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

Question

1562 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-10T06:43:56

Please see below.

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]}