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

Apr 28, 2017

Local Maximum/Minimum

Start by finding the first derivative.

$y ' = \frac{1 \left({x}^{2} - 9\right) - 2 x \left(x\right)}{{x}^{2} - 9} ^ 2$

$y ' = \frac{{x}^{2} - 9 - 2 {x}^{2}}{{x}^{2} - 9} ^ 2$

$y ' = - \frac{{x}^{2} + 9}{{x}^{2} - 9} ^ 2$

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

The derivative is undefined at $x = \pm 3$, 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.

$y ' ' = \frac{2 x \left({x}^{2} + 27\right)}{{x}^{2} - 9} ^ 3$

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

Test Point: $x = - 1$

$y ' ' \left(- 1\right) = \frac{2 \left(- 1\right) \left({\left(- 1\right)}^{2} + 27\right)}{{\left(- 1\right)}^{2} - 9} ^ 3$

$y ' ' \left(- 1\right) = \text{positive}$

Therefore, $y$ is concave up on $\left(- \infty , 0\right)$ and concave down on $\left(0 , \infty\right)$.

Asymptotes

This will have horizontal and vertical asymptotes.

Vertical

We factor the denominator:

${x}^{2} - 9 = \left(x + 3\right) \left(x - 3\right)$

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

Horizontal

$y = {\lim}_{x \to \infty} \frac{\frac{x}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{9}{x} ^ 2}$

$y = {\lim}_{x \to \infty} \frac{\frac{1}{x}}{1 - \frac{9}{x} ^ 2}$

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

By the identity ${\lim}_{x \to \infty} \frac{1}{x} = 0$, we have:

$y = \frac{0}{1 - 0}$

$y = 0$

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

Intercepts

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

$0 = \frac{x}{{x}^{2} - 9}$

$0 = x$

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

$y = \frac{0}{{0}^{2} - 9}$

$y = 0$

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!