What is the vertices ((x+3)^2)/4+((y-2)^2)/(16)=1 ?

The vertices ((x+3)^2)/4+((y-2)^2)/(16)=1 is (-3,6),(-3,-2)

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vertices ((x+3)^2)/4+((y-2)^2)/(16)=1

vertices

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{16} = 1

(- 3, 6), (- 3, - 2)

Solution steps

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

Calculate ellipse properties

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (- 3, 2), b = 4, a = 2

(- 3, 2 + 4), (- 3, 2 - 4)

Refine

(- 3, 6), (- 3, - 2)

\frac{x ^{2}}{49} + \frac{y ^{2}}{9} = 1

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

x^{2} + 9 y^{2} = 36

\frac{x ^{2}}{121} + (\frac{y}{1})^{2} = 1

\frac{( x + 4 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{16} = 1

What is the vertices ((x+3)^2)/4+((y-2)^2)/(16)=1 ?
The vertices ((x+3)^2)/4+((y-2)^2)/(16)=1 is (-3,6),(-3,-2)