Question

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

Answer 1

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)`