How do you graph #(6x)/(x^2-36)# using asymptotes, intercepts, end behavior?

1 Answer
May 29, 2017

Answer:

#color(brown)("Very detailed explanation given.")#
Vertical asymptotes at #x=+-6#
As #x->-oo: y->0^-#
As #x->+oo: y->0^+#

Explanation:

Let a minute amount of #x# be designated by the symbol #deltax#
I have deliberately used this symbol as in introduction to calculus without actually using calculus

Set to #" "y=(6x)/(x^2-36)#

Notice that the denominator is the difference of two squares. So we have:

#y=(6x)/((x-6)(x+6))#

Note that the use of #-># in this explanation is used as meaning 'tends to'.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the vertical asymptotes")#

This becomes 'undefined if the denominator is zero thus we have vertical asymptotes at those points.

Vertical asymptotes at #x=+-6#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the behaviour either side of the asymptotes")#

#color(brown)("Set "x=+6)#

Let #deltax >0# but minute

Set #y=(6(x+deltax))/((color(white)(.)[x+deltax]-6)color(white)(.)(color(white)(.)[x+deltax]+6))#

Now #deltax# is so small that it is almost not there so
#(color(white)(.)[6+delta6]-6 )# is positive but so small then it almost zero. Consequently the whole denominator is only just bigger than 0 but on the positive side. Dividing this into #6(6+delta6)# yields a number that is tending to #+oo #

...................................................................................................
Let #deltax < 0# but minute

Following the same methodology of thinking we find that in this case the denominator is only just smaller than 0 so when divide into the numerator the whole is tending to #-oo#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Set "x=-6)#

As in the processes above:

When #deltax >0# we have #(6(6+delta6))/(0^+)->+oo #
When #deltax <0# we have #(6(6+delta6))/(0^+)->-oo #
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the behaviour at the extremities "x->+-oo)#

As #x# becomes increasingly larger and larger the -36 has less and less effect. So the equation behaves as if it where #y=6x/x^2 = 6/x#

The net consequence is that #y->+-oo# in keeping the sign of #x#

For #x<0: y->-oo#
For #x>0: y->+oo#
Tony B