get the x-intercept set f(x)=0, then


{{{
0=(x-1)^2-2
}}}

{{{
2=(x-1)^2
}}}



{{{
abs(x-1)=2
}}}


so that

x-1=2  or  x-1=-2
x=0  or  x=-1


x-intercepts:  {0, -1}


:)