#f(x)= x+ (5 x)/(x-1)= (x^2-x+5 x)/(x-1)# or
# f(x)=(x^2+4 x)/(x-1)# Vertical asymptote occur when
denominator is zero.
# :. x-1=0 :. x= 1; lim(x->1^-) ; y -> -oo #
#lim (x->1^+); y -> +oo # numerator's degree is greater
(by a margin of 1), then,we have a slant asymptote which is
found by long division.
# f(x)=(x^2+4 x)/(x-1)= (x+5) + 5/(x-1)# , therefore, slant
asymptote is #y= x+5#
#y# intercept: Putting #x=0# in the equation we get,
#y=0 , x # intercepts: Putting #y=0# in the equation
we get , #x^2+4 x or x(x+4)=0 or x =0 , x=-4 #
End behavior: #y-> -oo # as #x -> -oo# and
#y-> +oo # as #x -> oo#
graph{x+ 5x/(x-1) [-80, 80, -40, 40]} [Ans]
