How do you graph #f(x)=(2x)/(3x-1)# using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Feb 24, 2018

#x=1/3# is the vertical asymptote.

#y=2/3# is the horizontal asymptote.

#x=0# is the x-intercept.

#y=0# is the y intercept.

Explanation:

The vertical asymptote is found when #f(x)# tends to infinity. #f(x)# normally tends to infinity when the denominator tends to #0#.

So here:

#3x-1=0#

#3x=1#

#x=1/3# is the vertical asymptote.

For the horizontal asymptote, we use the degrees of the numerator and the denominator. Say #m# is the former and #n# the latter. If:

#m>n#, then there is no horizontal asymptote, only a slant.
#m=n#, the horizontal asymptote is at the quotient of the leading coefficient of the numerator and denominator
#m<##n#, the asymptote is at #y=0#.

Here, #m=1# and #n=1#. So #m=n#.

We must divide the leading coefficients of the numerator #(2)# and the denominator (#3#).

#y=2/3# is the horizontal asymptote.

The x-intercept is found when #f(x)=0#. Here,

#(2x)/(3x-1)=0#

#2x=0#

#x=0# is the x-intercept.

The y-intercept is the answer to #f(0)#. Inputting:

#(2*0)/(3*0+1)#

#0/1#

#y=0# is the y intercept.