(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.  

Free points baby. free points baby. free points baby. hell yeah!