How do you find the vertical, horizontal or slant asymptotes for #f(x)=(x+2)/sqrt(6x^2+5x+4) #?

Note that:

#6x^2+5x+4#

is in standard quadratic form with #a=6#, #b=5# and #c=4#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (color(blue)(5))^2-4(color(blue)(6))(color(blue)(4)) = 25-96 = -71#

Since #Delta < 0#, this quadratic has no real zeros and since its leading coefficient is positive, it always takes positive values.

So the denominator #sqrt(6x^2+5x+4)# of #f(x)# is always well defined and non-zero.

So #f(x)# has no vertical asymptotes.

Note that:

#lim_(x->+oo) f(x) = lim_(x->+oo) (x+2)/sqrt(6x^2+5x+4)#

#color(white)(lim_(x->+oo) f(x)) = lim_(x->+oo) (1+2/x)/sqrt(6+5/x+4/x^2)#

#color(white)(lim_(x->+oo) f(x)) = 1/sqrt(6)#

#color(white)(lim_(x->+oo) f(x)) = sqrt(6)/6#

#lim_(x->-oo) f(x) = lim_(x->-oo) (x+2)/sqrt(6x^2+5x+4)#

#color(white)(lim_(x->-oo) f(x)) = lim_(x->+oo) (-x+2)/sqrt(6x^2-5x+4#

#color(white)(lim_(x->-oo) f(x)) = lim_(x->+oo) (-1+2/x)/sqrt(6-5/x+4/x^2)#

#color(white)(lim_(x->-oo) f(x)) = -1/sqrt(6)#

#color(white)(lim_(x->-oo) f(x)) = -sqrt(6)/6#

So #f(x)# has horizontal asymptotes #y = +-sqrt(6)/6#

That leaves no opportunity for slant asymptotes.

graph{(x+2)/sqrt(6x^2+5x+4) [-22, 22, -2.52, 2.48]}