SOLUTION: Verticies @ (1/2, -3), (-9/2, -3); Asymptotes y+3= =+&- 6/5 (x+2) Find the equation and show steps :)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Verticies @ (1/2, -3), (-9/2, -3); Asymptotes y+3= =+&- 6/5 (x+2) Find the equation and show steps :)      Log On


   

Question 872077: Verticies @ (1/2, -3), (-9/2, -3); Asymptotes y+3= =+&- 6/5 (x+2)
Find the equation and show steps :)
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
Verticies (1/2, -3), (-9/2, -3) along y = -3, C( -2, -3)
Hyperbola opening right and left
+%28x%2B2%29%5E2%2Fa%5E2+-+%28y%2B3%29%5E2%2Fb%5E2+=+1
m = ± b/a = ± 6/5
+%28x%2B2%29%5E2%2F5%5E2+-+%28y%2B3%29%5E2%2F6%5E2+=+1
+%28x%2B2%29%5E2%2F25+-+%28y%2B3%29%5E2%2F36+=+1