Question

How do you graph `f( x ) = x ^ { 3} - 5x ^ { 2} + 9x + 6`?

Answer 1

See explanation...

Explanation:

Given:

`f(x) = x^3-5x^2+9x+6`

`color(white)(f(x)) = (x-5/3)^3+2/3(x-5/3)+317/27`

Hence `f(x)` is rotationally symmetric about the point `(5/3, 317/27)`

`f'(x) = 3x^2-10x+9`

which has no real zeros

`f'(5/3) = 3(5/3)^2-10(5/3)+9 = 25/3-50/3+27/3 = 2/3`

So at the point of inflexion at `(5/3, 317/27)` the cubic curve has slope `2/3`.

The `y` intercept of `f(x)` is at `(0, 6)`, being `(0, f(0))`.

The `x` intercept is somewhere in the interval `(-1, 0)`, since:

`f(-1) = -1-5-9+6 = -9 < 0`

Linearly interpolating, the `x` intercept is approximately at `x = -0.4` being `-6/(9+6)`

Let's get a better approximation with one step of Newton's method:

`-0.4-(f(-0.4))/(f'(0.4)) = -0.4-1.536/13.48 ~~ -0.51`

Hence we find the graph looks somewhat like this...

graph{(y-(x^3-5x^2+9x+6)) = 0 [-10, 10, -50, 50]}