Question

y = (ex + e-x)/2.
What type of symmetry does the catenary curve have?
a. w/ respect to x-axis
b. w/ respect to y-axis
c. polar symmetry ??????? (whaaaat???)

Answer 1

\[\Large\rm \frac{1}{2}(e^x+e^{-x})=\cosh x\]Hmm I'm picturing the function in my head.
It has symmetry about x=0, the y-axis.

I'm not really sure how to explain how I know that though +_+

Answer 2

hahaha! Oh my gosh, how do you just picture that in your head? XD

Answer 3

Ummm ok ok here is one thing you can do I guess :)
If you graph e^x and e^{-x}...