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Question 486326: (x-3)^2/9-(y-2)^2/16=1
I have tried everything and cant not figure out a couple of my college algebra questions. I am looking for the CENTER and VERTICES of this hyperbola. Please help me!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! (x-3)^2/9-(y-2)^2/16=1
I have tried everything and cant not figure out a couple of my college algebra questions. I am looking for the CENTER and VERTICES of this hyperbola.
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Given equation is that of a hyperbola with horizontal transverse axis of the standard form:
(x-h)^2/a^2-(y-k)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
Center: (3,2)
a^2=9
a=3
length of horizontal transverse axis=2a=6
Vertices are end points of transverse axis=(3±a,2)=(3±3,2)=(6,2) and (0,2)
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