Question 741394
I will tell you just this much:


It appears to not have any hole, since the numerator is not perfectly divisible by the denominator.  There is a vertical asymptote at {{{x=-3}}}, because that is where the function is undefined, since denominator would be zero.


Try performing the division.  There will be a remainder.  The quotient portion is the slant asymptote.  Use synthetic division:


-3____|____1_____-2_____5
______|___________-3____15
__________________________
______|____1_____-5_____20
_________ ___ x-5_________remainder 20


The slant asymptote would be {{{ y=x-5}}}