How do you graph #y=4/(x-3)+2# using asymptotes, intercepts, end behavior?

1 Answer
Jul 13, 2018

Answer:

Vertical asymptote at #x=3#, horizontal asymptote at
#y=2#, x intercept is at #(1,0)#, y intercept is at #(0,2/3)#.
End behavior : # x-> -oo , y-> 2 and x-> oo, y-> 2#

Explanation:

# y= 4/(x-3)+2#

Vertical asymptote at #x-3=0 or x =3 #

When # x-> 3^-, y-> -oo and x-> 3^+, y-> oo#

Since denominator's degree is higher than that of numerator,

horizontal asymptote is at #y=0+2 or y=2 #

y intercept is at # x=0 :. y= 4/(0-3)+2=-4/3+2 or y=2/3#

x intercept is at #y=0 :. 0 = 4/(x-3)+2 or -2 =4/(x-3)# or

#-2(x-3)=4 or 2 x = 6-4 or 2 x =2 or x=1#

End behavior : When # x-> -oo , y-> 2 and x-> oo, y-> 2#

graph{(4/(x-3))+2 [-40, 40, -20, 20]} [Ans]