Question 586866: the vertices of the hyperbola
(x+2)^2 over 9 - (y-3)^2 over 25 = 1 are___
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the vertices of the hyperbola
(x+2)^2 over 9 - (y-3)^2 over 25 = 1
**
Standard form of equation for a hyperbola with horizontal transverse axis:
(x+h)^2/a^2-(y-k)^2/b^2=1, (h,k) being the (x,y) coordinates of the center.
For given equation:
(x+2)^2/9-(y-3)^2/25=1
center:(-2,3)
a^2=9
a=√9=3
Vertices=(-2±a,3)=(-2±3,3)=(-5,3) and (1,3)
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