How do you find the vertical asymptotes and holes of #f(x)=1/(x^2+5x+6)#?

1 Answer
Jun 17, 2018

Answer:

#"vertical asymptotes at "x=-3" and "x=-2#
#"horizontal asymptote at "y=0#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

#"solve "x^2+5x+6=0rArr(x+3)(x+2)=0#

#x=-3" and "x=-2" are the asymptotes"#

#"horizontal asymptotes occur as"#

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

#"divide terms on numerator/denominator by the "#
#"highest power of x that is "x^2#

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

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

#y=0" is the asymptote"#

#"holes occur when a common factor is removed from"#
#"the numerator/denominator. This is not the case here"#
#"hence there are no holes"#
graph{1/(x^2+5x+6) [-10, 10, -5, 5]}