Question

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

Answer 1

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!