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

The vertices ((x+3)^2)/4+((y-3)^2)/(16)=1 is (-3,7),(-3,-1)

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

vertices

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

(- 3, 7), (- 3, - 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

(- 3, 7), (- 3, - 1)

\frac{( x - 1 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{25} = 1

2 \frac{( x - 6 ) ^{2}}{3} + 7 \frac{( y + 4 ) ^{2}}{4} = 1

\frac{x ^{2}}{96} + \frac{y ^{2}}{100} = 1

\frac{( x + 4 ) ^{2}}{24} + \frac{( y + 2 ) ^{2}}{49} = 1

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

What is the vertices ((x+3)^2)/4+((y-3)^2)/(16)=1 ?
The vertices ((x+3)^2)/4+((y-3)^2)/(16)=1 is (-3,7),(-3,-1)