# Thread: a little stuck on this one

1. ## a little stuck on this one

I'm having some trouble figuring out how to find the symmetries, horizontal asymptote, max and min and inflection points on y=x2/1-x2.

Iv'e already found out the horizontal asymptote and the x and y intercepts, but am stuck on the rest. Can anybody give some pointers on how to figure this one out. (I have to find all of this so I can graph it).

Thank you in advance.

2. Hello, math619!

Find the symmetries, horizontal asymptote, max and min and inflection points of $\displaystyle y = \frac{x^2}{1-x^2}$

I've already found out the horizontal asymptote and the x and y intercepts.

If we replace $\displaystyle x$ with $\displaystyle -x$, we get: .$\displaystyle y \:=\:\frac{(-x)^2}{1 - (-x)^2} \:=\:\frac{x^2}{1-x^2}$ . . . the same function

Therefore, the graph is symmetric to the $\displaystyle y$-axis.

Vertical asymptotes occur where the denominator is zero.
. . Hence, $\displaystyle x = 1$ and $\displaystyle x = -1$ are vertical asymptotes.

For max and min, you need the derivative (quotient rule).

. . $\displaystyle y'\;=\;\frac{(1-x^2)\cdot2x - x^2(-2x)}{(1-x^2)^2} \;=\;\frac{2x}{(1-x^2)^2}$

The derivative equals 0 when its numerator is 0: .$\displaystyle x = 0$

To test it, we'll use the Second Derivative Test.

. . $\displaystyle y'' \;=\;\frac{(1-x^2)^2\cdot2 - 2x\cdot2(1-x^2)(-2x)}{(1-x^2)^4} \;= \;\frac{2(1-x^2)^2 + 8x^2(1-x^2)}{(1-x^2)^4}$

Factor: .$\displaystyle y''\;=\;2(1-x^2)\,\frac{1 - x^2 + 8x^2}{(1-x^2)^4} \;= \;\frac{2(1 + 7x^2)}{(1-x^2)^3}$ [1]

When $\displaystyle x = 0:\;\;y'' \:=\:\frac{2(1 + 0)}{(1-0)^3}\:=\:+2$ . . . positive, concave up: $\displaystyle \cup$

. . Therefore, a minimum at $\displaystyle (0,\,0).$

Inflection points occur when the second derivative equals zero.
. . But we see from [1], that $\displaystyle y''$ cannot equal 0.
Therefore, there are no inflection points.

3. Soroban you are a lifesaver. Thx so much for the help. It seems so much easy when you see the solution.

4. Originally Posted by math619
Soroban you are a lifesaver. Thx so much for the help. It seems so much easy when you see the solution.
If you wish to thank Soroban please use the thank button near the bottom of his post.

RonL