What are the asymptotes of #f(x)=-x/((2x-3)(x-7)) #?

Question

What are the asymptotes of #f(x)=-x/((2x-3)(x-7))  #?

Charles Anctil · Accepted Answer

#"vertical asymptotes at "x=3/2" and "x=7#

#"horizontal asymptote at " y=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 values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes. #"solve " (2x-3)(x-7)=0# #rArrx=3/2" and " x=7" are the asymptotes"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" ( a constant)"# #f(x)=-x/(2x^2-17x+21)# divide terms on numerator/denominator by the highest power of x, that is #x^2# #f(x)=-(x/x^2)/((2x^2)/x^2-(17x)/x^2+(21)/x^2)=-(1/x)/(2-17/x+(21)/x^2# as #xto+-oo,f(x)to-0/(2-0+0)# #rArry=0" is the asymptote"# graph{-(x)/((2x-3)(x-7)) [-10, 10, -5, 5]}

Adam Adkins · Answer

undefined The asymptotes of f(x)=-x/((2x-3)(x-7)) are vertical asymptotes at x=3/2 and x=7, and a horizontal asymptote at y=0.