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

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

(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)

\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

\frac{( y - 5 ) ^{2}}{4} - \frac{( x + 3 ) ^{2}}{9} = 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)