Question

identify the vertex and the axis of symmetry of the graph for the function y=3(x+2)^2
a.vertex(2,-3
axis of symmetry x=2
b.vertex(-2,-3)
axis of symmetry x=-2
c.vertex(2,3)
axis of symmetry x=2
d.vertex(-2,3)
axis of symmetry x=-2

identify the maximum or minimum value and the domain and range of the graph of the function y=2(x-3)^2-4.
a.minimum value -4
domain all real numbers
range all real numbers_>_-4
b.max value 4
domain all real numbers
range all real numbers_<_4
c. max value -4
domain all real numbers_<_-4
range all real numbers
d. minimum value 4
domain all real numbers_>_4
range all real numbers

Answer 1

for y = 3(x+2)^2
the vertex is (-2,0) and the axis of symmetry is x = -2
According to the choices, you probably forgot to add/subtract at the end.
if it is y = 3(x+2)^2 + c
then the vertex is (-2,c)

for the 2nd, y = 2(x-3)^2 - 4
the vertex is (3,-4) and it opens upwards,
So the minimum value is -4
the domain is any real number
and the range is y ≥-4

Here is how Wolfram graphed the 2nd, all you have to do is change the equation to graph the first.

http://www.wolframalpha.com/input/?i=y%3D2%28x-3%29%5E2-4