How do you sketch the curve #y=x^3+6x^2+9x# by finding local maximum, minimum, inflection points, asymptotes, and intercepts?
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1 Answer
Explanation:
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Polynomial graphs have no asymptotes. Roxio secure burn 1 2016.
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.
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#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.
Later 1 0 2. 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.