Question 698163
the equation for a parabola can also be written in "vertex form":

{{{y = a(x -h)^2 + k}}}

In this equation, the vertex of the parabola is the point ({{{h}}}, {{{k}}}).


if ({{{h}}}, {{{k}}})=({{{-6}}}, {{{2}}}), than


{{{y = a(x-(-5))^2 + 2}}}


{{{y = a(x +5)^2 + 2}}}.........the coefficient of {{{x}}} here is {{{-2ah}}}

{{{y = k + sqrt(-4h)}}}... to find the {{{y-intercept}}}, set {{{x = 0}}} 

{{{y = 2 + sqrt(-4(-6))}}}

{{{y = 2 + sqrt(24)}}}

{{{y = 2 + 4.9}}}

{{{y = 6.9}}}


next, we find {{{a}}}


{{{x = h + k^2/4a }}}


{{{x = -6 + 2^2/4a}}}

{{{x =a-6}}}.......since {{{y = 6.9}}} plug both in {{{y = a(x-h)^2 + k}}} and solve for {{{a}}}


{{{6.9= a(a-6 +6)^2 + 2}}}

{{{6.9= a*a^2 + 2}}}

{{{6.9-2= a^3}}}

{{{4.9= a^3}}}

{{{root(3,4.9)= a}}}

{{{a= 2.214}}}

equation will be:

{{{y = 2.214(x +6)^2 + 2}}}

{{{y = 2.214(x^2+12x +36) + 2}}}

{{{y = 2.214x^2+26.568x +79.704 + 2}}}

{{{y = 2.214x^2+26.568x +81.704 }}}


check it on a graph:



{{{drawing(600,600,-30, 10, -3,90,grid(0), locate(-6,2.5-.2,"V(-6,2)"),circle(-6,2,0.2),graph(600,600,-30, 10, -3,90,2.214*(x +6)^2 + 2))}}}