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

Jun 6, 2017

It is all the ordered pairs on the line of and outside of the plot of $y = {x}^{2} + 8 x + 16$. The line is solid as we have $\ge$

#### Explanation:

$\textcolor{b l u e}{\text{Determine the key features of } y = {x}^{2} + 8 x + 16}$

As the ${x}^{2}$ term is positive the graph is of form $\cup$ thus the vertex is a minimum

Note that ${4}^{2} = 16 \mathmr{and} 4 + 4 = 8$

so $y = 0 = \left(x + 4\right) \left(x + 4\right)$ This is called duality

Thus the graph is such that the x-axis is tangential thus
${y}_{\text{vertex}} = 0$

Consider the $x$ term: we have $+ 8 x$

Compare to the equation standardised form of $y = a {x}^{2} + b x + c$
From this ${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a}$

Thus x_("vertex")=(-1/2)xx8/1=-4" " as in keeping with the above.

Vertex$\to \left(x , y\right) = \left(- 4 , 0\right)$

y-intercept is the constant 16$\to \left(x , y\right) = \left(0 , 16\right)$