How do you graph #y=1/(x-7)+3# using asymptotes, intercepts, end behavior?

Question

1428 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-12T20:22:56

The vertical asymptote is #x=7#
The horizontal asymptote is #y=0#
There are no oblique asymptotes.
The intercepts are #(0,20/7)# and #(20/3,0)#

As you cannot divide by #0#, a vertical asymptote is #x=7#

#lim_(x->-oo)y=lim_(x->-oo)1/x=0^(-)#

#lim_(x->+oo)y=lim_(x->+oo)1/x=0^(+)#

Therefore #y=0# is a horizontal asymptote.

#lim_(x->7^-)y=-oo#

#lim_(x->7^+)y=+oo#

There are no oblique intercept as the degree of the numerator is #<# the degree of the denominator.

Intercepts, when #x=0#, #y=-1/7+3=20/7#

When #y=0# #=>##1/(x-7)+3=0# #=>##x=20/3#
graph{3+1/(x-7) [-14.87, 25.66, -8.18, 12.09]}