Question 671829
<font face="Times New Roman" size="+2">


Substitute 0 for *[tex \LARGE x]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{3(0)}{(0)^2\ +\ 81}\ =\ \frac{0}{81}\ =\ 0]


Hence both the *[tex \LARGE x] and *[tex \LARGE y] intercepts are the origin, *[tex \LARGE \left(0,0\right)]


A function is symmetric with respect to the *[tex \LARGE x] axis iff *[tex \LARGE f(a)\ =\ -f(a)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3x}{x^2\ +\ 81}\ \neq\ -\frac{3x}{x^2\ +\ 81}]


Hence, <b>not</b> symmetrical with respect to the *[tex \LARGE x] axis.


A function is symmetric with respect to the *[tex \LARGE y] axis iff *[tex \LARGE f(a)\ =\ f(-a)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3x}{x^2\ +\ 81}\ \neq\ \frac{3(-x)}{(-x)^2\ +\ 81}\ =\ -\frac{3x}{x^2\ +\ 81}]


Hence, <b>not</b> symmetrical with respect to the *[tex \LARGE y] axis.


A function is symmetric with respect to the origin iff *[tex \LARGE f(a)\ =\ -f(-a)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3x}{x^2\ +\ 81}\ =\ -\frac{3(-x)}{(-x)^2\ +\ 81}\ =\ \frac{3x}{x^2\ +\ 81}]


Hence, the function <b>is</b> symmetrical with respect to the origin.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>