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

1 Answer
Apr 9, 2016

Answer:

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 #

#rArr x = -1" and " x = -2" are the asymptotes "#

Horizontal asymptotes occur as #lim_(xto+-oo) f(x) 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]}