How do you graph y<=x^2+8x+16?

1 Answer
A. S. Adikesavan
Aug 24, 2016

Draw the parabola (x+4)^2 = y >=0 that has the vertex at (-4, 0) and focus at (-4, 1/4). Along with, shade the region outside the curve and its enclosure for y<=x^2+8x+16.

Explanation:

The given inequality represents the region outside the parabola

x^2+8x+16=(x+4)^2=y>=0.

The vertex of this boundary is at (-4, 0) and focus is at #(-4, 1/4)).

For any point (x, y) on and beneath this parabola, y<=x^2+8x+16.

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