How do you graph #y=(-5x+2)/(4x+5)# using asymptotes, intercepts, end behavior?

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Answer 1 · 2017-03-04T18:17:27

graph{(-5x+2)/(4x+5) [-10, 10, -5, 5]}

vertical asymptote at #x= -1.25#

#y#-intercept: #(0, 0.4)#

#x#-intercept: #(-0.4, 0)#

as #x to oo, y to -1.25#

#y = (-5x+2)/(4x+5)#

asymptote:

#n/0 =# undefined

#therefore# there is a vertical asymptote when #4x+5=0#

#4x+5 = 0#
#4x=-5#
#x= -5/4 or -1.25#

#y#-intercept:

#y = (-5x+2)/(4x+5)#

when #x=0#, #y = 0-2/0+5#
#=2/5#
#=0.4#

#therefore y#-intercept = #(0, 0.4)#

#x#-intercept:

#0/n = 0#

for #y# to be #0#, #x= 0/n#

#therefore -5x+2 = 0#

#5x+2=0#
#5x=-2#
#x=-2/5 or -0.4#

#therefore x#-intercept #= (-0.4, 0)#

end behaviour:

levels out
as #x to oo, y to -1.25#