How do you solve $-x^2+2x>=0$ ?

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Answer 1 · 2017-01-01T15:40:19

$0<=x<=2$ .
See graphical depiction, for both 1-D {(x)} and 2-D {(x, y)}

$-x^2+2x=-(x-1)^2+1>=0$ . So,

$-(x-1)^2>=-1 to (x-1)^2<=1$ , And so,

$|x-1|<=1.$

Thus, $-1<=x-1<=1$ , giving

$0<=x<=2$

graph{-x^2+2x>=0 [-2, 2, -1, 1]}

How do you solve -x^2+2x>=0-x^2+2x>=0?