SOLUTION: I need help putting this equation into standard form, and if it is a parabola I have to identify the vertex and focus. If its and ellipse or hyperbola I have to identify the vertic

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: I need help putting this equation into standard form, and if it is a parabola I have to identify the vertex and focus. If its and ellipse or hyperbola I have to identify the vertic      Log On


   

Question 895586: I need help putting this equation into standard form, and if it is a parabola I have to identify the vertex and focus. If its and ellipse or hyperbola I have to identify the vertices and foci.
X^2+12x+y+41=0
I have gotten to (x-6)^2= (y+5) I need help completing the rest of it and identifying all of the things above please.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2B12x%2By%2B41=0
x%5E2%2B12x%2B36%2By%2B41=36
%28x%2B6%29%5E2%2By%2B41=36
y=-%28x%2B6%29%5E2-5
Parabola with vertex (-6,-5) opens downwards.
Focus
y%2B5=-%28x%2B6%29%5E2
-%28y%2B5%29=%28x%2B6%29%5E2
4p=-1
p=-1%2F4
Focus : (-6,-5-1/4)=(-6,-21/4)
graph%28300%2C300%2C-10%2C2%2C-10%2C2%2C-%28x%2B6%29%5E2-5%29