Question

How do you find the asymptote and holes for `y=(x-6)/(x^2+5x+6)`?

Answer 1

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))`

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"`

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]}