SOLUTION: Find the vertices of the hyperbola defined by this equation: (x+3)^2/1 - (y-2)^2/16 = 1. Pleas help How do I solve this?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the vertices of the hyperbola defined by this equation: (x+3)^2/1 - (y-2)^2/16 = 1. Pleas help How do I solve this?      Log On


   

Question 391346: Find the vertices of the hyperbola defined by this equation: (x+3)^2/1 - (y-2)^2/16 = 1. Pleas help How do I solve this?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi,         



Standard Form of an Equation of an Hyperbola opening left and right is  

%28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1 

where Pt(h,k) is a center  with vertices 'a' units right and left of center.

 (x+3)^2/1 - (y-2)^2/16 = 1

 Center is (-3,2)

 Vertices  (-4,2) and )-2,2)