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

##### 1 Answer

**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

The derivative is undefined at

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

**Test Point: #x= -1#**

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

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

Therefore,

**Asymptotes**

This will have horizontal and vertical asymptotes.

Vertical

We factor the denominator:

This means there are vertical asymptotes at

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

#y = 0/(1 - 0)#

#y = 0#

Therefore, there will be a horizontal asymptote at

**Intercepts**

We find the x-intercepts by setting

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

#0 = x#

We find the y-intercepts by setting

#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!