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

1 Answer
Dec 6, 2016

Answer:

Horizontal: #y=-2/3#. The graph is real for #(-sqrt89+3)/8<=x<=(sqrt89+3)/8#.

Explanation:

To make y real, #(-sqrt89+3)/8<=x<=(sqrt89+3)/8#.

#y =(-4+3/x)/(6+sqrt(4-3/x-1/x^2)) to -2/3#, as #x to +-oo#.

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

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