1. ## graph of y=e^2x

how does the graph of $y=e^(2x)$

I have tried it using a math software but it shows a straight line with 0 as the y - intercept which seems wrong to me as $x=0 , y = e^(2(0)) , y = e^0 , y = 1$

I would be delighted if you can show me the graph of it and is there any asymptote? is the asymptote y = 0? , let x = -n , where n is an integer , $y = e^(2(-n)) y = 1 / e^(2n)$ , the y will be getting closer and closer to 0 but will never reach it

2. did you graph

y = e^2x

or

y = e^(2x)

?

3. You probably graphed ax where a is e^2 instead of e^(2x). I attached the graph of $y = e^{2x}$.

And e^{2x} approaches 0 as x approaches negative infinity.

4. Originally Posted by skeeter
did you graph

y = e^2x

or

y = e^(2x)

?
sry it is y=e^(2x)

5. Originally Posted by Chop Suey
You probably graphed ax where a is e^2 instead of e^(2x). I attached the graph of $y = e^{2x}$.

And e^{2x} approaches 0 as x approaches negative infinity.

Thanks a lot!

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# graph of y^2-e^2x

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