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

1 Answer
Oct 14, 2017

No holes.
Vertical asymptote: #x=1#
Horizontal asymptote: #y=0#
No #x# intercepts.
#y#-intercept: #-2#

Explanation:

Denote #f(x)# as #(n(x))/(d(x)#

There are no holes since there are no common factors.

To find the vertical asymptote,
Solve #d(x)=0#
#rArr##x-1=0#
#x=1#

Therefore the vertical asymptote is #x=1#.

To find the horizontal asymptote,
Compare the leading degree of #n(x)# and #d(x)#.

For #n(x)#, the degree is #0#, because #x^0*2# gives #2#. Denote this as #color(turquoise)n#
For #d(x)#, the degree is #1# (since #x^1#). Denote this as #color(magenta)m#

When #n < m#, the #x#-axis (that is, #y=0#) is the horizontal asymptote.

To find the #x# intercept, plug in #0# for #y# and solve for #x#.
#rArr##0=2/(x-1)#
There are no #x# intercepts.

To find the #y# intercept, plug in #0# of #x# and solve for #y#.
#rArr##f(x)=2/(0-1)#
#f(x)=-2#
The #y#-intercept is #-2#.

graph{2/(x-1 [-100, 100, -5, 5]}