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

Mar 25, 2016

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

#### Explanation:

Vertical asymptotes occur when the denominator of a rational function tends to zero. To find the equation/s ,equate the denominator to zero.

solve:  4x^2 + 7x - 2 = 0 → (x + 2 )(4x - 1 ) = 0

$\Rightarrow x = - 2 , x = \frac{1}{4} \text{ are the asymptotes }$

Horizontal asymptotes occur as  lim_(x→±∞) f(x) → 0

If the degree of the numerator is less than the degree of the denominator as is the case here , then the equation of the asymptote is always y = 0 .

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here , hence there are no oblique asymptotes.

Here is the graph of the function.
graph{x/(4x^2+7x-2) [-10, 10, -5, 5]}