Question

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

Answer 1

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]}