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

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

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

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 1, 1), b = 3, a = 2

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

Refine

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

\frac{x ^{2}}{6.25} + \frac{y ^{2}}{4.6875} = 1

\frac{x ^{2}}{15} + \frac{y ^{2}}{45} = 1

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

\frac{x ^{2}}{25} + \frac{( y + 3 ) ^{2}}{49} = 1

3 x^{2} + 2 y^{2} + 6 x - 4 y + 5 = 18

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