What is the vertices ((x+3)^2)/(16)-((y+2)^2)/(36)=1 ?

The vertices ((x+3)^2)/(16)-((y+2)^2)/(36)=1 is (1,-2),(-7,-2)

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vertices ((x+3)^2)/(16)-((y+2)^2)/(36)=1

vertices

\frac{( x + 3 ) ^{2}}{16} - \frac{( y + 2 ) ^{2}}{36} = 1

(1, - 2), (- 7, - 2)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 3 ) ^{2}}{16} - \frac{( y + 2 ) ^{2}}{36} = 1 :

Right-left Hyperbola with

(h, k) = (- 3, - 2), a = 4, b = 6

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

Refine

(1, - 2), (- 7, - 2)

y^{2} - x^{2} = \frac{1}{36}

\frac{x ^{2}}{25} - \frac{y ^{2}}{81} = 1

\frac{( x - 3 ) ^{2}}{9} - \frac{( y + 4 ) ^{2}}{4} = 1

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

- 4 x^{2} + y^{2} = 4

What is the vertices ((x+3)^2)/(16)-((y+2)^2)/(36)=1 ?
The vertices ((x+3)^2)/(16)-((y+2)^2)/(36)=1 is (1,-2),(-7,-2)