# How do you find the vertical, horizontal or slant asymptotes for (x )/(4x^2+7x-2) ?

Nov 29, 2016

The vertical asymptotes are $x = - 2$ and $x = \frac{1}{4}$
No slant asymptote
The horizontal asymptote is $y = 0$

#### Explanation:

Let's factorise the denominator

$4 {x}^{2} + 7 x - 2 = \left(4 x - 1\right) \left(x + 2\right)$

Let $f \left(x\right) = \frac{x}{4 {x}^{2} + 7 x - 2} = \frac{x}{\left(4 x - 1\right) \left(x + 2\right)}$

The domain of $f \left(x\right)$is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 2 , \frac{1}{4}\right\}$

As we cannot divide by $0$, $x \ne - 2$ and $x \ne \frac{1}{4}$

So, the vertical asymptotes are $x = - 2$ and $x = \frac{1}{4}$

As the degree of the numerator is $<$ than the degree of the denominator, there is no slant asymptote.

To calculate the limits as x tends to $\infty$, we take the terms of highest degree in the numerator and the denominator.

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{x}{4 {x}^{2}} = {\lim}_{x \to + \infty} \frac{1}{4 x} = {0}^{+}$

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{x}{4 {x}^{2}} = {\lim}_{x \to - \infty} \frac{1}{4 x} = {0}^{-}$

So, the horizontal asymptote is, $y = 0$
graph{x/(4x^2+7x-2) [-6.244, 6.243, -3.12, 3.123]}