f(x)=(3x+5)/(2x-11)f(x)=3x+52x−11
D(f)=(-oo,11/2)uuu(11/2,oo)D(f)=(−∞,112)⋃(112,∞)
Since this function isn't defined in x=11/2x=112 It has an asymptote there.
Y intercept: only when x=0
y=(3*0+5)/(2*0-11)=-5/11y=3⋅0+52⋅0−11=−511
[0,-5/11][0,−511]
X intercept> onlz when y=0
0=(3x+5)/(2x-11)0=3x+52x−11
0*2x-11=(3x+5)/(2x-11)*2x-110⋅2x−11=3x+52x−11⋅2x−11
0=3x+50=3x+5
-5/3=x−53=x
[-5/3,0][−53,0]
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
