How do you identify all asymptotes or holes for #f(x)=x/(-4x^2-16x)#?

1 Answer
Jul 15, 2017

Answer:

#"vertical asymptote at "x=-4#
#"horizontal asymptote at "y=0#

Explanation:

#"simplify f(x) by factorising the denominator"#

#f(x)=cancel(x)/(-4cancel(x)(x+4))=-1/(4(x+4))#

#"the removal of the factor x from the numerator/denominator"#
#"indicates a hole at x = 0"#

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

#"solve "4(x+4)=0rArrx=-4" is the asymptote"#

#"horizontal asymptotes occur as "#

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

#"divide terms on numerator/denominator by x"#

#f(x)=-(1/x)/((4x)/x+16/x)=-(1/x)/(4+16/x)#

as #xto+-oo,f(x)to0/(4+0)#

#rArry=0" is the asymptote"#
graph{x/(-4x^2-16x) [-10, 10, -5, 5]}