(look at picture and show all work)) and do all 3 !

I'd feel more comfortable if you'd already done some of the work involved here and had then shared your own efforts.

To find the x-intercept(s) of a graph, set y = 0 and solve the resulting equation for x.  In f(x) = (x+3)^2 - 1, setting this = to 0 results in (x+3)^2 = 1, or 
x+3 = plus or minus 1.  Then either x = -2 or x = -4.  The x-intercepts are (-4,0) and (-2,0).  Any questions about this?

To find the y-intercept(s):  Let x=0 and solve the resulting eqn for y:
f(x) = (x+3)^2 - 1  =>  (3)^2 -1 = 8.  Then the y-int. is (0,8).

This function has a parabolic graph.  The vertex is (-3, -1) (notice how these values come straight from the function   f(x) = (x+3)^2 - 1  ).

Because the coefficient of (x+3)^2 is positive, the graph opens up and we have a minimum at the vertex, (-3,-1).

The second question involves some of the same concepts as does the first.  Do what you can and share your results.