SOLUTION: I need to know the vertex, axis of symmetry, roots, domain, range for the following problem Y=2(x+2)^2-5 i also need to know if it opens up or down and if you can graph it that w

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: I need to know the vertex, axis of symmetry, roots, domain, range for the following problem Y=2(x+2)^2-5 i also need to know if it opens up or down and if you can graph it that w      Log On


   

Question 911616: I need to know the vertex, axis of symmetry, roots, domain, range for the following problem
Y=2(x+2)^2-5
i also need to know if it opens up or down and if you can graph it that would be greatly appreciated.
Found 2 solutions by Fombitz, ewatrrr:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
It's in vertex form.
(-2,-5) is the vertex.
The axis of symmetry is x=-2.
Domain is all real x values, (-infinity,infinity).
Since 2%3E0, the parabola opens upwards and the vertex value is the minimum.
The range is [-5,infinity).

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
Need to Know: the vertex form of a Parabola opening up(a>0) or down(a<0):
y=a%28x-highlight%28h%29%29%5E2+%2Bk where(h,k) is the vertex and x = h is the Line of Symmetry
Y= highlight_green%282%29(x+2)^2-5
Upward a = 2 > 0
V(-2,-5), axis of symmetry: x = -2