What are the vertical and horizontal asymptotes of #f(x)=5/((x+1)(x-3))#?

1 Answer
Mar 5, 2018

Answer:

#"vertical asymptotes at "x=-1" and "x=3#
#"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+1)(x-3)=0#

#rArrx=-1" and "x=3" 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)=(5/x^2)/(x^2/x^2-(2x)/x^2-3/x^2)=(5/x^2)/(1-2/x-3/x^2)#

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

#rArry=0" is the asymptote"#
graph{5/((x+1)(x-3)) [-10, 10, -5, 5]}