How do you graph #y=4/(3x-6)+5# using asymptotes, intercepts, end behavior?

1 Answer
LM
Dec 20, 2016

graph{(4/(3x-6))+5 [-17.58, 22.42, -6.32, 13.68]}

asymptote: #x = 2#, vertical

#y#-intercept is at #(0,4.33)#

#x#-intercept is at #(1.73,0)#

#y<5# when #x<2#
#y>5# when #x>2#

Explanation:

asymptotes:

#n/0# = undefined

#n/0 +5# =undefined

asymptote: #4/(3x-6) = 4/0#

#3x-6 = 0#

#3x = 6#

#x = 2#

intercepts:

#y#:
#y#-intercept: #x = 0#

#y = 4/(3x-6) +5#

#= 4/-6 +5#

#=4 1/3 or 4.33 (3s.f.)#

#x#:
#x#-intercept: #y=0#

#4/(3x-6)+5=0#

#4/(3x-6) = -5#

#3x-6 = -0.8#

#3x = -0.8 + 6 = 5.2#

#x = 5.2/3 = 1.73(3s.f.)#

end behaviour:

#y<5# when #x<2#

e.g. #x =1.5#:

#4/(3x-6) +5 = 4/-1.5 +5#

#=8/3 + 5#

#= 7 2/3#

#y>5# when #x>2#

e.g. #x=2.1#:

#4/(3x-6)+5 = 4/0.3 +5#

#=40/3 + 5#

#=18 1/3#

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