Question

How do you graph the inequality `y>=x^3-6x^2+12x-8`?

Answer 1

Please see below.

Explanation:

Note that considering the equality `x=2`, `f(x)=x^3-6x^2+12x-8=0`, hence `(x-2)` is a factor of `x^3-6x^2+12x-8`

`x^3-6x^2+12x-8`

= `(x-2)(x^2-4x+4)`

= `(x-2)^3`

Observe that for `x=2`, we have `y=0` and hence the curve `y=(x-2)^3` just moves the function `y=x^3`, `2` points to the right.

The graph of `(x-2)^3=0` appears as follows:

graph{(x-2)^3 [-10, 10, -5, 5]}

This divides the plane in three parts. One on the curve, which satisfies the equality and hence lies on the graph of `y=(x-2)^3`.

Other portions are to the left and right of the curve. Let us pick two points on either side say `(0,0)` and `(5,0)`.

At `(0,0)`, `y>=(x-2)^3=>0>=-8` which is true

but at `(5,0)`, we have `0>=27`, which is not true.

So, it is only on the left hand side of the curve that points satisfy the inequality `y>=(x-2)^3`

Hence graph appears as follows:

graph{(y-x^3+6x^2-12x+8)>=0 [-10, 10, -5, 5]}