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

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

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

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (1, - 3), a = 2, b = 4

(1 + 2, - 3), (1 - 2, - 3)

Refine

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

\frac{( y + 1 ) ^{2}}{25} - \frac{( x + 4 ) ^{2}}{64} = 1

x^{2} - y^{2} - 6 x = 16

\frac{( x + 4 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

16 y^{2} - x^{2} + 2 x + 64 y + 47 = 0

\frac{x ^{2}}{45} - \frac{y ^{2}}{360} = 1

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