How do you find the vertical, horizontal or slant asymptotes for #f(x)= (x-5)/(x^2+6x+5)#?

1 Answer
Jun 6, 2016

Answer:

vertical asymptotes x = -5 , x = -1
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: #x^2+6x+5=0rArr(x+5)(x+1)=0#

#rArrx=-5,x=-1" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by #x^2#

#(x/x^2-5/x^2)/(x^2/x^2+(6x)/x^2+5/x^2)=(1/x-5/x^2)/(1+6/x+5/x^2#

as #xto+-oo,f(x)to0/(1+0+0)#

#rArry=0" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1 , denominator-degree 2 ). Hence there are no slant asymptotes.
graph{(x-5)/(x^2+6x+5) [-8.89, 8.89, -4.444, 4.445]}