graph: y= e^x^2 ; -3≤x≤3?

Answer 1

i cannot do this for you, but if it is e^(x^2) then it is going to be symmetric along the y axis right?

so take the graph of e^u, where u=x^2, and do it from 0 to 9 (the square of 3) and then reflect that on the y axis and you have your answer. At u=0 the graph is going to equl to (0,1) and at 9 it will look like (9, e^9). If you wanted to go from e^(x^2) from our e^u then it will just be squished (closer to the y axis at all points except 0), but you will still get a general feeling from what i said.

also i think it is possible you left out a negative in front of the x in which case the graph will look like the normal curve: http://www.uow.edu.au/student/attributes/statlit/modules/images/normal-curve-1.gif

Answer 2

The question is ambiguous. Do you mean (e^x)^2, or e^(x^2)?

This looks a bit like a school question to me, so I won't answer it. However, you should know that (e^x)^2 = (e^x)*(e^x) = e^2x != e^(x^2). Also, x^2 = (-x)^2. So you have a hint, regardless of how the ambiguity is resolved!

Answer 3

Plot points. (0,1), (1,e),(-1,e),...
It should be symmetric across the y axis and should always be above the x axis.