How do you graph #y= x/(x-4)#?

1 Answer
Sep 17, 2015

Have a look:

Explanation:

Let us study our function in steps:

1) Domain (#=x# values allowed)
we need the denominator different from zero:
#x-4!=0#
#x!=4#
#x=4# will be a vertical asymptote.

2) Intercepts:
set: #x=0# then #y=0#
set: #y=0# then #x=0#

3) Limits:
#lim_(x->4^+)y=+oo#
#lim_(x->4^-)y=-oo#
#lim_(x->+oo)y=1#
#lim_(x->+oo)y=1#
#y=1# will be a horizontal asymptote.

4) Derivatives:
FIRST: #y'=(x-4-x)/(x-4)^2=-4/(x-4)^2# (Quotient Rule)
set: #y'=0# then #-4=0# never so we have NO max or min.
SECOND: #y''=(-2(-4)(x-4))/(x-4)^4=8/(x-4)^3#
set #y''>0#
you get:
#8>0# always
#(x-4)^3>0# when #x>4#
graphically:
enter image source here

graph{x/(x-4) [-10, 10, -5, 5]}