Ok, so if I set and solve for , I should be able to see where , therefore where the graph is either equal to zero or crossing the x-axis. From there, I would need to examine two points on either side of the x value that produces a .

Is that the general idea? I see that you made a point of saying "not just where the graph is zero", so I assume that it would not hurt to include those points in my evaluation. Is there a faster method to examine this?

If I use the substitution:

from

And then...

u=\frac{t}{2}

du=\frac{1}{2}dt

I find that I get the following (coefficients omitted):

What do I do with this? If my was not being raised to the 3/2 power, it would be pretty easy. But I'm not sure where this is supposed to go...

Did I make a larger mistake that has me way off target?