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

Sep 10, 2017

#### Explanation:

Note that considering the equality $x = 2$, $f \left(x\right) = {x}^{3} - 6 {x}^{2} + 12 x - 8 = 0$, hence $\left(x - 2\right)$ is a factor of ${x}^{3} - 6 {x}^{2} + 12 x - 8$

${x}^{3} - 6 {x}^{2} + 12 x - 8$

= $\left(x - 2\right) \left({x}^{2} - 4 x + 4\right)$

= ${\left(x - 2\right)}^{3}$

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

The graph of ${\left(x - 2\right)}^{3} = 0$ 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 $y = {\left(x - 2\right)}^{3}$.

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

At $\left(0 , 0\right)$, $y \ge {\left(x - 2\right)}^{3} \implies 0 \ge - 8$ which is true

but at $\left(5 , 0\right)$, we have $0 \ge 27$, which is not true.

So, it is only on the left hand side of the curve that points satisfy the inequality $y \ge {\left(x - 2\right)}^{3}$

Hence graph appears as follows:

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