How do you find the asymptotes for y = (3-4x )/( 6x + sqrt(4x^2-3x-5))y=3−4x6x+√4x2−3x−5?

1 Answer
A. S. Adikesavan
Dec 6, 2016

Horizontal: y=-2/3y=−23. The graph is real for (-sqrt89+3)/8<=x<=(sqrt89+3)/8−√89+38≤x≤√89+38.

Explanation:

To make y real, (-sqrt89+3)/8<=x<=(sqrt89+3)/8−√89+38≤x≤√89+38.

y =(-4+3/x)/(6+sqrt(4-3/x-1/x^2)) to -2/3y=−4+3x6+√4−3x−1x2→−23, as x to +-oox→±∞.

Look at the dead ends x = (3+-sqrt 89)/8= 1.554 and -0.804x=3±√898=1.554and−0.804, nearly.

graph{y(6x+sqrt(4x^2-3x-5))+4x-3=0 [-5, 5, -2.5, 2.5]}

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