# How do you find the Vertical, Horizontal, and Oblique Asymptote given x /( 4x^2+7x-2)?

Sep 15, 2016

vertical asymptotes at x = -2 , x $= \frac{1}{4}$
horizontal asymptote at y = 0

#### Explanation:

The denominator of the function cannot be zero as this would make the function undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: $4 {x}^{2} + 7 x - 2 = 0 \Rightarrow \left(4 x - 1\right) \left(x + 2\right) = 0$

$\Rightarrow x = - 2 \text{ and " x=1/4" are the asymptotes}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by the highest power of x that is ${x}^{2}$

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

as $x \to \pm \infty , f \left(x\right) \to \frac{0}{4 + 0 - 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1 ,denominator-degree 2 ) Hence there are no oblique asymptotes.
graph{x/(4x^2+7x-2) [-10, 10, -5, 5]}