What is the vertices ((x-3)^2)/9-((y+4)^2)/(25)=1 ?

The vertices ((x-3)^2)/9-((y+4)^2)/(25)=1 is (6,-4),(0,-4)

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vertices ((x-3)^2)/9-((y+4)^2)/(25)=1

vertices

\frac{( x - 3 ) ^{2}}{9} - \frac{( y + 4 ) ^{2}}{25} = 1

(6, - 4), (0, - 4)

Solution steps

(h + a, k), (h - a, k)

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{9} - \frac{( y + 4 ) ^{2}}{25} = 1 :

Right-left Hyperbola with

(h, k) = (3, - 4), a = 3, b = 5

(3 + 3, - 4), (3 - 3, - 4)

Refine

(6, - 4), (0, - 4)

10 y^{2} - 5 x^{2} = 50

\frac{x ^{2}}{9} - \frac{y ^{2}}{36} = 1

4 x^{2} - y^{2} = 4

2 x^{2} - 2 y^{2} - 20 x + 16 y - 16 = 0

- 9 x^{2} + 4 y^{2} = 36

What is the vertices ((x-3)^2)/9-((y+4)^2)/(25)=1 ?
The vertices ((x-3)^2)/9-((y+4)^2)/(25)=1 is (6,-4),(0,-4)