Question 409995
  <pre><font size = 3 color = "indigo"><b>
Hi
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
Sketching the first-quadrant portions of y=x^2+8x+16
 y = x^2+8x+16
 y= (x+4)^2 - 16 + 16
 y = (x+4)^2  |When x = 0, then y = 16
Cannot demonstrate here, however it would be the portion of the parabola
in the first-quadrant
{{{drawing(300,300,   -30,30,-30,30, grid(1),
graph( 300, 300, -30,30,-30,30,x^2+8x+16
))}}}
{{{drawing(300,300, -6, 6, -6, 6, grid(1),
circle(-3, 5,0.2),
locate(4.3,4.6,"I"),
locate(-4.3,4.6,"II"),
locate(-4.3,-4.6,"III"),
locate(4.3,-4.6,"IV"),
graph( 300, 300,-6,6,-6,6)) }}}