Question 755848
First examine what happens as x approaches but does not reach the value of -1.  x cannot be -1 because the denominator of the function would be zero, and that value is not allowed.  The graph has a vertical asymptote at {{{x=-1}}}.


Now perform the polynomial division which represents the function.  You produce a remainder.


x+1___|__x______-2
______|________________________
______|x^2______-x______-4
_______x^2______x
___________________
________0_______-2x______-4
________________-2x______-2
_______________________________
_______________________-6_____the remainder.


Therefore the equivalent function is {{{y=x-2-(6/(x+1))}}}
NOW, what happens to y as x tends toward infinity and what happens to y as x tends toward negative infinity?  The remainder approaches zero, and y approaches x-2.
***
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***Therefore, the SLANT ASYMPTOTE is {{{y=x-2}}}.


Please be aware, the division shown performed above is normal polynomial division, NOT synthetic division.  Easier for me to keep thoughts straight this way sometimes.