Question 1039476
The horizontal asymptote requires looking the degree of the first/degree of the second, here 2/3. The horizontal asymptote is 0.  As x gets very large, the leading term drives both the numerator and the denominator.  This becomes x^2/x^2, or 1/x, and as x gets large, this polynomial goes to 0.
Vertical asymptotes occur where the denominator = 0
x^3+5x^2-x-5
1===+5===-1===-5, synthetic division
1===6=====5====0
x=-1 is a root, so 1 is a potential vertical asymptote.
The quadratic quotient is x^2+6x+5, and that factors into (x+5)(x+1). Set those equal to zero, and -1 and -5 are potential asymptotes.  We need to be sure the numerator isn't 0 also at any of these points.
factor numerator and denominator 
(x+5)(x+3)/(x-1)(x+5)(x+1)
The x+5 divide out, so the function is (x+3)(x-1)(x+1), and there are asymptotes at +/- 1 and a hole at x=-5


{{{graph(300,200,-10,10,-10,10,(x^2+8x+15)/(x^2+5x^2-x-5))}}}