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:
#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!