Question 391027
  <pre><font size = 3 color = "indigo"><b>
Hi,  
Parabola with vertex at pt(-6,-1) and x-intercepts are (-9,0) and (-3,0)       
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
 y = a(x+6)^2 - 1 Using Pt(-3,0) to solve for a
 0 = a *3^2 - 1
 1 = 9a
 1/9 = a
y = 1/9(x+6)^2 - 1
{{{drawing(300,300, -10,10,-10,10,
 grid(1),
circle(-9, 0,0.4),
circle(-3, 0,0.4),
circle(-6, -1,0.4),

graph( 300, 300, -10,10,-10,10,(1/9)(x+6)^2 - 1
))}}}