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?
August 19th 2009, 08:14 AM
correct it has no asymptotes. Also the reason why the looks diff is because you have to zoom out and then it will look like a U.