# How do you find the asymptotes for R (x) = (7x) /( 4x^2 -1)?

Dec 9, 2016

The vertical asymptotes are $x = - \frac{1}{2}$ and $x = \frac{1}{2}$
No slant asypmtote.
The horizontal asymptote is $y = 0$

#### Explanation:

we use ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

Let's factorise the denominator

$4 {x}^{2} - 1 = \left(2 x + 1\right) \left(2 x - 1\right)$

The domain of $R \left(x\right)$ is ${D}_{R} \left(x\right) = \mathbb{R} - \left\{- \frac{1}{2} , \frac{1}{2}\right\}$

As we cannot divide by $0$, $x \ne - \frac{1}{2}$ and $x \ne \frac{1}{2}$

The vertical asymptotes are $x = - \frac{1}{2}$ and $x = \frac{1}{2}$

The degree of the numerator is $<$ than the degree of the denominator, there is no slant asymptotes.

To calculate the limits as $x \to \pm \infty$, we take the terms of highets degree in the denominator

${\lim}_{x \to - \infty} R \left(x\right) = {\lim}_{x \to - \infty} \frac{7 x}{4 {x}^{2}} = {\lim}_{x \to - \infty} \frac{7}{4 x} = {0}^{-}$

${\lim}_{x \to + \infty} R \left(x\right) = {\lim}_{x \to + \infty} \frac{7 x}{4 {x}^{2}} = {\lim}_{x \to + \infty} \frac{7}{4 x} = {0}^{+}$

The horizontal asymptote is $y = 0$

graph{7x/(4x^2-1) [-6.24, 6.244, -3.12, 3.12]}