Solving Cubic equation to graph

Yep it doesn't factorize nicely. That means you can't easily find the x-intercept(s), but don't let that stop you (there may not even be any x-intercepts in your region of interest).

Go ahead and find the start/end points plus any max/min and POIs in the given region.

 

Have you never, ever, drawn a graph before?

RGV

 

I did not ask about "transformations". I asked you if you had ever drawn a graph. Is the answer 'yes' or 'no'?

If you have already done similar things before, just do the same things for this problem.

If you have not drawn a graph before (and if your book does not explain how to do it) there are numerous web pages
that explain what to do. For example, see
http://cstl.syr.edu/fipse/grapha/unit2/unit2.html [Broken] and
http://www.cimt.plymouth.ac.uk/projects/mepres/book8/bk8i14/bk8_14i3.htm .
In your case, however, the curve y = f(t) is not a straight line, so all you can do is make a table of some(t,y) values, and
hand-draw a smooth curve that passes through them.

RGV

 

Plot a graph of p vs t for some values of t in your range of interest.
Stand back and look what shape it is trying to be and where it seems to be heading for the t axis. This will tell you when you are getting close to the roots.

Particularly - plot p for t=0 and t=13 (your endpoints). If your endpoints are on opposite sides of the t axis, then there is at least one root. If p is on opposite sides of the t axis for two adjacent values of t, then there is at least one root between them.

Use Newton-Raphson's method to get the rest of the way.

Looks like you can find and characterize the turning points of the graph OK ... that will also give you clues.

Or you can look up the general formula:
http://en.wikipedia.org/wiki/Cubic_function

 

If you shift the graph down by 2 you can factor that.

y = P(t) - 2 = -0.2t³+2t²+8t .