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

The vertices ((x+1)^2)/9-((y-3)^2)/(16)=1 is (2,3),(-4,3)

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vertices ((x+1)^2)/9-((y-3)^2)/(16)=1

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (- 1, 3), a = 3, b = 4

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

Refine

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

\frac{x ^{2}}{49} - y^{2} = 1

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

\frac{x ^{2}}{81} - \frac{y ^{2}}{144} = 1

\frac{y ^{2}}{36} - \frac{x ^{2}}{16} = 1

64 x^{2} - 100 y^{2} = 6400

What is the vertices ((x+1)^2)/9-((y-3)^2)/(16)=1 ?
The vertices ((x+1)^2)/9-((y-3)^2)/(16)=1 is (2,3),(-4,3)