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

Jan 2, 2016

You evaluate the limits of the function as x approaches infinities ($- \infty$ and $\infty$). 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

${\lim}_{x \to \infty} f \left(x\right) = {k}_{1}$

${\lim}_{x \to - \infty} f \left(x\right) = {k}_{2}$

If ${k}_{1}$ or ${k}_{2}$ are not real numbers then there are no asymptotes.

Solution

$f \left(x\right) = \frac{3 {x}^{2} + 2 x - 1}{x + 1}$

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 $x = \frac{1}{3}$ and $x = - 1$

So the numerator can be written as follows:

$3 {x}^{2} + 2 x - 1 = 3 \left(x - \frac{1}{3}\right) \left(x + 1\right)$

The two limits are now easier to find:

${k}_{1} = {\lim}_{x \to \infty} f \left(x\right) = {\lim}_{x \to \infty} \frac{3 {x}^{2} + 2 x - 1}{x + 1} =$
$= {\lim}_{x \to \infty} \frac{3 \left(x - \frac{1}{3}\right) \left(x + 1\right)}{x + 1} = {\lim}_{x \to \infty} 3 \left(x - \frac{1}{3}\right) = \infty$

${k}_{2} = {\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{3 {x}^{2} + 2 x - 1}{x + 1} =$
$= {\lim}_{x \to - \infty} \frac{3 \left(x - \frac{1}{3}\right) \left(x + 1\right)}{x + 1} = {\lim}_{x \to - \infty} 3 \left(x - \frac{1}{3}\right) = - \infty$

Since neither ${k}_{1}$ nor ${k}_{2}$ are real numbers, there are no horizontal asymptotes.