How do you graph $y<-1/2x^2-2x+4$ ?

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Answer 1 · 2016-12-15T14:08:18

See the graph and the explanation

The given equation represents the parabolic boundary in the graph

upon which $y=-x^2/2-2x+4$ .

For the (shaded ) interior points (x, y),

$y < -x^2/2-2x+4$

graph{y<-x^2/2-x-2x+4 [-27.35, 27.34, -13.67, 13.68]}

How do you graph y<-1/2x^2-2x+4y<-1/2x^2-2x+4?