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

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

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

vertices

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

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

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

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

3 x^{2} + 7 y^{2} - 21 = 0

\frac{( x - 3 ) ^{2}}{25} + (y - 2)^{2} = 1

x^{2} + 4 y^{2} + 2 x - 12 y + 6 = 0

\frac{x ^{2}}{144} + \frac{y ^{2}}{225} = 1

\frac{( x + 6 ) ^{2}}{7} + \frac{( y - 3 ) ^{2}}{16} = 1

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