How do you graph #y=(5x+3)/(-x+10)# using asymptotes, intercepts, end behavior?

1 Answer
Aug 6, 2018

Answer:

Vertical asymptote: #x=10#, horizontal asymptote: #y=-5#
x intercept: #x = -0.6# , y intercept: #x = 0.3#, end behavior: #y-> -5 # as #x -> -oo and y-> -5 # as #x -> oo#

Explanation:

#y= (5 x+3)/(-x+10# , Vertical asymptote occur when denominator

is zero. #-x+10=0 :. x= 10; lim(x->10^(-) y -> oo #

#lim (x->10^+) y - > -oo #. Vertical asymptote is #x=10#

Horizontal asymptote: #lim (x->-oo) ; y =-5/1=-5 #

#y= (5+(3/x))/(-1+(10/x)) , x -> +- oo , y -> -5 #

Horizontal asymptote is at # y=-5#

x intercept: Putting #y=0# in the equation we get,

#5 x +3= 0 or 5 x =-3 or x = -0.6 or (-0.6,0)# or

y intercept: Putting #x=0# in the equation we get,

#y=3/10= 0.3# or (0,0.3)#

End behavior: #y-> -5 # as #x -> -oo# and

#y-> -5 # as #x -> oo#

graph{(5 x+3)/(-x+10) [-80, 80, -40, 40]}[Ans]