Question 1176572

 the quadratic is written in the form 

{{{y = a(x- h)^2 + k}}}, then the vertex is the point ({{{h}}}, {{{k)

if a vertex is at  ​({{{-4}}}​,{{{4}}}​), then {{{h=-4}}} and {{{k=4}}}

{{{y = a(x- (-4))^2 + 4}}}

{{{y = a(x+4)^2 + 4}}}.........if passese through the point ​({{{-8​}}},{{{0}}}), we have 

{{{0= a(-8+4)^2 + 4}}}........solve for {{{a}}}

{{{0-4= a(-4)^2 }}}

{{{-4= a*16}}}

{{{a=-4/16}}}

{{{a=-1/4 }}}

and your equation is:

{{{y = -(1/4)(x+4)^2 + 4}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-4,4,.12),locate(-4,4,V(-4,4)),
circle(-8,0,.13),locate(-8,0.5,p(-8,0)),
 graph( 600, 600, -10, 10, -10, 10, -(1/4)(x+4)^2 + 4)) }}}