How do you sketch the curve #y=x^3+6x^2+9x# by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

Question

9934 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-18T03:57:28

See graph and explanation

Polynomial graphs have no asymptotes.

As x to +-oo, y = x^3(1+3/x)^2 to +-oo, showing end behavior of

#uarr and darr#, without limit.

So, there are no global extrema.

#y=x(x+3)^2=0#, at #x = 0 and -3#.

x-intercepts: #0 and -3#.

#y'=3(x+1)(x+3)=0#, at #x = -3 and -1#

Turning points or points of inflexion at (-1, -4) and (-3, 0)

#y''=6x+12=0#, at #x = -2#.

#y'''=6 ne 0#.

At #(-2, -2), y''=0 and y''' ne 0#. So, it is the point of inflexion.

This POI is marked, in the graph.

At # (-1, -4)), y'=0 and y''=6> 0#. So, local minimum #y = -4#.

At #(-3, 0), y'=0 and y''= -6 #. So,#0# is a local maximum.

graph{(x(x+3)^2-y)((x+2)^2+(y+2)^2-.01)=0 [-10, 10, -10, 5]}