Question 840520
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Hi,
the vertex form of a Parabola opening up(a>0) or down(a<0), {{{y=a(x-h)^2 +k}}} 
where(h,k) is the vertex  and  x = h  is the Line of Symmetry
vertex (6, &#8722;5), x-intercept &#8722;1 , P(-1,0)
{{{y=a(x-6)^2 -5}}}   |using P(-1,0) to solve for 'a'
  0 = 49a - 5
{{{ 5/49 = a}}}
{{{y = (5/49)(x-6)^2 -5 }}}
{{{drawing(300,300,  -20,20,-20,20,
 grid(1),
circle(-1, 0,0.6),
graph( 300, 300, -20,20,-20,20,0,(5/49)(x-6)^2 -5 ))}}}