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.
