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

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

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

vertices

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

(6, 0), (0, 0)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Refine

(6, 0), (0, 0)

\frac{y ^{2}}{49} - \frac{x ^{2}}{64} = 1

4 x^{2} - 3 y^{2} - 8 x - 8 = 0

\frac{x ^{2}}{2} - \frac{y ^{2}}{3} = 1

- 4 x^{2} + y^{2} + 16 x + 12 y + 16 = 0

\frac{y ^{2}}{81} - \frac{x ^{2}}{64} = 1

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