Question

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

Answer 1

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.