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

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

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

vertices

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

Right-left Hyperbola with

(h, k) = (3, - 4), a = 3, b = 2

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

Refine

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

+ \frac{( y - 2 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{16} = 1

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

4 x^{2} - y^{2} - 16 x - 6 y - 9 = 0

4 x^{2} - 9 y^{2} = 144

f (x) f (x) = 3 (x + 6) (x + 3)

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