Question

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

Answer 1

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`.