What is the vertices (x^2)/4-((y+1)^2)/(16)=1 ?

The vertices (x^2)/4-((y+1)^2)/(16)=1 is (2,-1),(-2,-1)

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vertices (x^2)/4-((y+1)^2)/(16)=1

vertices

\frac{x ^{2}}{4} - \frac{( y + 1 ) ^{2}}{16} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{4} - \frac{( y + 1 ) ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (0, - 1), a = 2, b = 4

(0 + 2, - 1), (0 - 2, - 1)

Refine

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

2 x^{2} - 2 y^{2} - 20 x + 16 y + 16 = 0

\frac{x ^{2}}{144} - \frac{y ^{2}}{25} = 1

9 y^{2} - x^{2} + 2 x + 54 y + 62 = 0

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 5 ) ^{2}}{9} = 1

9 x^{2} - 4 y^{2} - 54 x - 8 y + 113 = 0

What is the vertices (x^2)/4-((y+1)^2)/(16)=1 ?
The vertices (x^2)/4-((y+1)^2)/(16)=1 is (2,-1),(-2,-1)