How do you find the asymptotes for #(4x)/(x-3) #?

1 Answer
Apr 24, 2018

Answer:

#"vertical asymptote at "x=3#
#"horizontal asymptote at "y=4#

Explanation:

#"let " f(x)=(4x)/(x-3)#

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 "x-3=0rArrx=3" is the asymptote"#

#"Horizontal asymptotes occur as"#

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

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

#f(x)=((4x)/x)/(x/x-3/x)=4/(1-3/x)#

#"as "xto+-oo,f(x)to4/(1-0)#

#rArry=4" is the asymptote"#
graph{(4x)/(x-3) [-20, 20, -10, 10]}