# How do you identify all asymptotes or holes and intercepts for #f(x)=x/(3x^2+5x)#?

##### 1 Answer

Jan 26, 2018

#### Answer:

#### Explanation:

#"factorise the denominator"#

#f(x)=cancel(x)/(cancel(x)(3x+5))=1/(3x+5)#

#"since the factor x has been cancelled from the numerator/"#

#"denominator this indicates a hole at"(0,1/5)# 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 "3x+5=0rArrx=-5/3" is the asymptote"#

#"horizontal asymptotes occur as"#

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

#"divide numerator/denominator by x"#

#f(x)=(1/x)/((3x)/x+5/x)=(1/x)/(3+5/x)#

#"as "xto+-oo,f(x)to0/(3+0)#

#rArry=0" is the asymptote"#

graph{1/(3x+5) [-10, 10, -5, 5]}