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

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

Algebra.Com

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) (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

m = � b/a = � 6/5

RELATED QUESTIONS
Find the equation (hyperbola) if the asymptotes y - 3 = √13/6 (x-2) and focus at (9, (answered by greenestamps)

Find the equation (hyperbola) asymptotes y - 3 = √13/6 (x-2) and focus at (9, 3) (answered by MathLover1)

Find the equation asymptotes y - 3 = ± square root of 13/6 (𝑥 − 2) and... (answered by MathLover1)

x^2-y^2+6x+10y-17=0 A.) Write the equation in standard form. B.) Find the Center,... (answered by lwsshak3)

Find the coordinate of the vertices and foci and the equation of the asymptotes for the... (answered by Edwin McCravy,ikleyn)

Can someone please help me find the foci and asymptotes to (y-1)^2/9-(x-3)^2/4=1 (answered by lwsshak3)

Sketch the hyperbolas. Identify the center, asymptotes, vertices, and foci. (x-3)^2/9 -... (answered by lwsshak3)

1. Find the center 2. Slopes of Asymptotes Hyperbola (x-3)^2/9-(y+4)^2/16=1... (answered by lwsshak3)

Find the coordinate of the vertices and foci and the equation of the asymptotes for the... (answered by Edwin McCravy)