# How do you find the asymptotes for  (2x-10)/(x^2+3x+2)?

Apr 9, 2016

vertical asymptotes x = -1 , x =- 2
horizontal asymptote y = 0

#### Explanation:

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

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

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

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

When the degree of the numerator < degree of denominator , as is the case here then the equation is always y = 0

Here is the graph of the function.
graph{(2x-10)/(x^2+3x+2) [-10, 10, -5, 5]}