What is the vertices ((x+1)^2)/9-((y-1)^2)/8 =1 ?

The vertices ((x+1)^2)/9-((y-1)^2)/8 =1 is (2,1),(-4,1)

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vertices ((x+1)^2)/9-((y-1)^2)/8 =1

vertices

(2, 1), (- 4, 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (- 1, 1), a = 3, b = 22

(- 1 + 3, 1), (- 1 - 3, 1)

Refine

(2, 1), (- 4, 1)

What is the vertices ((x+1)^2)/9-((y-1)^2)/8 =1 ?
The vertices ((x+1)^2)/9-((y-1)^2)/8 =1 is (2,1),(-4,1)