How do you sketch the curve #y=x^3-3x^2-9x+5# by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

1 Answer
Nov 29, 2016

graph{x^3-3x^2-9x+5 [-145.9, 172.6, -85.6, 73.6]} FIrst determine the interval of definition, then the behavior of first and second derivatives and the behavior of the function as #x#approaches #+-oo#

Explanation:

1) The function is a polynomial and is defined for #x in (-oo,+oo)#

2) The highest monomial is of odd order so:

#lim_(x->-oo) y(x) = -oo#
#lim_(x->+oo) y(x) = +oo#

Also #(y(x))/x# does not have a finite limit, so the function has no asymptotes.

3) Calculate the first and second derivative:

#y'(x) = 3x^2-6x-9#
#y''(x) = 6(x-1)#

The points where #y'(x) = 0# are:

#x=(3+-sqrt(9+27))/3 = (3+-6)/3#

#x_1=-1, x_2=3#

In both points the second derivative is non null, so these are local extrema and not inflection points.

If we analyze the sign of #y'(x)# we see that:

#y'(x) > 0# for #x in (-oo, -1) and x in (3, +oo)#
#y'(x) < 0# for #x in (-1,3)#

So, #y(x)# starts from #-oo#, grows util #x=-1# where it reaches a local maximum, decreases until #x=3# where it reaches a local minimum, that starts growing again.

4) Analyze the value of the function in the local extrema:

#y(x)_(x=-1) =(-1)^3-3(-1)^2-9(-1)+5 =-1-3+9+5=10#

#y(x)_(x=3) =3^3-3*3^2-9*3+5 =27-27-27+5=-22#

So there will be three real roots: one in #(-oo, -1)#, one in #(-1,3)# and one in #(3, +oo)#