Horizontal asymptote means that it approches a certain value of #y# and tends to make a horizontal line as it approaches (limit involvement) infinite values of x (whether positive or negative). Therefore, the two possible asymptotes of the function must be constant functions and can be calculated with the following way:
Possible horizontals
#lim_(x->oo)f(x)=k_1#
#lim_(x->-oo)f(x)=k_2#
If #k_1# or #k_2# are not real numbers then there are no asymptotes.
Solution
#f(x)=(3x^2+2x-1)/(x+1)#
Solving the numerator:
#Δ=b^2-4*a*c=2^2-4*3*(-1)=16#
#x=(-b+-sqrt(Δ))/(2*a)=(-2+-sqrt(16))/(2*3)=(-2+-4)/6#
The two solutions are #x=1/3# and #x=-1#
So the numerator can be written as follows:
#3x^2+2x-1=3(x-1/3)(x+1)#
The two limits are now easier to find:
#k_1=lim_(x->oo)f(x)=lim_(x->oo)(3x^2+2x-1)/(x+1)=#
#=lim_(x->oo)(3(x-1/3)(x+1))/(x+1)=lim_(x->oo)3(x-1/3)=oo#
#k_2=lim_(x->-oo)f(x)=lim_(x->-oo)(3x^2+2x-1)/(x+1)=#
#=lim_(x->-oo)(3(x-1/3)(x+1))/(x+1)=lim_(x->-oo)3(x-1/3)=-oo#
Since neither #k_1# nor #k_2# are real numbers, there are no horizontal asymptotes.
