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

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 $\displaystyle x=0 , y = e^(2(0)) , y = e^0 , y = 1$

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 , $\displaystyle y = e^(2(-n)) y = 1 / e^(2n)$ , the y will be getting closer and closer to 0 but will never reach it