How do you graph #y=-3/(x-4)-1# using asymptotes, intercepts, end behavior?

1 Answer
Apr 12, 2018

Answer:

vertical asymptote at #x=4#
horizontal asymptote at #y=-1#
#(0,-1/4)# and #(1,0)# are intercepts

Explanation:

Vertical Asymptote at #x=4#

We can see that there is a vertical asymptote at #x=4#.

When #x# is just slightly less than 4 (eg. 3.9999),

#x-4# is a very small negative number which makes

#1/(x-4)# a very large negative number which makes

#-1/(x-4)# a very large positive number which makes

#y# a very large positive number.

When #x# is just slightly more than 4 (eg. 4.0001),

#x-4# is a very small positive number which makes

#1/(x-4)# a very large positive number which makes

#-1/(x-4)# a very large negative number which makes

#y# a very large negative number.

End Behavior

When #x# is largely negative (eg. #-10^10#), #-3/(x-4)# becomes very close to 0, so #y# is close to #-1#.

When #x# is largely positive (eg. #10^10#), #-3/(x-4)# becomes very close to 0, so #y# is close to #-1#.

So the plot approaches a horizontal asymptote at #y=-1# as its end behavior.

#y#-intercept

When #x#=0, #y=-3/(0-4)-1=-1/4#.

There is a #y#-intercept at #(0, -1/4)#

#x#-intercept

When #y#=0,

#0=-3/(x-4)-1#.

Solve for #x# to get the #x#-intercept.

#x-4=-3#

#x=1#

There is an #x#-intercept at #(1, 0)#

graph{-3/(x-4)-1 [-5, 15, -20, 20]}