# How do you find vertical, horizontal and oblique asymptotes for (x+4)/(3x^2+5x-2)?

Apr 18, 2016

vertical asymptotes x = -2 , x $= \frac{1}{3}$
horizontal asymptote y = 0

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve :  3x^2 + 5x - 2 = 0 → (3x-1)(x+2) = 0

$\Rightarrow x = - 2 , x = \frac{1}{3} \text{ are the asymptotes }$

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} f \left(x\right) \to 0$

When the degree of the numerator < degree of the denominator , the equation is always y = 0 .

Oblique asymptotes occur when the degree of the numerator > degree of the denominator . This is not the case here hence there are no oblique asymptotes.
graph{(x+4)/(3x^2+5x-2) [-10, 10, -5, 5]}