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

Aug 24, 2016

Draw the parabola ${\left(x + 4\right)}^{2} = y \ge 0$ 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 $y \le {x}^{2} + 8 x + 16$.

#### Explanation:

The given inequality represents the region outside the parabola

${x}^{2} + 8 x + 16 = {\left(x + 4\right)}^{2} = y \ge 0$.

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

For any point (x, y) on and beneath this parabola, $y \le {x}^{2} + 8 x + 16$.