You can put this solution on YOUR website! The equation of hyperbola H is
(x+4)^2/64 - (y+3)^3/16 = 1
What are the asymptotes?
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Given hyperbola has a horizontal transverse axis.
Its standard form of equation:
For given hyperbola:
center: (-4,-3)
a^2=64
a=8
b^2=16
b=4:
asymptotes are equations of straight lines of standard form: y=mx+b, m=slope, b=y-intercept
slopes=±b/a=±4/8=±1/2
asymptotes go thru center(-4,-3)
..
Equation of asymptote with slope<0
y=-x/2+b
solve for b using coordinates of center
-3=-(-4)/2+b
b=-5
equation:y=-x/2-5
..
Equation of asymptote with slope>0
y=x/2+b
solve for b using coordinates of center
-3=-4)/2+b
b=-1
equation:y=x/2-1
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