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

The asymptotes of (x^2)/4-((y+1)^2)/(16)=1 is y=2x-1,\quad y=-2x-1

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asymptotes of (x^2)/4-((y+1)^2)/(16)=1

asymptotes

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

y = 2 x - 1, y = - 2 x - 1

Solution steps

y = \pm \frac{b}{a} (x - h) + k

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, - 1), a = 2, b = 4

y = \frac{4}{2} (x - 0) + (- 1), y = - \frac{4}{2} (x - 0) + (- 1)

Refine

y = 2 x - 1, y = - 2 x - 1

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 5 ) ^{2}}{9} = 1

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

y^{2} - x^{2} = 1

16 y^{2} - 32 y - 9 x^{2} - 18 x = 137

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

What is the asymptotes of (x^2)/4-((y+1)^2)/(16)=1 ?
The asymptotes of (x^2)/4-((y+1)^2)/(16)=1 is y=2x-1,\quad y=-2x-1