this is a curve sketching question. I dont quite know how to start it. y=x^2(x+1)(3-2x)

Question

this is a curve sketching question. I dont quite know how to start it.   y=x^2(x+1)(3-2x)

anonymous · Answer

Since it already looks factored, I would start finding the zeros and then marking them down on your graph paper, before we move on to finding critical values, 2nd dervs,

laddiusmaximus · Answer

ok well i got as far  as (x+1)=0 that would make x=-1

laddiusmaximus · Answer

and 3-2x=0 would make x= 3/2 right?

anonymous · Answer

sorry, I got dc'ed.  I think you missed a zero as well.

anonymous · Answer

next you would take the 1st derivative and then solve IT for zero (not the previous function).  That will give you the locations where the tangent = 0.  We still don't know if that is a min or max yet though.  For that I think it is easiest to take the 2nd derivative and then plug those critical values (where tangent = 0) into it.