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

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

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

vertices

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

(2, - 3), (- 2, - 3)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, - 3), a = 2, b = 3

(0 + 2, - 3), (0 - 2, - 3)

Refine

(2, - 3), (- 2, - 3)

4 (y - 3)^{2} - 9 (x - 2)^{2} = 36

\frac{x ^{2}}{36} - \frac{( y - 10 ) ^{2}}{64} = 1

4 x^{2} - y^{2} + 32 x + 10 y + 38 = 0

4 x^{2} - y^{2} + 12 x + 5 = 0

\frac{y ^{2}}{36} - \frac{x ^{2}}{4} = 1

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