How do you find the horizontal asymptote for #f(x) = (3x^2 + 2x - 1) / (x + 1)#?

1 Answer
Jan 2, 2016

Answer:

You evaluate the limits of the function as x approaches infinities (#-oo# and #oo#). If one of them is a real number #k#, than #y=k# is a horizontal asymptote.

There are no horizontal asymptotes in this function.

Explanation:

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.