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?

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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)   (Show Source): You can put this solution on YOUR website!

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

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)


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