Asymptotes for hyperbolas q2 problem :S

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SUMMARY

The equation (x+1)^2 - 4y^2 = 0 describes two straight lines intersecting at the point (-1, 0), not a hyperbola. The correct interpretation reveals that the equation can be factored into (x + 1 - 2y)(x + 1 + 2y) = 0, leading to the straight line equations x + 1 - 2y = 0 and x + 1 + 2y = 0, which serve as the asymptotes. In contrast, the equation (x+1)^2 - 4y^2 = 1 would represent a hyperbola with its vertex at (-1, 0).

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Does the equation (x+1)^2-4y^2 = 0 have asymptotote??
i graphed it and from the graph it does not look like a hyperbola because it seems to intersect at the point x = -1 y = 0 :frown:


Thanks HallsofIvy for helpin me with the previous asymptote hyperbola prob
 
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Chadlee88 said:
Does the equation (x+1)^2-4y^2 = 0 have asymptotote??
i graphed it and from the graph it does not look like a hyperbola because it seems to intersect at the point x = -1 y = 0 :frown: Thanks HallsofIvy for helpin me with the previous asymptote hyperbola prob

The equation you wrote doesn't describe a hyperbola but two straight lines intersecting in the point (-1, 0). In fact you can write

[tex](x+1)^2- 4 y^2 = 0 \Rightarrow (x +1 - 2 y) (x +1 + 2 y) =0[/tex]

which implies

[tex]x+1 -2y=0[/tex]

or

[tex]x + 1 + 2 y =0[/tex]

(the stright line equations).

If the equation was

[tex](x+1)^2-4y^2=1[/tex]

it described a hyperbola shifted back along the x-axis with vertex in (-1,0).
The previously stright line equations are the asymptotes.
 

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