What is the vertices ((x+1)^2)/(25)-((y+4)^2)/(64)=1 ?

The vertices ((x+1)^2)/(25)-((y+4)^2)/(64)=1 is (4,-4),(-6,-4)

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vertices ((x+1)^2)/(25)-((y+4)^2)/(64)=1

vertices

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

(4, - 4), (- 6, - 4)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (- 1, - 4), a = 5, b = 8

(- 1 + 5, - 4), (- 1 - 5, - 4)

Refine

(4, - 4), (- 6, - 4)

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

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

What is the vertices ((x+1)^2)/(25)-((y+4)^2)/(64)=1 ?
The vertices ((x+1)^2)/(25)-((y+4)^2)/(64)=1 is (4,-4),(-6,-4)