Asymptotes for hyperbolas q2 problem :S

[Chadlee88]

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


Thanks HallsofIvy for helpin me with the previous asymptote hyperbola prob

 

[WigneRacah]

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 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

(x+1)^2- 4 y^2 = 0 \Rightarrow (x +1 - 2 y) (x +1 + 2 y) =0

which implies

x+1 -2y=0

or

x + 1 + 2 y =0

(the stright line equations).

If the equation was

(x+1)^2-4y^2=1

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