The equation of the graph is f(x)=X^4 -4x^3

The x intercepts are at (0,0) and (4,0)

The y-intercept is at (0,0)

I found f'(x) to be 4x^3-12x^2, meaning the critical points are at (0,0) and (3,-27).

Am I correct in saying that because its a quartic, there are no horizontal or vertical asymptotes?

If so, it doesnt make sense when I graph it, I get a slanted N which intersects x in three places (-4,0), (0,0), (4,0).

But when I enter the equation in an online graphing tool, [IMG]file:///C:/Users/NOORMA%7E1/AppData/Local/Temp/moz-screenshot.jpg[/IMG]I get a deep U.

Am i doing something wrong?