What is the vertices ((x+6)^2)/9 =((y-2)^2)/(16)+1 ?

The vertices ((x+6)^2)/9 =((y-2)^2)/(16)+1 is (-3,2),(-9,2)

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vertices ((x+6)^2)/9 =((y-2)^2)/(16)+1

vertices

\frac{( x + 6 ) ^{2}}{9} = \frac{( y - 2 ) ^{2}}{16} + 1

(- 3, 2), (- 9, 2)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 6 ) ^{2}}{9} = \frac{( y - 2 ) ^{2}}{16} + 1 :

Right-left Hyperbola with

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

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

Refine

(- 3, 2), (- 9, 2)

\frac{x ^{2}}{27} - \frac{y ^{2}}{72} = 1

\frac{y ^{2}}{3} - \frac{x ^{2}}{1} = 1

4 x^{2} - 5 y^{2} + 30 y = 65

f (x) f (x) = x^{2} + 5 x + 6

\frac{( x + 4 ) ^{2}}{9} - \frac{( y - 8 ) ^{2}}{49} = 1

What is the vertices ((x+6)^2)/9 =((y-2)^2)/(16)+1 ?
The vertices ((x+6)^2)/9 =((y-2)^2)/(16)+1 is (-3,2),(-9,2)