Question 933106
{{{y=a(x-h)^2+k}}},Vertex Form,  where {{{k=y-intercept}}} and y-coordinate of the vertex, and {{{h}}} is x-coordinate of the vertex

vertex is at (h, k)
given points: 
({{{2}}},{{{0}}}),----> x-intercept
({{{-4}}},{{{0}}})----> x-intercept
({{{0}}},{{{16}}})-> since {{{x=0}}}, this is y-intercept and {{{highlight(k=16)}}} , then vertex is at (h, 16)

since 
Vertex:
{{{x}}} coordinate of vertex is {{{h}}} and the x-coordinate of the vertex is halfway between the x-coordinates of the x-intercepts which is the distance from  {{{2}}} to {{{-4}}} divided by {{{2}}}: {{{(-4+2)/2=-1}}}

then {{{highlight(h=-1)}}}

so, we know  that {{{h=-1}}} and {{{k=16}}}; so, the coordinate of the vertex are ({{{-1}}},{{{16}}})

now find {{{a}}} using one of given points and vertex


{{{y=a(x-h)^2+k}}}...use ({{{2}}},{{{0}}})-> since {{{y=0}}}, this is {{{highlight(x-intercept)}}}

{{{0=a(2-(-1))^2+16}}}

{{{0=a(3)^2+16}}}

{{{0=9a+16}}}

{{{-16=9a}}}

{{{-16/9=a}}}

{{{highlight(-16/9=a)}}}

so, our equation is:

{{{y=-(16/9)(x+1)^2+16}}}


{{{ graph( 600,600, -10, 10, -10, 20, -(16/9)(x+1)^2+16) }}}