How do you sketch the curve y=x^3-3x^2-9x+5y=x33x29x+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 xxapproaches +-oo±

Explanation:

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

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)