How do you find the asymptote and holes for y=(x-6)/(x^2+5x+6)y=x−6x2+5x+6?
1 Answer
May 28, 2016
vertical asymptotes x = -3 , x = -2
horizontal asymptote y = 0
Explanation:
First step is to factorise the function.
rArr(x-6)/((x+2)(x+3))⇒x−6(x+2)(x+3) Since there are no common factors on numerator/denominator this function has no holes.
Vertical asymptotes occur as the denominator of a rational tends to zero. To find the equation/s set the denominator equal to zero.
solve: (x+2)(x+3) = 0 → x = -3 , x= -2
rArrx = -3" and "x=-2" are the asymptotes"⇒x=−3 and x=−2 are the asymptotes Horizontal asymptotes occur as
lim_(xto+-oo),yto0 When the degree of the numerator < degree of the denominator as is the case here (numerator-degree 1 , denominator-degree 2 ) then the equation of the asymptote is always y = 0
graph{(x-6)/(x^2+5x+6) [-10, 10, -5, 5]}
