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

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

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

vertices

(- 3, 4), (- 3, 0)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Refine

(- 3, 4), (- 3, 0)

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