How do you sketch the curve #y=x/(x^2-9)# by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

1 Answer
Apr 28, 2017

Local Maximum/Minimum

Start by finding the first derivative.

#y' = (1(x^2 - 9) - 2x(x))/(x^2 - 9)^2#

#y' = (x^2 - 9 - 2x^2)/(x^2 - 9)^2#

#y' = -(x^2 + 9)/(x^2 - 9)^2#

Critical numbers will occur whenever the derivative equals #0# or the derivative is undefined.

The derivative is undefined at #x = +- 3#, but this is undefined in the actual function as well, so these are actually not critical numbers. The derivative is never equal to #0# since #x^2 + 9 = 0# has no real solution.

Therefore, there are no critical numbers and thus no local max/min.

Points of inflection

Now find the second derivative. I used a derivative calculator for this, because it's very long to do by hand.

#y'' = (2x(x^2 + 27))/(x^2 - 9)^3#

This will have a point of inflection at #x= 0#. Let's now determine concavity.

Test Point: #x= -1#

#y''(-1) = (2(-1)((-1)^2 + 27))/((-1)^2 - 9)^3#

#y''(-1) = "positive"#

Therefore, #y# is concave up on #(-oo, 0)# and concave down on #(0, oo)#.

Asymptotes

This will have horizontal and vertical asymptotes.

Vertical

We factor the denominator:

#x^2 - 9 = (x + 3)(x - 3)#

This means there are vertical asymptotes at #x = -3# and #x = +3#.

Horizontal

#y = lim_(x->oo) (x/x^2)/(x^2/x^2 - 9/x^2)#

#y = lim_(x->oo) (1/x)/(1 - 9/x^2)#

#y = lim_(x->oo) (lim_(x->oo) (1/x))/((lim_(x->oo) (1) - lim_(x->oo) 9/x^2)#

By the identity #lim_(x-> oo) 1/x = 0#, we have:

#y = 0/(1 - 0)#

#y = 0#

Therefore, there will be a horizontal asymptote at #y= 0#.

Intercepts

We find the x-intercepts by setting #y# to #0#.

#0 = x/(x^2 - 9)#

#0 = x#

We find the y-intercepts by setting #x# to #0#.

#y = 0/(0^2 - 9)#

#y = 0#

We can now more or less trace the graph. Many people, including myself, like to prepare a table of values before graphing, but I'll leave that choice up to you.

graph{y = x/(x^2 - 9) [-28.87, 28.86, -14.43, 14.44]}

Hopefully this helps!