You can put this solution on YOUR website! Graph the hyperbola given by the equation 4x^2+8x-y^2-2y-1=0.
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4x^2+8x-y^2-2y-1=0
completing the squares:
4(x^2+2x+1)-(y^2+2y+1)=1+4-1=4
4(x+1)^2-(y+1)^2=4
(x+1)^2/1-(y+1)^2/4=1
This is a hyperbola with horizontal transverse axis. (opens sideways)
Standard form: (x-h)^2/a^2-(y-k)^2/b^2=1
For given hyperbola:
center: (-1,-1)
a^2=1
a=1
length of transverse axis=2a=2
Vertices: (-1±a,-1)=(-1±1,-1)
..
b^2=4
b=2
length of conjugate axis=2b=4
..
c^2=a^2+b^2=1+4=5
c√5=2.24
Foci: (-1±√5,-1)
..
Two Asymptotes:y=±mx+b
y=mx+b
m=b/a=2/1=2
Solving for b using (x,y) coordinates of center
-1=2(-1)+b
b=1
y=2x+1
..
y=-mx+b
-1=-2(-1)+b
b=-3
y=-2x-3
see graph as a visual check on answers above
..
y=±(-4+4(x+1)^2)^.5-1
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