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

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

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

vertices

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

(6, 2), (0, 2)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Refine

(6, 2), (0, 2)

\frac{( x + 1 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{8} = 1

\frac{x ^{2}}{60} - \frac{y ^{2}}{40} = 1

\frac{x ^{2}}{36} - \frac{y ^{2}}{25} = 1

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

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

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