How do you graph #y=(3x+5)/(2x-11)# using asymptotes, intercepts, end behavior?

1 Answer
Dec 9, 2017

Answer:

See below

Explanation:

#f(x)=(3x+5)/(2x-11)#

#D(f)=(-oo,11/2)uuu(11/2,oo)#
Since this function isn't defined in #x=11/2# It has an asymptote there.

Y intercept: only when x=0
#y=(3*0+5)/(2*0-11)=-5/11#
#[0,-5/11]#

X intercept> onlz when y=0
#0=(3x+5)/(2x-11)#
#0*2x-11=(3x+5)/(2x-11)*2x-11#
#0=3x+5#
#-5/3=x#
#[-5/3,0]#

End bahaviour in #+oo#
#Lim_(xrarroo)(3x+5)/(2x-11)*(1/x)/(1/x)=Lim_(xrarroo)(3+5/x)/(2-11/x)=(3+0)/(2-0)=3/2#

End bahaviour in #-oo#(should be the same)
#Lim_(xrarr-oo)(3x+5)/(2x-11)*(1/x)/(1/x)=Lim_(xrarr-oo)(3+5/x)/(2-11/x)=(3-0)/(2+0)=3/2#

so another asymptote is: #y=3/2#

Since we know it's Reciprocal Function. There's another way to graph this function which is to rewrite f(x) to this formula:
#y=k/(x-a)+b#

#f(x)= [((3x+5))/((2x-11))*2/3]*3/2=[(6x+10)/(6x-33)]*3/2=[(6x-33+33+10)/(6x-33)]*3/2=[(6x-33)/(6x-33)+43/(6x-33)]*3/2=[1+43/(3(2x-11))]*3/2=3/2+(43*cancel3)/(cancel3*2(2x-11))=3/2+43/(4x-22)#
#y=43/(4x-22)+3/2#

#k=43>0# This means it is in 1. and 3. quadrant
1. asymptote: #y=3/2#
2. asymptote: #x=22/4=11/2#