Question 634327
  <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi,
y+5=2(x+2)^2  (y = 0 &#8658;  (x+2)^2 = 5/2 and x-intercepts are: -2 ± {{{sqrt(5/2)}}}
y  =2(x+2)^2 - 5 ,              V(-2,-5)
the vertex form of a Parabola opening up(a>0) or down(a<0)is: {{{y=a(x-h)^2 +k}}} 
where*[tex \LARGE\ \ (h,k)] is the vertex  and  x = h  is the Line of Symmetry
{{{drawing(300,300,   -16, 6, -6, 6,  blue(line(-2,6,-2,-6))  , grid(1),
circle(-.4, 0,0.3),
circle(-3.6, 0,0.3),
circle(-2, -5,0.3),
graph( 300, 300, -16, 6, -6, 6, 0,2(x+2)^2 - 5  ))}}}