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

1 Answer
Aug 24, 2016

Answer:

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#.